## Eigenfunctions of position and momentum; unit operators

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education, Section 3.6.

Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014), Section 3.5.

There are a few results that will be used frequently in quantum theory that I think it’s worth collecting together and explaining in full.

First, we’ll revisit the eigenfunctions of the position and momentum operators. In the earlier post, we showed that the eigenfunctions of the position operator are delta functions and we wrote

$\displaystyle \left|x_{0}\right\rangle =\delta\left(x-x_{0}\right) \ \ \ \ \ (1)$

Strictly speaking this equation gives the position space representation of the eigenfunction. More precisely, we should just say that ${\left|x_{0}\right\rangle }$ is an eigenvalue of the position operator ${\hat{x}}$ and leave it at that. In order to write it as a ‘proper’ function (that is, a function we can use in calculations such as integrals), we need to specify the space we’re using and then write ${\left|x_{0}\right\rangle }$ in that space, as we did above for position space.

For momentum, we’ve seen that the eigenfunctions are

$\displaystyle \left|p_{0}\right\rangle =\frac{1}{\sqrt{2\pi\hbar}}e^{ip_{0}x/\hbar} \ \ \ \ \ (2)$

The normalizations of both the position and momentum eigenfunctions give us more delta functions:

 $\displaystyle \left\langle x_{1}\left|x_{0}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \int\delta\left(x-x_{1}\right)\delta\left(x-x_{2}\right)dx\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta\left(x_{1}-x_{2}\right)\ \ \ \ \ (4)$ $\displaystyle \left\langle p_{1}\left|p_{0}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi\hbar}\int e^{i\left(p_{0}-p_{1}\right)x/\hbar}dx\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta\left(p_{0}-p_{1}\right) \ \ \ \ \ (6)$

Given a complete basis set of states, we can define a set of projection operators each of which projects a function onto the basis vector that defines the projection operator. A projection operator has the form

$\displaystyle \hat{P}\equiv|\alpha\rangle\langle\alpha| \ \ \ \ \ (7)$

so that applying it to a state ${\left|\psi\right\rangle }$ gives

$\displaystyle \hat{P}\left|\psi\right\rangle =\left\langle \alpha\left|\psi\right.\right\rangle \left|\alpha\right\rangle \ \ \ \ \ (8)$

Note that this is a completely general expression; we can choose any basis states ${\left|\alpha\right\rangle }$ (they could be the eigenstates of position or momentum, or the discrete set of states for some system such as the states of the infinite square well or harmonic oscillator) and the projection operator gives the component of ${\left|\psi\right\rangle }$ ‘along’ that basis vector. In practice, to do calculations we usually express ${\left|\psi\right\rangle }$ in position or momentum space (or in matrix form if it’s a spin state) but in this formula, ${\left|\psi\right\rangle }$ is just an abstract symbol representing some arbitrary state.

For a complete set of discrete basis states we can define the unit operator

$\displaystyle 1\equiv\sum_{\alpha}\left|\alpha\right\rangle \left\langle \alpha\right| \ \ \ \ \ (9)$

or for a continuous set of basis states

$\displaystyle 1\equiv\int d\alpha\left|\alpha\right\rangle \left\langle \alpha\right| \ \ \ \ \ (10)$

This works because it’s just like expressing a 3-d vector as a sum of its components in some basis, such as rectangular coordinates

$\displaystyle \mathbf{v}=v_{x}\hat{\mathbf{x}}+v_{y}\hat{\mathbf{y}}+v_{z}\hat{\mathbf{z}} \ \ \ \ \ (11)$

Since the basis consisting of the states ${\left|\alpha\right\rangle }$ is complete, we can write any other state in terms of that basis set. We’re using the projection operator for each basis state to project out the new state onto each of the basis states in turn, then adding up the result:

$\displaystyle \left|\psi\right\rangle =\sum_{\alpha}\left|\alpha\right\rangle \left\langle \alpha\left|\psi\right.\right\rangle \ \ \ \ \ (12)$

or

$\displaystyle \left|\psi\right\rangle =\int d\alpha\left|\alpha\right\rangle \left\langle \alpha\left|\psi\right.\right\rangle \ \ \ \ \ (13)$

Example 1 Armed with these results, it’s worth looking at Example 3.6 in Lancaster & Blundell in a bit more detail. In that example, they extend the creation-annihilation operator representation to cases where the momentum (and hence the energy) states merge into a continuum. In that case, the commutation relation for the operators becomes

$\displaystyle \left[a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}\right]=\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right) \ \ \ \ \ (14)$

Somewhat confusingly (I think), L&B use the notation ${\left|\mathbf{p}\right\rangle }$ to represent an eigenstate of the hamiltonian with momentum ${\mathbf{p}}$ rather than an eigenfunction of the bare momentum operator, so I’ll use that notation here, with a caution not to confuse it with the momentum eigenstates above. They begin with a one-particle state

 $\displaystyle \left\langle \mathbf{p}\left|\mathbf{p}'\right.\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \hat{a}_{\mathbf{p}}^{\dagger}0\left|\hat{a}_{\mathbf{p}'}^{\dagger}\right.0\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|\hat{a}_{\mathbf{p}}\hat{a}_{\mathbf{p}'}^{\dagger}\right|0\right\rangle \ \ \ \ \ (16)$

We can now use the relation 14 to get

$\displaystyle \hat{a}_{\mathbf{p}}\hat{a}_{\mathbf{p}'}^{\dagger}=\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right)+\hat{a}_{\mathbf{p}'}^{\dagger}\hat{a}_{\mathbf{p}} \ \ \ \ \ (17)$

and since ${\hat{a}_{\mathbf{p}}\left|0\right\rangle =0}$ we get

 $\displaystyle \left\langle \mathbf{p}\left|\mathbf{p}'\right.\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right)\right|0\right\rangle \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right) \ \ \ \ \ (19)$

Now we want to get the position space version of the state ${\left|\mathbf{p}\right\rangle }$. From 1 (generalized to 3-d) we see that

 $\displaystyle \phi_{\mathbf{p}}\left(\mathbf{x}\right)$ $\displaystyle \equiv$ $\displaystyle \left\langle \mathbf{x}\left|\mathbf{p}\right.\right\rangle \ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}'\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{x}'\right)\left|\mathbf{p}\right\rangle \ \ \ \ \ (21)$

so if we can write ${\left|\mathbf{p}\right\rangle }$ as a function of ${\mathbf{x}'}$ then the expression ${\left\langle \mathbf{x}\left|\mathbf{p}\right.\right\rangle }$ merely picks out the precise position ${\mathbf{x}}$ that we’re interested in. Using a set of momentum basis states ${\left|\mathbf{q}\right\rangle }$ we can transform ${\left|\mathbf{x}\right\rangle }$ using the unit operator 13:

 $\displaystyle \left|\mathbf{x}\right\rangle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\left|\mathbf{q}\right\rangle \left\langle \mathbf{q}\left|\mathbf{x}\right.\right\rangle \ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\left|\mathbf{q}\right\rangle \left\langle \mathbf{x}\left|\mathbf{q}\right.\right\rangle ^*\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\phi_{\mathbf{q}}^*\left(\mathbf{x}\right)\left|\mathbf{q}\right\rangle \ \ \ \ \ (24)$

Therefore

 $\displaystyle \left\langle \mathbf{x}\right|$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\phi_{\mathbf{q}}\left(\mathbf{x}\right)\left\langle \mathbf{q}\right|\ \ \ \ \ (25)$ $\displaystyle \left\langle \mathbf{x}\left|\mathbf{p}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\phi_{\mathbf{q}}\left(\mathbf{x}\right)\left\langle \mathbf{q}\left|\mathbf{p}\right.\right\rangle \ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\phi_{\mathbf{q}}\left(\mathbf{x}\right)\delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{p}\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \phi_{\mathbf{p}}\left(\mathbf{x}\right) \ \ \ \ \ (28)$

Example 2 We can apply the same arguments to a 2-particle state. Start with

$\displaystyle \left\langle \mathbf{p}'\mathbf{q'}\left|\mathbf{p}\mathbf{q}\right.\right\rangle =\left\langle 0\left|\hat{a}_{\mathbf{p}'}\hat{a}_{\mathbf{q}'}\hat{a}_{\mathbf{q}}^{\dagger}\hat{a}_{\mathbf{p}}^{\dagger}\right|0\right\rangle \ \ \ \ \ (29)$

From commutation relations 14 we get

 $\displaystyle \hat{a}_{\mathbf{p}'}\hat{a}_{\mathbf{q}'}\hat{a}_{\mathbf{q}}^{\dagger}\hat{a}_{\mathbf{p}}^{\dagger}$ $\displaystyle =$ $\displaystyle \hat{a}_{\mathbf{p}'}\left(\delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{q}'\right)+\hat{a}_{\mathbf{q}}^{\dagger}\hat{a}_{\mathbf{q}'}\right)\hat{a}_{\mathbf{p}}^{\dagger}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{q}'\right)\left(\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right)+\hat{a}_{\mathbf{p}}^{\dagger}\hat{a}_{\mathbf{p}'}\right)+\hat{a}_{\mathbf{p}'}\hat{a}_{\mathbf{q}}^{\dagger}\hat{a}_{\mathbf{q}'}\hat{a}_{\mathbf{p}}^{\dagger}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{q}'\right)\left(\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right)+\hat{a}_{\mathbf{p}}^{\dagger}\hat{a}_{\mathbf{p}'}\right)+\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(\delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{p}'\right)+\hat{a}_{\mathbf{q}}^{\dagger}\hat{a}_{\mathbf{p}'}\right)\left(\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}'\right)+\hat{a}_{\mathbf{p}}^{\dagger}\hat{a}_{\mathbf{q}'}\right) \ \ \ \ \ (33)$

Applying ${\hat{a}_{\mathbf{p}}\left|0\right\rangle =0}$ we get

$\displaystyle \left\langle \mathbf{p}'\mathbf{q'}\left|\mathbf{p}\mathbf{q}\right.\right\rangle =\delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{q}'\right)\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right)+\delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{p}'\right)\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}'\right) \ \ \ \ \ (34)$

To convert to position coordinates, this time we have two independent positions, one for each particle, which we’ll call ${\mathbf{x}}$ and ${\mathbf{y}}$, so a position state is ${\left|\mathbf{xy}\right\rangle }$. To change basis like we did above we need to deal with two variables. I’m not entirely sure this is the right way to do it, but since the particles are independent, we should be able to represent the compound state as the product of two single-particle states:

$\displaystyle \left|\mathbf{p}'\mathbf{q}'\right\rangle =\left|\mathbf{p}'\right\rangle \left|\mathbf{q}'\right\rangle \ \ \ \ \ (35)$

Also, since the particles are identical, the state ${\left|\mathbf{p}'\mathbf{q}'\right\rangle }$ is the same as ${\left|\mathbf{q}'\mathbf{p}'\right\rangle }$.

In that case we can write

 $\displaystyle \left|\mathbf{xy}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2!}}\int d^{3}p'\int d^{3}q'\left|\mathbf{p}'\right\rangle \left|\mathbf{q}'\right\rangle \left\langle \mathbf{p}'\right|\left\langle \mathbf{q}'\right|\left|\mathbf{y}\right\rangle \left|\mathbf{x}\right\rangle \ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2!}}\int d^{3}p'\int d^{3}q'\left|\mathbf{p}'\right\rangle \left|\mathbf{q}'\right\rangle \left\langle \mathbf{q'}\left|\mathbf{y}\right.\right\rangle \left\langle \mathbf{p'}\left|\mathbf{x}\right.\right\rangle \ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2!}}\int d^{3}p'\int d^{3}q'\phi_{\mathbf{p}'}^*\left(\mathbf{x}\right)\phi_{\mathbf{q}'}^*\left(\mathbf{y}\right)\left|\mathbf{p}'\right\rangle \left|\mathbf{q}'\right\rangle \ \ \ \ \ (38)$ $\displaystyle \left\langle \mathbf{xy}\right|$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2!}}\int d^{3}p'\int d^{3}q'\phi_{\mathbf{p}'}\left(\mathbf{x}\right)\phi_{\mathbf{q}'}\left(\mathbf{y}\right)\left\langle \mathbf{p}'\right|\left\langle \mathbf{q}'\right|\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2!}}\int d^{3}p'\int d^{3}q'\phi_{\mathbf{p}'}\left(\mathbf{x}\right)\phi_{\mathbf{q}'}\left(\mathbf{y}\right)\left\langle \mathbf{p}'\mathbf{q}'\right| \ \ \ \ \ (40)$

The ${\frac{1}{\sqrt{2!}}}$ is there because the double integral extends over all values of both ${\mathbf{p}'}$ and ${\mathbf{q}'}$ so it counts the state ${\left|\mathbf{p}'\right\rangle \left|\mathbf{q}'\right\rangle }$ twice, once as ${\left|\mathbf{p}'\right\rangle \left|\mathbf{q}'\right\rangle }$ and once as ${\left|\mathbf{q}'\right\rangle \left|\mathbf{p}'\right\rangle }$. It’s a square root because we’re dealing with a raw wave function and it’s the square modulus of this that must be normalized.

With this, we get, using 34

 $\displaystyle \left\langle \mathbf{xy}\left|\mathbf{p}\mathbf{q}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2!}}\int d^{3}p'\int d^{3}q'\phi_{\mathbf{p}'}\left(\mathbf{x}\right)\phi_{\mathbf{q}'}\left(\mathbf{y}\right)\left\langle \mathbf{p}'\mathbf{q}'\left|\mathbf{p}\mathbf{q}\right.\right\rangle \ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[\phi_{\mathbf{p}}\left(\mathbf{x}\right)\phi_{\mathbf{q}}\left(\mathbf{y}\right)+\phi_{\mathbf{q}}\left(\mathbf{x}\right)\phi_{\mathbf{p}}\left(\mathbf{y}\right)\right] \ \ \ \ \ (42)$

This is the symmetrized wave function for two identical bosons. Following through the same argument using anticommutators for fermions gives the fermion result

$\displaystyle \left\langle \mathbf{xy}\left|\mathbf{p}\mathbf{q}\right.\right\rangle _{fermion}=\frac{1}{\sqrt{2}}\left[\phi_{\mathbf{p}}\left(\mathbf{x}\right)\phi_{\mathbf{q}}\left(\mathbf{y}\right)-\phi_{\mathbf{q}}\left(\mathbf{x}\right)\phi_{\mathbf{p}}\left(\mathbf{y}\right)\right] \ \ \ \ \ (43)$

## Temperature

References: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 1.1 – 1.6

Although we’re all familiar with temperature, it’s quite difficult to give a precise definition of it. To get started, we can look at the notion of thermal equilibrium. Two objects are in thermal equilibrium if, when they are in contact, there is no net energy transferred from one to the other. This allows us to define temperature in a relative sense: if there is a net spontaneous transfer of energy from object ${A}$ to object ${B}$, then the temperature ${T_{A}}$ of ${A}$ is higher than the temperature ${T_{B}}$ of ${B}$.

To attach units to temperature we can pick some common substance such as water and consider its freezing and boiling points at standard atmospheric pressure and assign some numbers to the temperature of water at these two points. The Celsius (centigrade) scale assigns 0 to the freezing point and 100 to the boiling point. We can then insert a thermometer, such as a mercury thermometer, into the water and mark the points 0 and 100 on the tube. We can then divide up the portion of the tube between these two points into 100 equally spaced intervals, thus defining the temperature of any other object into which we place the thermometer (extending the marks below 0 and above 100 as required). This technique has obvious limitations; mercury freezes at ${-38.8^{\circ}\mbox{C}}$ so the thermometer won’t be much use below that point. At the other extreme, the glass tube will become soft and deform at some point, and mercury boils at ${356.7^{\circ}\mbox{C}}$. We’ve also made the assumption that the expansion rate of mercury is constant over the range of the thermometer, and so on…

A thermometer can also be made using the expansion and contraction of a gas with temperature. Experimentally, gases obey the ideal gas law over a wide range of pressures and temperatures. The law says

$\displaystyle PV=nRT \ \ \ \ \ (1)$

where ${P}$ is the pressure, ${V}$ the volume, ${n}$ is a measure of the number of gas molecules in the container, ${T}$ is the temperature (in kelvin) and ${R}$ is the gas constant. The kelvin scale is obtained from Celsius scale by adding 273.15 to the latter:

$\displaystyle K=C+273.15 \ \ \ \ \ (2)$

Thus absolute zero is ${0\mbox{ K}=-273.15\mbox{ C}}$.

Example 1 The Fahrenheit scale defines the freezing point of water to be ${32^{\circ}\mbox{F}}$ and the boiling point to be ${212^{\circ}\mbox{F}}$. (The origin of these rather bizarre values, or more to the point, the reason why ${0^{\circ}\mbox{F }}$is where it is, seems rather obscure; see the Wikipedia article if you’re interested.) This means that there are 180 Fahrenheit degrees between the freezing and boiling points, so there are ${\frac{180}{100}=\frac{9}{5}}$ Fahrenheit degrees per Celsius degree. Thus to convert from F to C we first subtract 32, then multiply by ${\frac{5}{9}}$:

$\displaystyle C=\frac{5}{9}\left(F-32\right) \ \ \ \ \ (3)$

or, the other way round:

$\displaystyle F=\frac{9}{5}C+32 \ \ \ \ \ (4)$

Thus absolute zero is ${\frac{9}{5}\left(-273.15\right)+32=-459.67^{\circ}\mbox{F}}$.

An approximate formula that works fairly well for temperatures in the 0 to 30 C range is to double the C value then add 30 to get the F value (or, conversely, subtract 30 from F then divide by 2 to get C). The temperature of ${-40}$ is the same on both scales (so 40 below is darned cold no matter how you measure it!).

Example 2 The Rankine temperature scale uses Fahrenheit-sized degrees, but its zero is at absolute zero, so that

 $\displaystyle R$ $\displaystyle =$ $\displaystyle F+459.67\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{9}{5}C+491.67\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{9}{5}K \ \ \ \ \ (7)$

Room temperature (21 C, say) is thus 529.47 R.

Example 3 Some examples of kelvin temperatures are

 Celsius kelvin Human body temp 37 310.15 Boiling point of water 100 373.15 A very cold day (in Dundee, anyway) ${-10}$ 263.15 Boiling point of nitrogen ${-196}$ 77.15 Melting point of lead 327 600.15

Example 4 It makes sense to say one object is “twice as hot” as another if we’re using the kelvin (or Rankine) scale, since the absolute zero of temperature is zero on these scales. Saying that 20 C is twice as hot as 10 C is wrong, of course. That’s like measuring a person’s height by defining the zero height to be at an ‘absolute’ height of 150 cm. Using that definition, we wouldn’t say that someone with an absolute height of 160 cm is twice as tall as someone with a height of 155 cm.

Example 5 The relaxation time is, roughly speaking, the time required for two objects initially at different temperatures to come to thermal equilibrium when placed in contact. Mathematically, the temperature difference declines as ${\Delta T=\Delta T_{0}e^{-At}}$ for some constant ${A}$, so it’s more precise to define relaxation time as the time required for the temperature difference to decrease to specified fraction (say ${1/e}$) of its initial value ${\Delta T_{0}}$. When you take your temperature by putting a fever thermometer under your tongue, you typically have to wait around 2 minutes before taking a reading, so that’s the relaxation time between objects ${A}$ (your mouth) and ${B}$ (the thermometer).

Example 6 The human sense of touch is notoriously bad at being able to judge temperature. A common experiment involves placing one of your hands in a bowl of cold water and the other in a bowl of hot (not too hot!) water. Leave your hands there for a minute or two and then touch an object at room temperature with each hand in turn. The cold hand will sense the object as warm, while the hot hand will sense it as cold.

## Occupation number representation; delta function as a series

References: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014) – Problem 3.1.

We can write the hamiltonian for the harmonic oscillator in terms of the creation and annihilation operators as

$\displaystyle \hat{H}=\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\right) \ \ \ \ \ (1)$

 $\displaystyle a\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (2)$ $\displaystyle a^{\dagger}\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (3)$

so the combined operator ${a^{\dagger}a}$ acts as a number operator, giving the number of quanta in a state:

 $\displaystyle a^{\dagger}a\left|n\right\rangle$ $\displaystyle =$ $\displaystyle a^{\dagger}\sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{n}a^{\dagger}\left|n-1\right\rangle \ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n\left|n\right\rangle \ \ \ \ \ (6)$

We can generalize this to a collection of independent oscillators where oscillator ${k}$ has frequency ${\omega_{k}}$. In that case

$\displaystyle \hat{H}=\hbar\sum_{k}\omega_{k}\left(a_{k}^{\dagger}a_{k}+\frac{1}{2}\right) \ \ \ \ \ (7)$

where ${a_{k}^{\dagger}}$ and ${a_{k}}$ are the creation and annihilation operators for one quantum in oscillator ${k}$. For the harmonic oscillator, the energy levels are all equally spaced, with a spacing of ${\hbar\omega_{k}}$ so if we redefine the zero point of energy to be ${\frac{1}{2}\hbar\omega_{k}}$ for oscillator ${k}$, then the hamiltonian above can be rewritten as

$\displaystyle \hat{H}=\sum_{k}n_{k}\hbar\omega_{k} \ \ \ \ \ (8)$

where ${n_{k}}$ is the number of quanta in oscillator ${k}$. An eigenstate of this hamiltonian is a state containing ${N}$ oscillators with oscillator ${k}$ containing ${n_{k}}$ quanta, which we can write as ${\left|n_{1}n_{2}...n_{N}\right\rangle }$. This is called the occupation number representation since rather than writing out a complex wave function describing all ${N}$ oscillators, we just list the number of quanta contained within each oscillator.

The application of this to quantum field theory is that we can interpret each quantum in oscillator ${k}$ as a particle with a momentum ${p_{k}}$. We’re not saying that a particle is an oscillator; rather we’re noting that we can use the same notation to refer to both particles and oscillators. So if we have a number of momentum states ${p_{k}}$ available in our system, then we can define creation and annihilation operators ${a_{p_{k}}^{\dagger}}$ and ${a_{p_{k}}}$ for that momentum state and write the hamiltonian as

$\displaystyle \hat{H}=\sum_{k}E_{p_{k}}a_{p_{k}}^{\dagger}a_{p_{k}} \ \ \ \ \ (9)$

In order for creation operators to work properly when creating elementary particles, we need to recall that there are two fundamental types of particles: fermions and bosons. The wave function for two bosons is, in position space:

$\displaystyle \psi\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)=A\left[\psi_{1}\left(\mathbf{r}_{a}\right)\psi_{2}\left(\mathbf{r}_{b}\right)+\psi_{2}\left(\mathbf{r}_{a}\right)\psi_{1}\left(\mathbf{r}_{b}\right)\right] \ \ \ \ \ (10)$

If we interchange the two particles by swapping ${\mathbf{r}_{a}}$ and ${\mathbf{r}_{b}}$, the compound wave function ${\psi\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)}$ doesn’t change, so that ${\psi\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)=\psi\left(\mathbf{r}_{b},\mathbf{r}_{a}\right)}$

If we have two fermions, on the other hand, the wave function is

$\displaystyle \psi\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)=A\left[\psi_{1}\left(\mathbf{r}_{a}\right)\psi_{2}\left(\mathbf{r}_{b}\right)-\psi_{2}\left(\mathbf{r}_{a}\right)\psi_{1}\left(\mathbf{r}_{b}\right)\right] \ \ \ \ \ (11)$

and now if we swap the particles we get ${\psi\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)=-\psi\left(\mathbf{r}_{b},\mathbf{r}_{a}\right)}$.

If we use two creation operators operating on the vacuum state ${\left|0\right\rangle }$ to create a state containing two particles, the resulting state must behave properly under the exchange of the two particles. Another way of putting this is that if we swap the order in which the particles are created we must get exactly the same state if the particles are bosons, but the negative of the original state if the particles are fermions. That is, for bosons

$\displaystyle a_{p_{1}}^{\dagger}a_{p_{2}}^{\dagger}=a_{p_{2}}^{\dagger}a_{p_{1}}^{\dagger} \ \ \ \ \ (12)$

or in terms of commutators

$\displaystyle \left[a_{p_{1}}^{\dagger},a_{p_{2}}^{\dagger}\right]=0 \ \ \ \ \ (13)$

For fermions, we’ll use the symbols ${c^{\dagger}}$ and ${c}$ for creation and annihilation operators, and in this case we must have

$\displaystyle c_{p_{1}}^{\dagger}c_{p_{2}}^{\dagger}=-c_{p_{2}}^{\dagger}c_{p_{1}}^{\dagger} \ \ \ \ \ (14)$

For fermions we define an anticommutator as

$\displaystyle \left\{ c_{p_{1}}^{\dagger},c_{p_{2}}^{\dagger}\right\} \equiv c_{p_{1}}^{\dagger}c_{p_{2}}^{\dagger}+c_{p_{2}}^{\dagger}c_{p_{1}}^{\dagger} \ \ \ \ \ (15)$

so we have

$\displaystyle \left\{ c_{p_{1}}^{\dagger},c_{p_{2}}^{\dagger}\right\} =0 \ \ \ \ \ (16)$

For the harmonic oscillator, the creation and annihilation operators satisfied the commutation relation

$\displaystyle \left[a_{p_{1}},a_{p_{2}}^{\dagger}\right]=\delta_{p_{1}p_{2}} \ \ \ \ \ (17)$

That is, the annihilation operator commutes with the creation operator if they refer to different oscillators; otherwise the commutator is 1. To complete the analogy between particles and oscillators, we just define the commutation relations between creation and annihilation operators for particles as

 $\displaystyle \left[a_{p_{1}},a_{p_{2}}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \delta_{p_{1}p_{2}}\ \ \ \ \ (18)$ $\displaystyle \left\{ c_{p_{1}},c_{p_{2}}^{\dagger}\right\}$ $\displaystyle =$ $\displaystyle \delta_{p_{1}p_{2}} \ \ \ \ \ (19)$

Example The commutation relations can be inserted into a formula which gives a new form of the Dirac delta function. For two different momentum states ${\mathbf{p}}$ and ${\mathbf{q}}$ we have, for a pair of bosons

$\displaystyle \left[a_{p},a_{q}^{\dagger}\right]=\delta_{pq} \ \ \ \ \ (20)$

Suppose that the system is enclosed in a cube of side length ${L}$. Then we can construct the sum

$\displaystyle \frac{1}{\mathcal{V}}\sum_{p,q}e^{i\left(\mathbf{p}\cdot\mathbf{x}-\mathbf{q}\cdot\mathbf{y}\right)}\left[a_{p},a_{q}^{\dagger}\right]=\frac{1}{\mathcal{V}}\sum_{p}e^{i\mathbf{p}\cdot\left(\mathbf{x}-\mathbf{y}\right)} \ \ \ \ \ (21)$

What can we make of the sum on the RHS? To see what it is, suppose we have some function ${f\left(x\right)}$ defined for ${-\pi\le x\le\pi}$. We can expand it in a Fourier series as follows:

$\displaystyle f\left(x\right)=\sum_{n=-\infty}^{\infty}c_{n}e^{inx} \ \ \ \ \ (22)$

where the coefficients are

$\displaystyle c_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f\left(x\right)e^{-inx}dx \ \ \ \ \ (23)$

We can write the Fourier series for the function at a particular point ${x=a}$ as

 $\displaystyle f\left(a\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi}\sum_{n}e^{ina}\int_{-\pi}^{\pi}f\left(x\right)e^{-inx}dx\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\pi}^{\pi}f\left(x\right)\left[\frac{1}{2\pi}\sum_{n}e^{in\left(a-x\right)}\right]dx \ \ \ \ \ (25)$

The term in brackets in the last line behaves exactly like ${\delta\left(x-a\right)}$ so we can take it as another definition of the Dirac delta function

$\displaystyle \delta\left(x-a\right)=\frac{1}{2\pi}\sum_{n}e^{in\left(a-x\right)}=\frac{1}{2\pi}\sum_{n}e^{in\left(x-a\right)} \ \ \ \ \ (26)$

where we can change the exponent in the last term because the sum over ${n}$ extends from ${-\infty}$ to ${\infty}$ so we can replace ${n}$ by ${-n}$ and get the same sum.

Now if the function ${f\left(x\right)}$ extends from 0 to ${L}$ instead of from ${-\pi}$ to ${\pi}$ we can replace ${x}$ by ${\xi\equiv Lx/2\pi}$ (and ${a}$ by ${\xi_{a}\equiv La/2\pi}$) to get

 $\displaystyle f\left(\xi_{a}\right)$ $\displaystyle =$ $\displaystyle \int_{0}^{L}f\left(\xi\right)\left[\frac{1}{2\pi}\frac{2\pi}{L}\sum_{n}e^{i2\pi n\left(\xi_{a}-\xi\right)/L}\right]d\xi\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{L}f\left(\xi\right)\left[\frac{1}{L}\sum_{p}e^{ip\left(\xi-\xi_{a}\right)}\right]d\xi \ \ \ \ \ (28)$

where

$\displaystyle p\equiv\frac{2\pi n}{L} \ \ \ \ \ (29)$

Obviously, the same argument works for the ${y}$ and ${z}$ directions, so in 3-d

 $\displaystyle f\left(\mathbf{a}\right)$ $\displaystyle =$ $\displaystyle \int_{\mathcal{V}}f\left(\mathbf{r}\right)\left[\frac{1}{L^{3}}\sum_{p}e^{i\mathbf{p}\cdot\left(\mathbf{r}-\mathbf{a}\right)}\right]d^{3}\mathbf{r}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{\mathcal{V}}f\left(\mathbf{r}\right)\left[\frac{1}{\mathcal{V}}\sum_{p}e^{i\mathbf{p}\cdot\left(\mathbf{r}-\mathbf{a}\right)}\right]d^{3}\mathbf{r} \ \ \ \ \ (31)$

so the 3-d delta function is

$\displaystyle \delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{y}\right)=\frac{1}{\mathcal{V}}\sum_{p}e^{i\mathbf{p}\cdot\left(\mathbf{x}-\mathbf{y}\right)} \ \ \ \ \ (32)$

From 21 we get

$\displaystyle \frac{1}{\mathcal{V}}\sum_{p,q}e^{i\left(\mathbf{p}\cdot\mathbf{x}-\mathbf{q}\cdot\mathbf{y}\right)}\left[a_{p},a_{q}^{\dagger}\right]=\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{y}\right) \ \ \ \ \ (33)$

## Harmonic oscillator ground state from annihilation operator

References: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014) – Problem 2.4.

We can use the annihilation operator ${\hat{a}}$ in the harmonic oscillator to reclaim the position space form of the ground state wave function. The operator is

 $\displaystyle \hat{a}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega}}\left[i\hat{p}+m\omega\hat{x}\right] \ \ \ \ \ (1)$

Applying ${\hat{a}}$ to the ground state ${\left|0\right\rangle }$ we get 0 (that is, annihilating the ground state eliminates the wave function altogether), so

$\displaystyle \left[i\hat{p}+m\omega\hat{x}\right]\left|0\right\rangle =0 \ \ \ \ \ (2)$

The eigenfunction of position is found from

$\displaystyle \hat{x}\left|x_{0}\right\rangle =x_{0}\left|x_{0}\right\rangle \ \ \ \ \ (3)$

Since the operator ${\hat{x}}$ multiplies any function by the position ${x}$ and we want the eigenfunction ${\left|x_{0}\right\rangle }$ to represent a particular position ${x_{0}}$, ${\left|x_{0}\right\rangle }$ must pick out ${x_{0}}$ from all possible values of ${x}$, that is, it must be zero everywhere except ${x=x_{0}}$. This condition is satisfied if we take

$\displaystyle \left|x_{0}\right\rangle =\delta\left(x-x_{0}\right) \ \ \ \ \ (4)$

We then get

 $\displaystyle \left\langle x\left|\hat{p}\right|\psi\right\rangle$ $\displaystyle =$ $\displaystyle \int\delta\left(x'-x\right)\hat{p}\psi\left(x'\right)\; dx'\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\int\delta\left(x'-x\right)\frac{d}{dx'}\psi\left(x'\right)\; dx'\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\frac{d}{dx}\int\delta\left(x'-x\right)\psi\left(x'\right)\; dx'\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\frac{d}{dx}\left\langle x\left|\psi\right.\right\rangle \ \ \ \ \ (8)$

Also

 $\displaystyle \left\langle x\left|\hat{x}\right|\psi\right\rangle$ $\displaystyle =$ $\displaystyle \int\delta\left(x'-x\right)\hat{x}\psi\left(x'\right)\; dx'\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\int\delta\left(x'-x\right)x'\psi\left(x'\right)\; dx'\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar x\int\delta\left(x'-x\right)\psi\left(x'\right)\; dx'\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar x\left\langle x\left|\psi\right.\right\rangle \ \ \ \ \ (12)$

Therefore, from 2 we get

 $\displaystyle \left\langle x\left|\left[i\hat{p}+m\omega\hat{x}\right]\right|0\right\rangle$ $\displaystyle =$ $\displaystyle \hbar\frac{d}{dx}\left\langle x\left|0\right.\right\rangle +m\omega x\left\langle x\left|0\right.\right\rangle =0\ \ \ \ \ (13)$ $\displaystyle \hbar\frac{d}{dx}\left\langle x\left|0\right.\right\rangle$ $\displaystyle =$ $\displaystyle -m\omega x\left\langle x\left|0\right.\right\rangle \ \ \ \ \ (14)$

This is a differential equation for ${\left\langle x\left|0\right.\right\rangle }$ which has the solution

$\displaystyle \left\langle x\left|0\right.\right\rangle =Ae^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (15)$

where ${A}$ is found from normalization:

 $\displaystyle \int\left|\left\langle x\left|0\right.\right\rangle \right|^{2}dx$ $\displaystyle =$ $\displaystyle A^{2}\int_{-\infty}^{\infty}e^{-m\omega x^{2}/\hbar}dx=1\ \ \ \ \ (16)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (17)$

This is the same function that we got earlier.

## Coupled oscillators in terms of creation and annihilation operators; phonons

References: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014) – Problem 2.3.

Suppose we have a one-dimensional chain of ${N}$ equal masses ${m}$ connected by identical springs of rest length ${a}$ and spring constant ${k}$, so that mass ${j}$ has a rest position of ${ja}$, ${j=0...N-1}$. The masses can move only in the ${x}$ direction, and moving one mass extends the spring on one side while compressing the spring on the other side. The total energy of the system is the sum of the kinetic energies of the masses and the potential energies of the springs. The kinetic energy of mass ${j}$ is just ${\frac{1}{2}mv_{j}^{2}}$ or, in terms of the momentum operator

$\displaystyle T_{j}=\frac{\hat{p}_{j}^{2}}{2m} \ \ \ \ \ (1)$

The potential energy of the spring connecting masses ${j}$ and ${j+1}$ is ${\frac{1}{2}k\left(\Delta x\right)^{2}}$ where ${\Delta x}$ is the amount the spring is stretched (or compressed). ${\Delta x}$ is the difference between the amounts that the two masses on either end have moved from their equilibrium positions, so if we call ${x_{j}}$ the amount by which mass ${j}$ has moved from position ${ja}$, then

 $\displaystyle \Delta x_{j,j+1}$ $\displaystyle =$ $\displaystyle x_{j+1}-x_{j}\ \ \ \ \ (2)$ $\displaystyle V_{j}$ $\displaystyle =$ $\displaystyle \frac{1}{2}K\left(\hat{x}_{j+1}-\hat{x}_{j}\right)^{2} \ \ \ \ \ (3)$

where we’ve added hats to show that ${\hat{x}_{j}}$ is an operator. Therefore the hamiltonian of the system is

$\displaystyle \hat{H}=\sum_{j}\frac{\hat{p}_{j}^{2}}{2m}+\sum_{j}\frac{1}{2}K\left(\hat{x}_{j+1}-\hat{x}_{j}\right)^{2} \ \ \ \ \ (4)$

As it stands, the hamiltonian contains terms such as ${k\hat{x}_{j}\hat{x}_{j+1}}$ in which the spatial terms of two masses occur in a product, that is, it contains coupled terms. We can convert the hamiltonian to an uncoupled system in which it consists of a sum of terms where each term refers to only a single index. This is done by using discrete Fourier transforms. A discrete Fourier transform assumes that the raw data (the values of ${x_{j}}$ and ${p_{j}}$) are samples at equally spaced intervals and that the behaviour outside the observed range (that is, for ${j<0}$ and ${j\ge N}$) is periodic, so that it repeats the observed behaviour with a period of ${Na}$. This is equivalent to imposing periodic boundary conditions so that

 $\displaystyle x_{j+N}$ $\displaystyle =$ $\displaystyle x_{j}\ \ \ \ \ (5)$ $\displaystyle p_{j+N}$ $\displaystyle =$ $\displaystyle p_{j} \ \ \ \ \ (6)$

The discrete Fourier transform is then

 $\displaystyle x_{j}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{N}}\sum_{k}\tilde{x}_{k}e^{ikja}\ \ \ \ \ (7)$ $\displaystyle p_{j}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{N}}\sum_{k}\tilde{p}_{k}e^{ikja} \ \ \ \ \ (8)$

[This transform differs from the one in the earlier reference in that it has a factor of ${1/\sqrt{N}}$ in front. All that matters is that the product of this factor with the corresponding factor in front of the inverse transform is ${1/N}$.] The index ${k}$ is the frequency and because of the periodic boundary conditions, we must have

 $\displaystyle e^{ikja}$ $\displaystyle =$ $\displaystyle e^{ik\left(j+N\right)a}\ \ \ \ \ (9)$ $\displaystyle e^{ikNa}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (10)$ $\displaystyle k$ $\displaystyle =$ $\displaystyle \frac{2\pi m}{Na} \ \ \ \ \ (11)$

for an integer ${m}$ which is in a range such that ${ka}$ varies over ${2\pi}$. Any range of ${m}$ that satisfies this condition would do, but it turns out to be most convenient to choose ${-\frac{N}{2}. This gives ${-\pi.

We can now work out the hamiltonian 4 using the Fourier transformed variables ${\tilde{x}_{k}}$ and ${\tilde{p}_{k}}$. First, the kinetic energy term:

 $\displaystyle \sum_{j}p_{j}^{2}$ $\displaystyle =$ $\displaystyle \sum_{j}\left(\frac{1}{\sqrt{N}}\sum_{k}\tilde{p}_{k}e^{ikja}\right)\left(\frac{1}{\sqrt{N}}\sum_{k'}\tilde{p}_{k'}e^{ik'ja}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{N}\sum_{j}\sum_{k}\sum_{k'}\tilde{p}_{k}\tilde{p}_{k'}e^{i(k+k')ja} \ \ \ \ \ (13)$

We can now make use of the sum (derived from a geometric series):

$\displaystyle \sum_{j}e^{i2\pi mj/N}=N\delta_{m,0} \ \ \ \ \ (14)$

This means that

$\displaystyle \sum_{j}e^{i(k+k')ja}=N\delta_{k,-k'} \ \ \ \ \ (15)$

so

$\displaystyle \sum_{j}p_{j}^{2}=\sum_{k}\tilde{p}_{k}\tilde{p}_{-k} \ \ \ \ \ (16)$

For the potential energy term we have

 $\displaystyle \sum_{j}\left(\hat{x}_{j+1}-\hat{x}_{j}\right)^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{N}\sum_{j}\left(\sum_{k}\tilde{x}_{k}e^{ik\left(j+1\right)a}-\sum_{k}\tilde{x}_{k}e^{ikja}\right)\left(\sum_{k'}\tilde{x}_{k'}e^{ik'\left(j+1\right)a}-\sum_{k'}\tilde{x}_{k'}e^{ik'ja}\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{N}\sum_{j}\left(\sum_{k}\tilde{x}_{k}e^{ikja}\left(e^{ika}-1\right)\right)\left(\sum_{k'}\tilde{x}_{k'}e^{ik'ja}\left(e^{ik'a}-1\right)\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{N}\sum_{j}\sum_{k}\sum_{k'}\tilde{x}_{k}\tilde{x}_{k'}e^{i\left(k+k'\right)ja}\left(e^{ika}-1\right)\left(e^{ik'a}-1\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\tilde{x}_{k}\tilde{x}_{-k}\left(e^{ika}-1\right)\left(e^{-ika}-1\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\tilde{x}_{k}\tilde{x}_{-k}\left(2-\left(e^{ika}+e^{-ika}\right)\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\sum_{k}\tilde{x}_{k}\tilde{x}_{-k}\left(1-\cos\left(ka\right)\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\sum_{k}\tilde{x}_{k}\tilde{x}_{-k}\sin^{2}\frac{ka}{2} \ \ \ \ \ (23)$

It might not seem that we’ve made much progress, since now both the kinetic and potential energy terms appear to be coupled, involving products of ${+k}$ and ${-k}$ modes. However, the inverses of 7 and 8 are

 $\displaystyle \tilde{x}_{k}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{N}}\sum_{j}x_{j}e^{-ikja}\ \ \ \ \ (24)$ $\displaystyle \tilde{p}_{k}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{N}}\sum_{j}p_{j}e^{-ikja} \ \ \ \ \ (25)$

Since ${x_{j}}$ and ${p_{j}}$ are observables, they must be hermitian operators, so ${x_{j}^{\dagger}=x_{j}}$ and ${p_{j}^{\dagger}=p_{j}}$ so

 $\displaystyle \tilde{x}_{k}^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{N}}\sum_{j}x_{j}e^{ikja}=\tilde{x}_{-k}\ \ \ \ \ (26)$ $\displaystyle \tilde{p}_{k}^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{N}}\sum_{j}p_{j}e^{ikja}=\tilde{p}_{-k} \ \ \ \ \ (27)$

Therefore

 $\displaystyle \sum_{j}p_{j}^{2}$ $\displaystyle =$ $\displaystyle \sum_{k}\tilde{p}_{k}\tilde{p}_{-k}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\tilde{p}_{k}\tilde{p}_{k}^{\dagger}\ \ \ \ \ (29)$ $\displaystyle \sum_{j}\left(\hat{x}_{j+1}-\hat{x}_{j}\right)^{2}$ $\displaystyle =$ $\displaystyle 4\sum_{k}\tilde{x}_{k}\tilde{x}_{-k}\sin^{2}\frac{ka}{2}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\sum_{k}\tilde{x}_{k}\tilde{x}_{k}^{\dagger}\sin^{2}\frac{ka}{2}\ \ \ \ \ (31)$ $\displaystyle \hat{H}$ $\displaystyle =$ $\displaystyle \frac{1}{2m}\sum_{k}\tilde{p}_{k}\tilde{p}_{k}^{\dagger}+2K\sum_{k}\tilde{x}_{k}\tilde{x}_{k}^{\dagger}\sin^{2}\frac{ka}{2}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2m}\sum_{k}\tilde{p}_{k}\tilde{p}_{k}^{\dagger}+\frac{1}{2}m\sum_{k}\omega_{k}^{2}\tilde{x}_{k}\tilde{x}_{k}^{\dagger}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2m}\sum_{k}\tilde{p}_{k}\tilde{p}_{-k}+\frac{1}{2}m\sum_{k}\omega_{k}^{2}\tilde{x}_{k}\tilde{x}_{-k} \ \ \ \ \ (34)$

where

$\displaystyle \omega_{k}^{2}\equiv4\frac{K}{m}\sin^{2}\frac{ka}{2} \ \ \ \ \ (35)$

That is, we’ve managed to write the hamiltonian as the sum over uncoupled oscillators, where oscillator ${k}$ has frequency ${\omega_{k}}$. The catch is that the operators ${\tilde{p}_{k}}$ and ${\tilde{x}_{k}}$ are in frequency space, not ‘normal’ space, so they are the ‘momentum’ and ‘position’ operators for the modes of oscillation of the coupled set of oscillators.

The formal equivalence of the hamiltonian in mode space with the hamiltonian for a single oscillator in normal space means we can define creation and annihilation operators in the same way. That is (reverting to ‘hat’ notation to indicate operators):

 $\displaystyle \hat{a}_{k}^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega_{k}}}\left[-i\hat{p}_{k}^{\dagger}+m\omega_{k}\hat{x}_{k}^{\dagger}\right]\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega_{k}}}\left[-i\hat{p}_{-k}+m\omega_{k}\hat{x}_{-k}\right]\ \ \ \ \ (37)$ $\displaystyle \hat{a}_{k}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega_{k}}}\left[i\hat{p}_{k}+m\omega_{k}\hat{x}_{k}\right] \ \ \ \ \ (38)$

Inverting these equations, we get (note that ${\omega_{k}}$ is always the positive square root of 35):

 $\displaystyle \hat{a}_{-k}^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega_{k}}}\left[-i\hat{p}_{k}+m\omega_{k}\hat{x}_{k}\right]\ \ \ \ \ (39)$ $\displaystyle \hat{a}_{-k}^{\dagger}+\hat{a}_{k}$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{2\hbar m\omega_{k}}}m\omega_{k}\hat{x}_{k}\ \ \ \ \ (40)$ $\displaystyle \hat{x}_{k}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega_{k}}}\left(\hat{a}_{-k}^{\dagger}+\hat{a}_{k}\right)\ \ \ \ \ (41)$ $\displaystyle \hat{p}_{k}$ $\displaystyle =$ $\displaystyle i\sqrt{\frac{\hbar m\omega_{k}}{2}}\left(\hat{a}_{-k}^{\dagger}-\hat{a}_{k}\right) \ \ \ \ \ (42)$

Example We can express the original space coordinate ${x_{j}}$ in terms of the creation and annihilation operators. From 7

 $\displaystyle x_{j}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{N}}\sum_{k}\hat{x}_{k}e^{ikja}\ \ \ \ \ (43)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2mN}}\sum_{k}\frac{1}{\sqrt{\omega_{k}}}\left(\hat{a}_{-k}^{\dagger}+\hat{a}_{k}\right)e^{ikja}\ \ \ \ \ (44)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2mN}}\sum_{k}\frac{1}{\sqrt{\omega_{k}}}\left(\hat{a}_{k}^{\dagger}e^{-ikja}+\hat{a}_{k}e^{ikja}\right) \ \ \ \ \ (45)$

where the last line follows from the fact that we’re summing over ${k}$ over a range of values symmetric about ${k=0}$ so we can replace ${k}$ by ${-k}$ and get the same sum.

Inserting 41 and 42 into 34 we get

 $\displaystyle \hat{H}$ $\displaystyle =$ $\displaystyle -\frac{\hbar}{4}\sum_{k}\omega_{k}\left(\hat{a}_{-k}^{\dagger}-\hat{a}_{k}\right)\left(\hat{a}_{k}^{\dagger}-\hat{a}_{-k}\right)+\frac{\hbar}{4}\sum_{k}\omega_{k}\left(\hat{a}_{-k}^{\dagger}+\hat{a}_{k}\right)\left(\hat{a}_{k}^{\dagger}+\hat{a}_{-k}\right)\ \ \ \ \ (46)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\sum_{k}\omega_{k}\left(\hat{a}_{-k}^{\dagger}\hat{a}_{-k}+\hat{a}_{k}\hat{a}_{k}^{\dagger}\right) \ \ \ \ \ (47)$

Since the commutator is

$\displaystyle \left[\hat{a}_{k},\hat{a}_{k}^{\dagger}\right]=1 \ \ \ \ \ (48)$

we get

 $\displaystyle \hat{H}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\sum_{k}\omega_{k}\left(\hat{a}_{-k}^{\dagger}\hat{a}_{-k}+1+\hat{a}_{-k}^{\dagger}\hat{a}_{-k}\right)\ \ \ \ \ (49)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\hbar\omega_{k}\left(\hat{a}_{-k}^{\dagger}\hat{a}_{-k}+\frac{1}{2}\right)\ \ \ \ \ (50)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\hbar\omega_{k}\left(\hat{a}_{k}^{\dagger}\hat{a}_{k}+\frac{1}{2}\right) \ \ \ \ \ (51)$

That is, the hamiltonian is the sum of the hamiltonians for individual oscillators in terms of creation and annihilation operators. The ‘particles’ that are created or annihilated are the modes of oscillation in the chain of ‘real’ oscillators; these modes are called phonons, since they are reminiscent of sound waves passing through a medium.

## Berry’s phase: definition and value for a spin-1 particle in a magnetic field

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.6.

So far, when calculating phases in the adiabatic theorem we’ve assumed that there is only one parameter ${R}$ in the hamiltonian that is time-dependent. In that case, we can write the geometric phase as

 $\displaystyle \gamma_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle i\int_{0}^{t}\left\langle \psi_{n}\left(t'\right)\left|\frac{\partial}{\partial t'}\psi_{n}\left(t'\right)\right.\right\rangle dt'\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\int_{R_{i}}^{R_{f}}\left\langle \psi_{n}\left|\frac{\partial}{\partial R}\psi_{n}\right.\right\rangle dR \ \ \ \ \ (2)$

and if we take the system through a complete loop where we start at ${R=R_{i}}$ then take ${R}$ out to ${R_{f}}$ then back to ${R_{i}}$ again, the net phase is always zero because the two limits on the integral are the same once we’ve travelled the complete loop.

However, if the hamiltonian has two or more time-dependent parameters, then the chain rule for derivatives says that

$\displaystyle \frac{\partial\psi_{n}}{\partial t}=\frac{\partial\psi_{n}}{\partial R_{1}}\frac{dR_{1}}{dt}+\frac{\partial\psi_{n}}{\partial R_{2}}\frac{dR_{2}}{dt}+...+\frac{\partial\psi_{n}}{\partial R_{N}}\frac{dR_{N}}{dt} \ \ \ \ \ (3)$

If we treat the complete set of parameters ${R_{j}}$ as the components of an ${N}$-dimensional vector, we can define a gradient in the ${R_{j}}$ coordinate system as ${\nabla_{R}}$ and rewrite this derivative as

$\displaystyle \frac{\partial\psi_{n}}{\partial t}=\left(\nabla_{R}\psi_{n}\right)\cdot\frac{d\mathbf{R}}{dt} \ \ \ \ \ (4)$

giving the phase as

$\displaystyle \gamma_{n}\left(t\right)=i\int_{\mathbf{R}_{i}}^{\mathbf{R}_{f}}\left\langle \psi_{n}\left|\nabla_{R}\psi_{n}\right.\right\rangle \cdot d\mathbf{R} \ \ \ \ \ (5)$

If we now take the system around a closed loop in ${R}$-space in time ${T}$, we can write the phase change over that loop as a line integral around the path:

$\displaystyle \gamma_{n}\left(t\right)=i\oint\left\langle \psi_{n}\left|\nabla_{R}\psi_{n}\right.\right\rangle \cdot d\mathbf{R} \ \ \ \ \ (6)$

As this is the line integral of a vector around a closed path, if ${\mathbf{R}}$ consists of three parameters, we can convert it to a surface integral over the area enclosed by the path by using Stokes’s theorem:

$\displaystyle \gamma_{n}\left(t\right)=i\int\left(\nabla\times\left\langle \psi_{n}\left|\nabla_{R}\psi_{n}\right.\right\rangle \right)\cdot d\mathbf{a} \ \ \ \ \ (7)$

This phase is known as Berry’s phase and is not, in general, zero. Griffiths works out the classic example of calculating Berry’s phase for an electron in a precessing magnetic field and then, more generally, for an electron in a magnetic field of constant magnitude but varying in direction by sweeping out some closed path (of any shape). The results apply to any spin-1/2 particle, so here we’ll work out Berry’s phase for a particle of spin 1.

Ultimately, what we want is to work out 7 for some initial spin state of the particle. If we’re using spherical coordinates, then ${\nabla_{R}}$ is the usual gradient in spherical coordinates, so to complete the calculation, we need to know ${\psi_{n}}$. Suppose we start with the particle in the ${+1}$ spin state (for spin 1, the ${z}$ component can have values ${\pm\hbar}$ and 0, so a spin of ${+1}$ corresponds to ${+\hbar}$). The spin matrices are

 $\displaystyle S_{z}$ $\displaystyle =$ $\displaystyle \hbar\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array}\right)\ \ \ \ \ (8)$ $\displaystyle S_{x}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{\sqrt{2}}\left(\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array}\right)\ \ \ \ \ (9)$ $\displaystyle S_{y}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{\sqrt{2}}\left(\begin{array}{ccc} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end{array}\right) \ \ \ \ \ (10)$

so the component of ${\mathbf{S}}$ along a direction (which is taken to be the magnetic field’s instantaneous direction) given by

$\displaystyle \hat{\mathbf{r}}=\sin\theta\cos\phi\hat{\mathbf{x}}+\sin\theta\sin\phi\hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}} \ \ \ \ \ (11)$

is

 $\displaystyle \mathbf{S}\cdot\hat{\mathbf{r}}$ $\displaystyle =$ $\displaystyle \hbar\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array}\right)\cos\theta+\frac{\hbar}{\sqrt{2}}\left(\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array}\right)\sin\theta\cos\phi+\frac{\hbar}{\sqrt{2}}\left(\begin{array}{ccc} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end{array}\right)\sin\theta\sin\phi\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{\sqrt{2}}\left(\begin{array}{ccc} \sqrt{2}\cos\theta & \sin\theta e^{-i\phi} & 0\\ \sin\theta e^{i\phi} & 0 & \sin\theta e^{-i\phi}\\ 0 & \sin\theta e^{i\phi} & -\sqrt{2}\cos\theta \end{array}\right) \ \ \ \ \ (13)$

The eigenvalues of this matrix are ${\pm\hbar,0}$ as required and the normalized eigenvector corresponding to ${+\hbar}$ is

$\displaystyle \chi_{+1}=\frac{1}{2}\left[\begin{array}{c} e^{-2i\phi}\left(1+\cos\theta\right)\\ \sqrt{2}e^{-i\phi}\sin\theta\\ 1-\cos\theta \end{array}\right] \ \ \ \ \ (14)$

The gradient in spherical coordinates of ${\chi_{+1}}$ has components only in the ${\theta}$ and ${\phi}$ directions and we have

 $\displaystyle \nabla\chi_{+1}$ $\displaystyle =$ $\displaystyle \frac{1}{r}\frac{\partial\chi_{+1}}{\partial\theta}\hat{\boldsymbol{\theta}}+\frac{1}{r\sin\theta}\frac{\partial\chi_{+1}}{\partial\phi}\hat{\boldsymbol{\phi}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2r}\left[\begin{array}{c} -e^{-2i\phi}\sin\theta\\ \sqrt{2}e^{-i\phi}\cos\theta\\ \sin\theta \end{array}\right]\hat{\boldsymbol{\theta}}-\frac{i}{2r}\left[\begin{array}{c} \frac{e^{-2i\phi}\left(1+\cos\theta\right)}{\sin\theta}\\ \sqrt{2}e^{-i\phi}\\ 0 \end{array}\right]\hat{\boldsymbol{\phi}} \ \ \ \ \ (16)$

We can now work out

 $\displaystyle \left\langle \chi_{+1}\left|\nabla\chi_{+1}\right.\right\rangle$ $\displaystyle =$ $\displaystyle 0\hat{\boldsymbol{\theta}}-i\frac{1+\cos\theta}{r\sin\theta}\hat{\boldsymbol{\phi}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\frac{1+\cos\theta}{r\sin\theta}\hat{\boldsymbol{\phi}} \ \ \ \ \ (18)$

The curl of this has a component only in the ${r}$ direction:

 $\displaystyle \nabla\times\left\langle \chi_{+1}\left|\nabla\chi_{+1}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\left[-i\frac{1+\cos\theta}{r\sin\theta}\right]\hat{\mathbf{r}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i}{r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (20)$

To get the phase we need to integrate this over the surface enclosed by a complete loop traversed by the point of the ${\mathbf{B}}$ field, that is

$\displaystyle \gamma=i\int\frac{i}{r^{2}}\hat{\mathbf{r}}\cdot d\mathbf{a} \ \ \ \ \ (21)$

Since the magnetic field’s magnitude is constant, the traversed path is on the surface of a sphere with radius ${r}$ and ${d\mathbf{a}}$ subtends an element of solid angle on this sphere so that

$\displaystyle d\mathbf{a}=r^{2}\hat{\mathbf{r}}d\Omega \ \ \ \ \ (22)$

The integral thus comes out to

$\displaystyle \gamma=-\int d\Omega=-\Omega \ \ \ \ \ (23)$

## Geometric phase is always zero for real wave functions

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.5.

Here’s another example of calculating phases in the adiabatic theorem which says that if a system starts out in the ${n}$th state of a time-dependent hamiltonian, and the hamiltonian changes slowly compared to the internal period of the time-independent wave function (that is, the time scale over which the hamiltonian changes is much longer than ${\hbar/E_{n}}$), then after a time ${t}$ the system will end up in state

$\displaystyle \Psi_{n}\left(t\right)=e^{i\theta_{n}\left(t\right)}e^{i\gamma_{n}\left(t\right)}\psi_{n}\left(t\right) \ \ \ \ \ (1)$

where

 $\displaystyle \theta_{n}\left(t\right)$ $\displaystyle \equiv$ $\displaystyle -\frac{1}{\hbar}\int_{0}^{t}E_{n}\left(t'\right)dt'\ \ \ \ \ (2)$ $\displaystyle \gamma_{n}\left(t\right)$ $\displaystyle \equiv$ $\displaystyle i\int_{0}^{t}\left\langle \psi_{n}\left(t'\right)\left|\frac{\partial}{\partial t'}\psi_{n}\left(t'\right)\right.\right\rangle dt' \ \ \ \ \ (3)$

${\theta}$ is called the dynamic phase and ${\gamma}$ is called the geometric phase. If ${\psi_{n}}$ is real, then ${\gamma_{n}}$ is always zero, as we can see by differentiating the normalization condition:

 $\displaystyle \left\langle \psi_{n}\left|\psi_{n}\right.\right\rangle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (4)$ $\displaystyle \frac{d}{dt}\left\langle \psi_{n}\left|\psi_{n}\right.\right\rangle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \dot{\psi}_{n}\left|\psi_{n}\right.\right\rangle +\left\langle \psi_{n}\left|\dot{\psi}_{n}\right.\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \psi_{n}\left|\dot{\psi}_{n}\right.\right\rangle ^*+\left\langle \psi_{n}\left|\dot{\psi}_{n}\right.\right\rangle \ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\Re\left(\left\langle \psi_{n}\left|\dot{\psi}_{n}\right.\right\rangle \right) \ \ \ \ \ (8)$

That is, ${\left\langle \psi_{n}\left(t'\right)\left|\frac{\partial}{\partial t'}\psi_{n}\left(t'\right)\right.\right\rangle }$ must be purely imaginary, so if ${\psi_{n}}$ is real, the bracket must be zero. This also means that ${\gamma}$ is always real.

We can multiply the real wave function ${\psi_{n}}$ by a phase factor ${e^{i\phi_{n}}}$ where ${\phi_{n}}$ is a real function of whatever parameters are dependent on time in the hamiltonian (but ${\phi_{n}}$ is not a function of ${x}$). In that case we have a new wave function (we’ll drop the subscript ${n}$ to save time):

 $\displaystyle \psi'$ $\displaystyle =$ $\displaystyle e^{i\phi}\psi\ \ \ \ \ (9)$ $\displaystyle \dot{\psi}'$ $\displaystyle =$ $\displaystyle i\dot{\phi}e^{i\phi}\psi+e^{i\phi}\dot{\psi}\ \ \ \ \ (10)$ $\displaystyle \left\langle \psi'\left|\dot{\psi}'\right.\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|i\dot{\phi}\psi+\dot{\psi}\right.\right\rangle \ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\left\langle \psi\left|\dot{\phi}\psi\right.\right\rangle +\left\langle \psi\left|\dot{\psi}\right.\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\dot{\phi} \ \ \ \ \ (13)$

where in the last line we took ${\dot{\phi}}$ outside the bracket since it doesn’t depend on ${x}$ and used ${\left\langle \psi\left|\dot{\psi}\right.\right\rangle =0}$. The geometric phase for the modified wave function is therefore

 $\displaystyle \gamma'$ $\displaystyle =$ $\displaystyle i\int_{0}^{t}i\dot{\phi}dt'\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(\phi\left(t\right)-\phi\left(0\right)\right) \ \ \ \ \ (15)$

Putting this back into 1 we get

 $\displaystyle \Psi'\left(t\right)$ $\displaystyle =$ $\displaystyle e^{i\theta\left(t\right)}e^{-i\left(\phi\left(t\right)-\phi\left(0\right)\right)}\psi'\left(t\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{i\theta\left(t\right)}e^{-i\left(\phi\left(t\right)-\phi\left(0\right)\right)}e^{i\phi\left(t\right)}\psi\left(t\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{i\theta\left(t\right)}e^{i\phi\left(0\right)}\psi\left(t\right) \ \ \ \ \ (18)$

Although the wave function picks up a constant phase ${\phi\left(0\right)}$, there is no time-dependent geometric phase.

## Phases in the adiabatic theorem: delta function well

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.4.

Here’s another example of calculating phases in the adiabatic theorem which says that if a system starts out in the ${n}$th state of a time-dependent hamiltonian, and the hamiltonian changes slowly compared to the internal period of the time-independent wave function (that is, the time scale over which the hamiltonian changes is much longer than ${\hbar/E_{n}}$), then after a time ${t}$ the system will end up in state

$\displaystyle \Psi_{n}\left(t\right)=e^{i\theta_{n}\left(t\right)}e^{i\gamma_{n}\left(t\right)}\psi_{n}\left(t\right) \ \ \ \ \ (1)$

where

 $\displaystyle \theta_{n}\left(t\right)$ $\displaystyle \equiv$ $\displaystyle -\frac{1}{\hbar}\int_{0}^{t}E_{n}\left(t'\right)dt'\ \ \ \ \ (2)$ $\displaystyle \gamma_{n}\left(t\right)$ $\displaystyle \equiv$ $\displaystyle i\int_{0}^{t}\left\langle \psi_{n}\left(t'\right)\left|\frac{\partial}{\partial t'}\psi_{n}\left(t'\right)\right.\right\rangle dt' \ \ \ \ \ (3)$

${\theta}$ is called the dynamic phase and ${\gamma}$ is called the geometric phase.

The wave functions ${\psi_{n}\left(t\right)}$ are the solutions of the eigenvalue equation at a particular time ${t}$:

$\displaystyle H\left(t\right)\psi_{n}\left(t\right)=E_{n}\left(t\right)\psi_{n}\left(t\right) \ \ \ \ \ (4)$

That is, they aren’t a full solution of the time dependent SchrÃ¶dinger equation; rather they are the solutions of the time-independent SchrÃ¶dinger equation with whatever parameters are now time-dependent in the hamiltonian replaced by their time-dependent forms.

With a delta function well the potential is

$\displaystyle V\left(x\right)=-\alpha\delta\left(x\right) \ \ \ \ \ (5)$

and the time-independent wave function for the bound state is

$\displaystyle \psi\left(x\right)=\frac{\sqrt{m\alpha}}{\hbar}e^{-m\alpha\left|x\right|/\hbar^{2}} \ \ \ \ \ (6)$

If the strength of the delta function ${\alpha}$ varies with time, then

 $\displaystyle \gamma_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle i\int_{\alpha_{1}}^{\alpha_{2}}\left\langle \psi_{n}\left(t'\right)\left|\frac{\partial}{\partial\alpha}\psi_{n}\left(\alpha\right)\right.\right\rangle d\alpha\ \ \ \ \ (7)$ $\displaystyle \frac{\partial}{\partial\alpha}\psi_{n}\left(\alpha\right)$ $\displaystyle =$ $\displaystyle e^{-m\alpha\left|x\right|/\hbar^{2}}\frac{m\left(2m\alpha\left|x\right|-\hbar^{2}\right)}{2\hbar^{3}\sqrt{m\alpha}}\ \ \ \ \ (8)$ $\displaystyle \left\langle \psi_{n}\left(t'\right)\left|\frac{\partial}{\partial\alpha}\psi_{n}\left(\alpha\right)\right.\right\rangle$ $\displaystyle =$ $\displaystyle \frac{m}{2\hbar^{4}}\int_{-\infty}^{\infty}e^{-2m\alpha\left|x\right|/\hbar^{2}}\left(2m\alpha\left|x\right|-\hbar^{2}\right)dx\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{\hbar^{4}}\int_{0}^{\infty}e^{-2m\alpha x/\hbar^{2}}\left(2m\alpha x-\hbar^{2}\right)dx\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left.\frac{m}{2\hbar^{2}}xe^{-2m\alpha x/\hbar^{2}}\right|_{0}^{\infty}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (12)$

Therefore ${\gamma_{n}=0}$.

The bound state energy is

$\displaystyle E=-\frac{m\alpha^{2}}{2\hbar^{2}} \ \ \ \ \ (13)$

so the dynamic phase is

$\displaystyle \theta_{n}\left(t\right)=\frac{m}{2\hbar^{3}}\int_{0}^{t}\alpha^{2}\left(t'\right)dt' \ \ \ \ \ (14)$

If ${\alpha}$ changes at a constant rate, then ${d\alpha/dt=c}$ and ${\alpha\left(t\right)=\alpha_{1}+ct}$, so

 $\displaystyle \theta_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle \frac{m}{2\hbar^{3}}\int_{0}^{\left(\alpha_{2}-\alpha_{1}\right)/c}\left(\alpha_{1}+ct\right)^{2}dt\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\left(\alpha_{1}^{3}-\alpha_{2}^{3}\right)}{6c\hbar^{3}} \ \ \ \ \ (16)$

## Phases in the adiabatic approximation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.3.

The adiabatic theorem (see Griffiths, section 10.1 for a proof) says that if a system starts out in the ${n}$th state of a time-dependent hamiltonian, and the hamiltonian changes slowly compared to the internal period of the time-independent wave function (that is, the time scale over which the hamiltonian changes is much longer than ${\hbar/E_{n}}$), then after a time ${t}$ the system will end up in state

$\displaystyle \Psi_{n}\left(t\right)=e^{i\theta_{n}\left(t\right)}e^{i\gamma_{n}\left(t\right)}\psi_{n}\left(t\right) \ \ \ \ \ (1)$

where

 $\displaystyle \theta_{n}\left(t\right)$ $\displaystyle \equiv$ $\displaystyle -\frac{1}{\hbar}\int_{0}^{t}E_{n}\left(t'\right)dt'\ \ \ \ \ (2)$ $\displaystyle \gamma_{n}\left(t\right)$ $\displaystyle \equiv$ $\displaystyle i\int_{0}^{t}\left\langle \psi_{n}\left(t'\right)\left|\frac{\partial}{\partial t'}\psi_{n}\left(t'\right)\right.\right\rangle dt' \ \ \ \ \ (3)$

${\theta}$ is called the dynamic phase and ${\gamma}$ is called the geometric phase.

The wave functions ${\psi_{n}\left(t\right)}$ are the solutions of the eigenvalue equation at a particular time ${t}$:

$\displaystyle H\left(t\right)\psi_{n}\left(t\right)=E_{n}\left(t\right)\psi_{n}\left(t\right) \ \ \ \ \ (4)$

That is, they aren’t a full solution of the time dependent SchrÃ¶dinger equation; rather they are the solutions of the time-independent SchrÃ¶dinger equation with whatever parameters are now time-dependent in the hamiltonian replaced by their time-dependent forms.

For example, with an infinite square well whose right wall moves so that its position ${w}$ is a function of time ${w\left(t\right)}$, we have

 $\displaystyle \psi_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{w\left(t\right)}}\sin\frac{n\pi}{w\left(t\right)}x\ \ \ \ \ (5)$ $\displaystyle E_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle \frac{\left(n\pi\hbar\right)^{2}}{2mw^{2}\left(t\right)} \ \ \ \ \ (6)$

In this case, ${\psi_{n}}$ depends on only one time-dependent parameter, so we can use the chain rule to write

 $\displaystyle \gamma_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle i\int_{0}^{t}\left\langle \psi_{n}\left|\frac{\partial}{\partial w}\psi_{n}\right.\right\rangle \frac{dw}{dt'}dt'\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\int_{w_{1}}^{w_{2}}\left\langle \psi_{n}\left|\frac{\partial}{\partial w}\psi_{n}\right.\right\rangle dw \ \ \ \ \ (8)$

where the wall moves from ${w_{1}}$ to ${w_{2}}$ between times 0 and ${t}$. We get

 $\displaystyle \frac{\partial}{\partial w}\psi_{n}$ $\displaystyle =$ $\displaystyle -\frac{\sqrt{2}}{2w^{5/2}}\left[w\sin\frac{n\pi}{w}x+2n\pi x\cos\frac{n\pi}{w}x\right]\ \ \ \ \ (9)$ $\displaystyle \left\langle \psi_{n}\left|\frac{\partial}{\partial w}\psi_{n}\right.\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{w^{3}}\int_{0}^{w}\sin\frac{n\pi}{w}x\left[w\sin\frac{n\pi}{w}x+2n\pi x\cos\frac{n\pi}{w}x\right]dx\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sin^{2}n\pi}{w}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (12)$

In this case, there is no change in phase due to the geometric phase. In fact, we can see this is generally true for real wave functions ${\psi_{n}}$ since

 $\displaystyle \left\langle \psi_{n}\left|\psi_{n}\right.\right\rangle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (13)$ $\displaystyle \frac{d}{dt}\left\langle \psi_{n}\left|\psi_{n}\right.\right\rangle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \dot{\psi}_{n}\left|\psi_{n}\right.\right\rangle +\left\langle \psi_{n}\left|\dot{\psi}_{n}\right.\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\Re\left(\left\langle \psi_{n}\left|\dot{\psi}_{n}\right.\right\rangle \right) \ \ \ \ \ (16)$

That is, ${\left\langle \psi_{n}\left(t'\right)\left|\frac{\partial}{\partial t'}\psi_{n}\left(t'\right)\right.\right\rangle }$ must be purely imaginary, so if ${\psi_{n}}$ is real, the bracket must be zero. This also means that ${\gamma}$ is always real.

Thus ${\gamma}$ is zero as the wall moves from ${w_{1}}$ to ${w_{2}}$ and also as it moves back from ${w_{2}}$ to ${w_{1}}$.

The dynamic phase for the same journey is

 $\displaystyle \theta_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle -\frac{1}{\hbar}\int_{0}^{t}E_{n}\left(t'\right)dt'\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar\left(n\pi\right)^{2}}{2m}\int_{0}^{t}\frac{1}{w^{2}\left(t'\right)}dt' \ \ \ \ \ (18)$

If the speed of the wall is constant so that ${w=w_{1}+vt}$ we have

 $\displaystyle \theta_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle -\frac{\hbar\left(n\pi\right)^{2}}{2m}\int_{0}^{\left(w_{2}-w_{1}\right)/v}\frac{dt'}{\left(w_{1}+vt'\right)^{2}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar\left(n\pi\right)^{2}}{2mv}\frac{w_{1}-w_{2}}{w_{1}w_{2}} \ \ \ \ \ (20)$

## Electron in a precessing magnetic field

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.2.

For a spin 1/2 particle in a magnetic field ${\mathbf{B}}$, we’ve seen that the hamiltonian is

$\displaystyle \mathsf{H}=-\gamma\mathbf{B}\cdot\mathsf{S} \ \ \ \ \ (1)$

where ${\gamma}$ is the gyromagnetic ratio, which for an electron is ${-e/m}$. Now suppose that the magnetic field’s direction precesses around the ${z}$ axis (sweeps out a cone) with angular speed ${\omega}$, so that ${\mathbf{B}}$ makes an angle ${\alpha}$ with the ${z}$ axis. That is

$\displaystyle \mathbf{B}\left(t\right)=B_{0}\left[\sin\alpha\cos\left(\omega t\right)\hat{\mathbf{x}}+\sin\alpha\sin\left(\omega t\right)\hat{\mathbf{y}}+\cos\alpha\hat{\mathbf{z}}\right] \ \ \ \ \ (2)$

At time ${t}$, the component of ${\mathbf{S}}$ along ${\mathbf{B}}$ is given by

$\displaystyle \textsf{S}_{B}=\frac{\hbar}{2}\left(\begin{array}{cc} \cos\alpha & \sin\alpha e^{-i\omega t}\\ \sin\alpha e^{i\omega t} & -\cos\alpha \end{array}\right) \ \ \ \ \ (3)$

so the hamiltonian is

 $\displaystyle \mathsf{H}$ $\displaystyle =$ $\displaystyle \frac{\hbar\omega_{1}}{2}\left(\begin{array}{cc} \cos\alpha & \sin\alpha e^{-i\omega t}\\ \sin\alpha e^{i\omega t} & -\cos\alpha \end{array}\right)\ \ \ \ \ (4)$ $\displaystyle \omega_{1}$ $\displaystyle \equiv$ $\displaystyle \frac{eB_{0}}{m} \ \ \ \ \ (5)$

If we freeze the system at time ${t}$ and solve the time-independent SchrÃ¶dinger equation to get the eigenvalues and eigenspinors we get

$\displaystyle \chi_{+}=\left(\begin{array}{c} \cos(\alpha/2)\\ e^{i\omega t}\sin(\alpha/2) \end{array}\right) \ \ \ \ \ (6)$

$\displaystyle \chi_{-}=\left(\begin{array}{c} e^{-i\omega t}\sin(\alpha/2)\\ -\cos(\alpha/2) \end{array}\right) \ \ \ \ \ (7)$

with energies

$\displaystyle E_{\pm}=\pm\frac{\hbar\omega_{1}}{2} \ \ \ \ \ (8)$

Griffiths gives the exact solution to the time-dependent SchrÃ¶dinger equation for this problem as

$\displaystyle \chi\left(t\right)=\left[\begin{array}{c} \left(\cos\frac{\lambda t}{2}-i\frac{\omega_{1}-\omega}{\lambda}\sin\frac{\lambda t}{2}\right)\cos\frac{\alpha}{2}e^{-i\omega t/2}\\ \left(\cos\frac{\lambda t}{2}-i\frac{\omega_{1}+\omega}{\lambda}\sin\frac{\lambda t}{2}\right)\sin\frac{\alpha}{2}e^{i\omega t/2} \end{array}\right] \ \ \ \ \ (9)$

where

$\displaystyle \lambda\equiv\sqrt{\omega^{2}+\omega_{1}^{2}-2\omega\omega_{1}\cos\alpha} \ \ \ \ \ (10)$

To prove this, we need to show that

$\displaystyle \mathsf{H}\chi=i\hbar\frac{\partial\chi}{\partial t} \ \ \ \ \ (11)$

As usual, I’ll use Maple to help things along, although even Maple requires a bit of help here and there. We’ll start with ${\mathsf{H}\chi}$ which is the matrix product of 4 and 9. After multiplying out the terms and using the trig identities ${\cos\left(a\pm b\right)=\cos a\cos b\mp\sin a\sin b}$ we get

$\displaystyle \mathsf{H}\chi=\frac{\hbar\omega_{1}}{2\lambda}\left[\begin{array}{c} e^{-i\omega t/2}\cos\frac{\alpha}{2}\left[i\left(4\omega\cos^{2}\frac{\alpha}{2}-3\omega-\omega_{1}\right)\sin\frac{\lambda t}{2}+\lambda\cos\frac{\lambda t}{2}\right]\\ e^{i\omega t/2}\sin\frac{\alpha}{2}\left[i\left(4\omega\cos^{2}\frac{\alpha}{2}-\omega-\omega_{1}\right)\sin\frac{\lambda t}{2}+\lambda\cos\frac{\lambda t}{2}\right] \end{array}\right] \ \ \ \ \ (12)$

Now for the RHS. We get after collecting terms

$\displaystyle i\hbar\frac{\partial\chi}{\partial t}=\frac{\hbar}{2\lambda}\left[\begin{array}{c} e^{-i\omega t/2}\cos\frac{\alpha}{2}\left[i\left(\omega^{2}-\omega\omega_{1}-\lambda^{2}\right)\sin\frac{\lambda t}{2}+\lambda\omega_{1}\cos\frac{\lambda t}{2}\right]\\ e^{i\omega t/2}\sin\frac{\alpha}{2}\left[i\left(\omega^{2}+\omega\omega_{1}-\lambda^{2}\right)\sin\frac{\lambda t}{2}+\lambda\omega_{1}\cos\frac{\lambda t}{2}\right] \end{array}\right] \ \ \ \ \ (13)$

The two sides are equal if both the following are true:

 $\displaystyle \left(4\omega\cos^{2}\frac{\alpha}{2}-3\omega-\omega_{1}\right)\omega_{1}$ $\displaystyle =$ $\displaystyle \omega^{2}-\omega\omega_{1}-\lambda^{2}\ \ \ \ \ (14)$ $\displaystyle \left(4\omega\cos^{2}\frac{\alpha}{2}-\omega-\omega_{1}\right)\omega_{1}$ $\displaystyle =$ $\displaystyle \omega^{2}+\omega\omega_{1}-\lambda^{2} \ \ \ \ \ (15)$

Substituting from 10 both equations give the same condition:

 $\displaystyle 4\omega\omega_{1}\cos^{2}\frac{\alpha}{2}$ $\displaystyle =$ $\displaystyle 2\omega\omega_{1}\left(1+\cos\alpha\right)\ \ \ \ \ (16)$ $\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle 2\cos^{2}\frac{\alpha}{2}-1 \ \ \ \ \ (17)$

The last line is a trig identity, so the time-dependent SchrÃ¶dinger equation is satisfied.

We can also express 9 as a linear combination of 6 and 7. Griffiths gives the answer as

 $\displaystyle \chi\left(t\right)$ $\displaystyle =$ $\displaystyle \left[\cos\frac{\lambda t}{2}-i\frac{\left(\omega_{1}-\omega\cos\alpha\right)}{\lambda}\sin\frac{\lambda t}{2}\right]e^{-i\omega t/2}\chi_{+}\left(t\right)+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle i\left[\frac{\omega}{\lambda}\sin\alpha\sin\frac{\lambda t}{2}\right]e^{\omega t/2}\chi_{-}\left(t\right) \ \ \ \ \ (18)$

This can be verified by direct calculation. Take the top element first and use the cosine of difference of angles formula:

 $\displaystyle \chi_{1}$ $\displaystyle =$ $\displaystyle \left[\cos\frac{\lambda t}{2}-i\frac{\left(\omega_{1}-\omega\cos\alpha\right)}{\lambda}\sin\frac{\lambda t}{2}\right]e^{-i\omega t/2}\cos\frac{\alpha}{2}+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle i\left[\frac{\omega}{\lambda}\sin\alpha\sin\frac{\lambda t}{2}\right]e^{\omega t/2}e^{-i\omega t}\sin\frac{\alpha}{2}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-i\omega t/2}\left[\cos\frac{\lambda t}{2}\cos\frac{\alpha}{2}+\frac{i}{\lambda}\sin\frac{\lambda t}{2}\left(-\omega_{1}\cos\frac{\alpha}{2}+\omega\cos\frac{\alpha}{2}\right)\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\cos\frac{\lambda t}{2}-i\frac{\omega_{1}-\omega}{\lambda}\sin\frac{\lambda t}{2}\right)\cos\frac{\alpha}{2}e^{-i\omega t/2} \ \ \ \ \ (21)$

The bottom element works much the same way, using the sine of difference of angles formula:

 $\displaystyle \chi_{2}$ $\displaystyle =$ $\displaystyle \left[\cos\frac{\lambda t}{2}-i\frac{\left(\omega_{1}-\omega\cos\alpha\right)}{\lambda}\sin\frac{\lambda t}{2}\right]e^{-i\omega t/2}e^{i\omega t}\sin\frac{\alpha}{2}-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle i\left[\frac{\omega}{\lambda}\sin\alpha\sin\frac{\lambda t}{2}\right]e^{\omega t/2}\cos\frac{\alpha}{2}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\cos\frac{\lambda t}{2}-i\frac{\omega_{1}+\omega}{\lambda}\sin\frac{\lambda t}{2}\right)\sin\frac{\alpha}{2}e^{i\omega t/2} \ \ \ \ \ (23)$

Finally, writing ${\chi\left(t\right)=c_{+}\left(t\right)\chi_{+}+c_{-}\left(t\right)\chi_{-}}$ we can check that the coefficients are normalized.

 $\displaystyle \left|c_{+}\right|^{2}+\left|c_{-}\right|^{2}$ $\displaystyle =$ $\displaystyle \cos^{2}\frac{\lambda t}{2}+\frac{\left(\omega_{1}-\omega\cos\alpha\right)^{2}}{\lambda^{2}}\sin^{2}\frac{\lambda t}{2}+\left[\frac{\omega}{\lambda}\sin\alpha\sin\frac{\lambda t}{2}\right]^{2}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos^{2}\frac{\lambda t}{2}+\sin^{2}\frac{\lambda t}{2}\left[\frac{\left(\omega_{1}-\omega\cos\alpha\right)^{2}}{\lambda^{2}}+\left(\frac{\omega}{\lambda}\sin\alpha\right)^{2}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos^{2}\frac{\lambda t}{2}+\sin^{2}\frac{\lambda t}{2}\left[\frac{\omega^{2}+\omega_{1}^{2}-2\omega\omega_{1}\cos\alpha}{\lambda^{2}}\right]\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos^{2}\frac{\lambda t}{2}+\sin^{2}\frac{\lambda t}{2}\left[\frac{\omega^{2}+\omega_{1}^{2}-2\omega\omega_{1}\cos\alpha}{\omega^{2}+\omega_{1}^{2}-2\omega\omega_{1}\cos\alpha}\right]\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (28)$

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