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.
Strictly speaking this equation gives the position space representation of the eigenfunction. More precisely, we should just say that is an eigenvalue of the position operator 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 in that space, as we did above for position space.
For momentum, we’ve seen that the eigenfunctions are
The normalizations of both the position and momentum eigenfunctions give us more delta functions:
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
so that applying it to a state gives
Note that this is a completely general expression; we can choose any basis states (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 ‘along’ that basis vector. In practice, to do calculations we usually express in position or momentum space (or in matrix form if it’s a spin state) but in this formula, is just an abstract symbol representing some arbitrary state.
For a complete set of discrete basis states we can define the unit operator
or for a continuous set of basis states
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
Since the basis consisting of the states 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:
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
Somewhat confusingly (I think), L&B use the notation to represent an eigenstate of the hamiltonian with momentum 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
We can now use the relation 14 to get
and since we get
Now we want to get the position space version of the state . From 1 (generalized to 3-d) we see that
so if we can write as a function of then the expression merely picks out the precise position that we’re interested in. Using a set of momentum basis states we can transform using the unit operator 13:
Example 2 We can apply the same arguments to a 2-particle state. Start with
From commutation relations 14 we get
Applying we get
To convert to position coordinates, this time we have two independent positions, one for each particle, which we’ll call and , so a position state is . 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:
Also, since the particles are identical, the state is the same as .
In that case we can write
The is there because the double integral extends over all values of both and so it counts the state twice, once as and once as . 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
This is the symmetrized wave function for two identical bosons. Following through the same argument using anticommutators for fermions gives the fermion result