Tag Archives: transmission coefficient

Transmission coefficients from water through glass into air

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.36.

Here’s another simple example of the transmission coefficient for waves passing from medium 1 through medium 2 into medium 3. Suppose we have an aquarium filled with water (index of refraction {n_{1}=\frac{4}{3}}). Light from the aquarium passes (at normal incidence) through a sheet of glass ({n_{2}=\frac{3}{2}}) and into air ({n_{3}=1}).

The transmission coefficient is given by

\displaystyle  T^{-1}=\frac{1}{4n_{1}n_{3}}\left[\left(n_{1}+n_{3}\right)^{2}+\sin^{2}\left(\frac{n_{2}\omega d}{c}\right)\frac{\left(n_{1}^{2}-n_{2}^{2}\right)\left(n_{3}^{2}-n_{2}^{2}\right)}{n_{2}^{2}}\right] \ \ \ \ \ (1)


The minimum value of {T} occurs when the sine is 1:

\displaystyle   T_{min}^{-1} \displaystyle  = \displaystyle  \frac{1}{4n_{1}n_{3}}\left[\left(n_{1}+n_{3}\right)^{2}+\frac{\left(n_{1}^{2}-n_{2}^{2}\right)\left(n_{3}^{2}-n_{2}^{2}\right)}{n_{2}^{2}}\right]\ \ \ \ \ (2)
\displaystyle  T_{min} \displaystyle  = \displaystyle  0.935 \ \ \ \ \ (3)

The maximum is when the sine is zero:

\displaystyle   T_{max} \displaystyle  = \displaystyle  \frac{4n_{1}n_{3}}{\left(n_{1}+n_{3}\right)^{2}}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  0.980 \ \ \ \ \ (5)

It doesn’t matter much what the frequency of the light is or how thick the glass is; most of the light makes it through in any case. Since 1 is symmetric in {n_{1}} and {n_{3}} a fish inside the aquarium can see out as easily as we can see in.

Transmission coefficient for a wave passing through 3 media

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.34.

We can extend the analysis of reflection and transmission of waves at a boundary by considering the case of an electromagnetic wave starting out in medium 1 (with wave speed {v_{1}}, wave number {k_{1}} and index of refraction {n_{1}=c/v_{1}}), then passing at normal incidence to medium 2 at {z=-d} and then to medium 3 at {z=0}. (We’ve changed the origin from that stated in Griffiths’s problem to make the analysis a bit easier, as we’ll see). We’d like to find the transmission coefficient between mediums 1 and 3, that is, we’d like to see how much of the wave’s energy gets transmitted all the way through the middle medium. We’ll assume the mediums are all homogeneous and linear, and that {\mu=\mu_{0}} in all of them.

The analysis is much the same as the earlier method, but a bit more complicated. We have a wave {\tilde{\mathbf{E}}_{1R}} travelling in towards the right in medium 1. At the boundary with medium 2, it gives rise to a reflected wave {\tilde{\mathbf{E}}_{1L}} travelling to the left in medium 1 and a trasmitted wave {\tilde{\mathbf{E}}_{2R}} travelling to the right in medium 2. When this wave hits the boundary with medium 3, there is a reflected wave {\tilde{\mathbf{E}}_{2L}} travelling to the left and a transmitted wave {\tilde{\mathbf{E}}_{3R}} travelling to the right. There are, of course, corresponding magnetic waves {\tilde{\mathbf{B}}_{1L}} and so on. We can then apply the boundary conditions to work out the amplitudes. [Actually, the wave reflected back to the left from the 2-3 boundary will hit the 1-2 boundary and be reflected and transmitted there too, so that there is, in principle, an infinite number of reflected and transmitted waves resulting from the wave bouncing back and forth between the two boundaries. However, we can subsume all the left-moving waves into {\tilde{\mathbf{E}}_{1L}} and {\tilde{\mathbf{E}}_{2L}} and all the right moving waves into {\tilde{\mathbf{E}}_{1R}}, {\tilde{\mathbf{E}}_{2R}} and {\tilde{\mathbf{E}}_{3R}}. The important thing is that these waves must satisfy the boundary conditions.]

We can take the electric component to be polarized along the {x} direction, so that the magnetic component is then along the {y} direction. The waves are

\displaystyle   \tilde{\mathbf{E}}_{1L} \displaystyle  = \displaystyle  E_{1L}e^{i\left(-k_{1}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (1)
\displaystyle  \tilde{\mathbf{E}}_{1R} \displaystyle  = \displaystyle  E_{1R}e^{i\left(k_{1}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (2)
\displaystyle  \tilde{\mathbf{E}}_{2L} \displaystyle  = \displaystyle  E_{2L}e^{i\left(-k_{2}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (3)
\displaystyle  \tilde{\mathbf{E}}_{2R} \displaystyle  = \displaystyle  E_{2R}e^{i\left(-k_{2}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (4)
\displaystyle  \tilde{\mathbf{E}}_{3R} \displaystyle  = \displaystyle  E_{3R}e^{i\left(k_{3}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (5)
\displaystyle  \tilde{\mathbf{B}}_{1L} \displaystyle  = \displaystyle  -\frac{1}{v_{1}}E_{1L}e^{i\left(-k_{1}z-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (6)
\displaystyle  \tilde{\mathbf{B}}_{1R} \displaystyle  = \displaystyle  \frac{1}{v_{1}}E_{1R}e^{i\left(k_{1}z-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (7)
\displaystyle  \tilde{\mathbf{B}}_{2L} \displaystyle  = \displaystyle  -\frac{1}{v_{2}}E_{2L}e^{i\left(-k_{2}z-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (8)
\displaystyle  \tilde{\mathbf{B}}_{2R} \displaystyle  = \displaystyle  \frac{1}{v_{2}}E_{2R}e^{i\left(k_{2}z-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (9)
\displaystyle  \tilde{\mathbf{B}}_{3R} \displaystyle  = \displaystyle  \frac{1}{v_{3}}E_{3R}e^{i\left(k_{3}z-\omega t\right)}\hat{\mathbf{y}} \ \ \ \ \ (10)

The negative signs for the left-moving magnetic waves are to keep the Poynting vector pointing to the left. The coefficients {E_{1L}} and so on are actually complex numbers, but we’ve dropped the tilde (and the subscript 0 that Griffiths uses) to make the notation simpler.

Because the medium 2-3 boundary consists of an incident right-moving wave, a reflected left-moving wave and a transmitted wave, it is identical to the case we treated earlier, provided we take {z=0} at this point (which we’ve done). We can therefore write down the results:

\displaystyle   E_{2L} \displaystyle  = \displaystyle  \frac{v_{3}-v_{2}}{v_{2}+v_{3}}E_{2R}\ \ \ \ \ (11)
\displaystyle  E_{3R} \displaystyle  = \displaystyle  \frac{2v_{3}}{v_{2}+v_{3}}E_{2R} \ \ \ \ \ (12)

Now for the medium 1-2 boundary at {z=-d}. From the boundary condition {\mathbf{E}_{1}^{\parallel}=\mathbf{E}_{2}^{\parallel}} and (since {\mu=\mu_{0}} everywhere) {\mathbf{B}_{1}^{\parallel}=\mathbf{B}_{2}^{\parallel}} we get

\displaystyle   E_{2R}e^{-ik_{2}d}+E_{2L}e^{ik_{2}d} \displaystyle  = \displaystyle  E_{1R}e^{-ik_{1}d}+E_{1L}e^{ik_{1}d}\ \ \ \ \ (13)
\displaystyle  \frac{1}{v_{2}}\left(E_{2R}e^{-ik_{2}d}-E_{2L}e^{ik_{2}d}\right) \displaystyle  = \displaystyle  \frac{1}{v_{1}}\left(E_{1R}e^{-ik_{1}d}-E_{1L}e^{ik_{1}d}\right) \ \ \ \ \ (14)

These 4 equations are linear in the {E} coefficients so it’s straightforward (although tedious) to solve them. It’s easiest to let Maple handle this part, and we get (since we’re interested only in expressing {E_{3R}} in terms of {E_{1R}}):

\displaystyle   E_{1R} \displaystyle  = \displaystyle  E_{3R}\frac{e^{ik_{1}d}}{4v_{2}v_{3}}\left[e^{ik_{2}d}\left(v_{2}v_{3}+v_{1}v_{2}-v_{2}^{2}-v_{1}v_{3}\right)+e^{-ik_{2}d}\left(v_{2}v_{3}+v_{1}v_{2}+v_{2}^{2}+v_{1}v_{3}\right)\right]\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  E_{3R}\frac{e^{ik_{1}d}}{4v_{2}v_{3}}\left[2\cos\left(k_{2}d\right)\left(v_{2}v_{3}+v_{1}v_{2}\right)-2i\sin\left(k_{2}d\right)\left(v_{2}^{2}+v_{1}v_{3}\right)\right] \ \ \ \ \ (16)

The intensity of a wave is

\displaystyle  I=\frac{\left|E\right|^{2}}{2\mu v} \ \ \ \ \ (17)

and the transmission coefficient is

\displaystyle  T=\frac{I_{3R}}{I_{1R}} \ \ \ \ \ (18)

so we get

\displaystyle   T^{-1} \displaystyle  = \displaystyle  \frac{v_{3}}{v_{1}}\frac{4\cos^{2}\left(k_{2}d\right)\left(v_{2}v_{3}+v_{1}v_{2}\right)^{2}+4\sin^{2}\left(k_{2}d\right)\left(v_{2}^{2}+v_{1}v_{3}\right)^{2}}{16v_{2}^{2}v_{3}^{2}}\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{4v_{1}v_{3}}\left[\frac{\left(v_{2}v_{3}+v_{1}v_{2}\right)^{2}}{v_{2}^{2}}\left(1-\sin^{2}\left(k_{2}d\right)\right)+\frac{\left(v_{2}^{2}+v_{1}v_{3}\right)^{2}}{v_{2}^{2}}\sin^{2}\left(k_{2}d\right)\right]\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{4v_{1}v_{3}}\left[\left(v_{3}+v_{1}\right)^{2}+\sin^{2}\left(k_{2}d\right)\frac{\left(v_{2}^{2}+v_{1}v_{3}\right)^{2}-\left(v_{2}v_{3}+v_{1}v_{2}\right)^{2}}{v_{2}^{2}}\right]\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{4v_{1}v_{3}}\left[\left(v_{3}+v_{1}\right)^{2}+\sin^{2}\left(k_{2}d\right)\frac{\left(v_{1}^{2}-v_{2}^{2}\right)\left(v_{3}^{2}-v_{2}^{2}\right)}{v_{2}^{2}}\right] \ \ \ \ \ (22)

We can express this in terms of the indexes of refraction by noting that since {n_{i}=c/v_{i}} and the power of the {v}s is the same in numerator and denominator so the factors of {c} cancel out and we can replace {v_{i}} with {1/n_{i}}. We can then multiply the first term top and bottom by {n_{1}^{2}n_{3}^{2}} and the second term top and bottom by {n_{1}^{2}n_{3}^{2}n_{2}^{4}} to get

\displaystyle  T^{-1}=\frac{1}{4n_{1}n_{3}}\left[\left(n_{1}+n_{3}\right)^{2}+\sin^{2}\left(k_{2}d\right)\frac{\left(n_{1}^{2}-n_{2}^{2}\right)\left(n_{3}^{2}-n_{2}^{2}\right)}{n_{2}^{2}}\right] \ \ \ \ \ (23)

Finally we note that the wave speed in medium 2 is {v_{2}=\omega/k_{2}=c/n_{2}} so {k_{2}=n_{2}\omega/c} and we get

\displaystyle  T^{-1}=\frac{1}{4n_{1}n_{3}}\left[\left(n_{1}+n_{3}\right)^{2}+\sin^{2}\left(\frac{n_{2}\omega d}{c}\right)\frac{\left(n_{1}^{2}-n_{2}^{2}\right)\left(n_{3}^{2}-n_{2}^{2}\right)}{n_{2}^{2}}\right] \ \ \ \ \ (24)

Transfer matrix

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.53.

We’ve seen that in the general scattering problem, we can write the particle stream magnitudes on each side of the potential by using a scattering matrix. In general, the wave function in the left hand region where {V=0} is

\displaystyle  \psi_{l}=Ae^{ikx}+Be^{-ikx}

where

\displaystyle  k\equiv\frac{\sqrt{2mE}}{\hbar}

On the right,

\displaystyle  \psi_{r}=Fe^{ikx}+Ge^{-ikx}

We can relate the coefficients by using the matrix

\displaystyle  \left[\begin{array}{c} B\\ F \end{array}\right]=\left[\begin{array}{cc} S_{11} & S_{12}\\ S_{21} & S_{22} \end{array}\right]\left[\begin{array}{c} A\\ G \end{array}\right]

This expresses the outgoing particle streams on each side in terms of the incoming streams.

We can also express the streams on the right in terms of the streams on the left by using a transfer matrix. That is, we can write

\displaystyle  \left[\begin{array}{c} F\\ G \end{array}\right]=\left[\begin{array}{cc} M_{11} & M_{12}\\ M_{21} & M_{22} \end{array}\right]\left[\begin{array}{c} A\\ B \end{array}\right]

By solving the scattering matrix equation for {F} and {G} in terms of {A} and {B} we can express the transfer matrix in terms of the scattering matrix.

\displaystyle  \mathsf{M}=-\frac{1}{S_{12}}\left[\begin{array}{cc} S_{11}S_{22}-S_{12}S_{21} & -S_{22}\\ S_{11} & -1 \end{array}\right]

The element {M_{11}} is just the determinant of {\mathsf{S}} so we have

\displaystyle  \mathsf{M}=-\frac{1}{S_{12}}\left[\begin{array}{cc} \det\mathsf{S} & -S_{22}\\ S_{11} & -1 \end{array}\right]

Conversely, we can express the scattering matrix in terms of the transfer matrix:

\displaystyle  \mathsf{S}=\frac{1}{M_{22}}\left[\begin{array}{cc} -M_{21} & 1\\ \det\mathsf{M} & M_{12} \end{array}\right]

In the special case where the only incoming particles are from the left, {G=0} and from the scattering matrix, we have for the reflection coefficient

\displaystyle   R_{l} \displaystyle  = \displaystyle  \frac{\left|B\right|^{2}}{\left|A\right|^{2}}
\displaystyle  \displaystyle  = \displaystyle  \left|S_{11}\right|^{2}
\displaystyle  \displaystyle  = \displaystyle  \frac{\left|M_{21}\right|^{2}}{\left|M_{22}\right|^{2}}

For the transmission coefficient

\displaystyle   T_{l} \displaystyle  = \displaystyle  \frac{\left|F\right|^{2}}{\left|A\right|^{2}}
\displaystyle  \displaystyle  = \displaystyle  \left|S_{21}\right|^{2}
\displaystyle  \displaystyle  = \displaystyle  \frac{\left|\det\mathsf{M}\right|^{2}}{\left|M_{22}\right|^{2}}

If the incoming particles are from the right only, {A=0} and we get

\displaystyle   R_{r} \displaystyle  = \displaystyle  \frac{\left|F\right|^{2}}{\left|G\right|^{2}}
\displaystyle  \displaystyle  = \displaystyle  \left|S_{22}\right|^{2}
\displaystyle  \displaystyle  = \displaystyle  \frac{\left|M_{12}\right|^{2}}{\left|M_{22}\right|^{2}}

\displaystyle   T_{l} \displaystyle  = \displaystyle  \frac{\left|B\right|^{2}}{\left|G\right|^{2}}
\displaystyle  \displaystyle  = \displaystyle  \left|S_{12}\right|^{2}
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\left|M_{22}\right|^{2}}

Now suppose we have a potential which is non-zero in only two isolated regions. For example, we could have two finite square wells separated by a gap, or a double delta function well. To the left of the leftmost non-zero region, the wave function is

\displaystyle  \psi_{1}=Ae^{ikx}+Be^{-ikx}

In between the two regions, we have

\displaystyle  \psi_{2}=Ce^{ikx}+De^{-ikx}

and to the right of the second region we have

\displaystyle  \psi_{3}=Fe^{ikx}+Ge^{-ikx}

We cannot say what the wave function within either region is unless we specify the potential there, of course.

In terms of the transfer matrix, the transition from region 1 to region 2 is given by

\displaystyle  \left[\begin{array}{c} C\\ D \end{array}\right]=\mathsf{M}_{1}\left[\begin{array}{c} A\\ B \end{array}\right]

Between regions 2 and 3, we have

\displaystyle  \left[\begin{array}{c} F\\ G \end{array}\right]=\mathsf{M}_{2}\left[\begin{array}{c} C\\ D \end{array}\right]=\mathsf{M}_{2}\mathsf{M}_{1}\left[\begin{array}{c} A\\ B \end{array}\right]

Thus the overall transfer matrix is the product of the two individual ones:

\displaystyle  \mathsf{M}=\mathsf{M}_{2}\mathsf{M}_{1}

This result generalizes to any potential that consists of a number of distinct regions where it is non-zero.

As an example, we can consider the delta function well again, except this time we’ll position the well at some arbitrary location {x=a}, so we have

\displaystyle  V(x)=-\alpha\delta(x-a)

In this case, we have two regions, so we can write

\displaystyle  \psi(x)=\begin{cases} Ae^{ikx}+Be^{-ikx} & x<a\\ Fe^{ikx}+Ge^{-ikx} & x>a \end{cases}

Following the same analysis as in the original delta function at the origin, we have a couple of boundary conditions at {x=a}. From continuity of the wave function we have

\displaystyle  Ae^{ika}+Be^{-ika}=Fe^{ika}+Ge^{-ika}

The first derivative is discontinuous, and we get the condition

\displaystyle   \Delta\psi' \displaystyle  = \displaystyle  -\frac{2m\alpha}{\hbar^{2}}\psi(a)
\displaystyle  ik\left[\left(F-A\right)e^{ika}-\left(G-B\right)e^{-ika}\right] \displaystyle  = \displaystyle  -\frac{2m\alpha}{\hbar^{2}}\left[Ae^{ika}+Be^{-ika}\right]

Solving these two equations in terms of {A} and {B} we can read off the transfer matrix

\displaystyle  \mathsf{M}=\frac{1}{2k}\left[\begin{array}{cc} 2k+iz & ize^{-2ika}\\ ize^{2ika} & -2k+iz \end{array}\right]

where

\displaystyle  z\equiv\frac{2m\alpha}{\hbar^{2}}

We can now reconsider the double delta well problem by applying the product of transfer matrixes above. The potential is

\displaystyle  V(x)=-\alpha\left[\delta(x+a)+\delta(x-a)\right]

We already have the transfer matrix for the {\delta(x-a)} part of the potential, which we’ll call {\mathsf{M}_{2}} since it’s on the right hand side. We can get the other transfer matrix by substituting {-a} for {a}:

\displaystyle  \mathsf{M}_{2}=\frac{1}{2k}\left[\begin{array}{cc} 2k+iz & ize^{2ika}\\ ize^{-2ika} & -2k+iz \end{array}\right]

The transfer matrix for the combined potential is then {\mathsf{M}=\mathsf{M}_{2}\mathsf{M}_{1}} so we get

\displaystyle  \mathsf{M}=\frac{1}{4k^{2}}\left[\begin{array}{cc} z^{2}\left(e^{-4ika}-1\right)+4k^{2}+4ikz & i\left[4k\cos\left(2ka\right)-2z\sin\left(2ka\right)\right]\\ -i\left[4k\cos\left(2ka\right)-2z\sin\left(2ka\right)\right] & z^{2}\left(e^{4ika}-1\right)+4k^{2}-4ikz \end{array}\right]

As a check, we can work out the transmission coefficient for the case {G=0} and compare it with our earlier result. From above, we have

\displaystyle  T_{l}=\frac{\left|\det\mathsf{M}\right|^{2}}{\left|M_{22}\right|^{2}}

The determinant conveniently works out to

\displaystyle  \det\mathsf{M}=1

so we get

\displaystyle  T_{l}=\frac{8k^{4}}{8k^{4}+4k^{2}z^{2}+z^{4}-4kz^{3}\sin\left(4ka\right)+z^{2}\cos\left(4ka\right)\left[4k^{2}-z^{2}\right]}

which is the same as the result we got previously.

Scattering matrix

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.52.

In a system with a localized potential (that is, a potential that is non-zero only for some finite range, such as the delta function or finite square well) we can always analyze the scattering problem, in which particles come in from either right or left (or both) and get transmitted or reflected. In general, the wave function in the left hand region where {V=0} is

\displaystyle  \psi_{l}=Ae^{ikx}+Be^{-ikx}

where

\displaystyle  k\equiv\frac{\sqrt{2mE}}{\hbar}

On the right,

\displaystyle  \psi_{r}=Fe^{ikx}+Ge^{-ikx}

In the middle, where {V(x)\ne0}, we can’t say what the wave function will be until {V(x)} is specified. In the cases we’ve analyzed so far, the only incident particles have been from the left, so we’ve always taken {G=0}. However, it’s not too difficult to generalize the results we’ve obtained for the delta function and finite square well to the case where we have incident particles from both directions.

Since a particle coming in from the left will be either transmitted (continue to the right past the potential region) or reflected (travel back to the left), particles incident from the left cannot affect the particle stream travelling to the left on the right side of the potential region. By symmetry, particles incident from the right cannot affect the particle stream travelling to the right on the left side of the potential. That is, we can always specify {A} and {G} in the wave functions above, and express {B} and {F} in terms of them. We can write this dependence as a matrix equation

\displaystyle  \left[\begin{array}{c} B\\ F \end{array}\right]=\left[\begin{array}{cc} S_{11} & S_{12}\\ S_{21} & S_{22} \end{array}\right]\left[\begin{array}{c} A\\ G \end{array}\right]

where the matrix {S} is called the scattering matrix.

For the delta function, with potential {V(x)=-\alpha\delta(x)} we’ve seen that in the case where {G=0},

\displaystyle   B \displaystyle  = \displaystyle  \frac{i\beta}{1-i\beta}A
\displaystyle  F \displaystyle  = \displaystyle  \frac{1}{1-i\beta}A
\displaystyle  \beta \displaystyle  \equiv \displaystyle  \frac{m\alpha}{\hbar^{2}k}

By symmetry, if {A=0} so that particles come in only from the right,

\displaystyle   F \displaystyle  = \displaystyle  \frac{i\beta}{1-i\beta}G
\displaystyle  B \displaystyle  = \displaystyle  \frac{1}{1-i\beta}G

If both {A\ne0} and {G\ne0}, we can just add up the contributions from the two cases, since they don’t interfere with each other, and we get

\displaystyle  \left[\begin{array}{c} B\\ F \end{array}\right]=\frac{1}{1-i\beta}\left[\begin{array}{cc} i\beta & 1\\ 1 & i\beta \end{array}\right]\left[\begin{array}{c} A\\ G \end{array}\right]

For the finite square well of depth {V_{0}}, with {G=0} we had

\displaystyle   B \displaystyle  = \displaystyle  \frac{e^{-2ika}\left(k^{2}-\mu^{2}\right)\sin\left(2\mu a\right)}{\sin\left(2\mu a\right)\left(k^{2}+\mu^{2}\right)+2i\mu k\cos\left(2\mu a\right)}A
\displaystyle  F \displaystyle  = \displaystyle  \frac{2i\mu ke^{-2ika}}{\sin\left(2\mu a\right)\left(k^{2}+\mu^{2}\right)+2i\mu k\cos\left(2\mu a\right)}A

where

\displaystyle  \mu=\frac{\sqrt{2m\left(E+V_{0}\right)}}{\hbar}

Here, the general scattering matrix is, by symmetry

\displaystyle  \left[\begin{array}{c} B\\ F \end{array}\right]=\frac{e^{-2ika}}{\sin\left(2\mu a\right)\left(k^{2}+\mu^{2}\right)+2i\mu k\cos\left(2\mu a\right)}\left[\begin{array}{cc} \left(k^{2}-\mu^{2}\right)\sin\left(2\mu a\right) & 2i\mu k\\ 2i\mu k & \left(k^{2}-\mu^{2}\right)\sin\left(2\mu a\right) \end{array}\right]\left[\begin{array}{c} A\\ G \end{array}\right]

Reflectionless potential

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.51.

An interesting potential is

\displaystyle V(x)=-\frac{\hbar^{2}a^{2}}{m}\text{sech}^{2}(ax)

where {a} is a constant and {\mathrm{sech}(ax)} is the hyperbolic secant, which is defined as {\mathrm{sech}(ax)\equiv1/\cosh(ax)}. The general shape of this potential is as shown in the figure.

We can verify by direct substitution that the function

\displaystyle  \psi_{0}(x)=A\text{sech}(ax)

is a solution. We get

\displaystyle   -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi_{0}}{dx^{2}}+V\psi_{0} \displaystyle  = \displaystyle  -\frac{\hbar^{2}}{2m}Aa^{2}\mathrm{sech}\left(ax\right)
\displaystyle  \displaystyle  = \displaystyle  -\frac{\hbar^{2}a^{2}}{2m}\psi_{0}

Thus the energy of this state is

\displaystyle  E_{0}=-\frac{\hbar^{2}a^{2}}{2m}

We can normalize {\psi_{0}} to find {A}:

\displaystyle   \int_{-\infty}^{\infty}\psi_{0}^{2}dx \displaystyle  = \displaystyle  1
\displaystyle  A \displaystyle  = \displaystyle  \sqrt{a/2}

A plot of {\psi_{0}(x)} looks like this:

For positive energies, we can verify that

\displaystyle  \psi_{k}(x)=B\left(\frac{ik-a\tanh(ax)}{ik+a}\right)e^{ikx}

is a solution of the Schrodinger equation for any energy by direct substitution. Here, as usual, {k\equiv\sqrt{2mE}/\hbar}.

We get

\displaystyle   -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi_{k}}{dx^{2}}+V\psi_{k} \displaystyle  = \displaystyle  \frac{\hbar^{2}k^{2}B}{2m}\left(\frac{ik-a\tanh\left(ax\right)}{ik+a}\right)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{2}k^{2}}{2m}\psi_{k}
\displaystyle  \displaystyle  = \displaystyle  E\psi_{k}

The asymptotic behaviour of {\psi_{k}} can be found from the limit {\lim_{x\rightarrow\infty}\tanh(ax)=1}. We therefore get:

\displaystyle   \lim_{x\rightarrow\infty}\psi_{k}(x) \displaystyle  = \displaystyle  B\frac{ik-a}{ik+a}e^{ikx}
\displaystyle  \displaystyle  = \displaystyle  -B\frac{(ik-a)^{2}}{a^{2}+k^{2}}e^{ikx}

For large negative {x} {\lim_{x\rightarrow-\infty}\tanh(ax)=-1} so we get

\displaystyle   \lim_{x\rightarrow-\infty}\psi_{k}(x) \displaystyle  = \displaystyle  B\frac{ik+a}{ik+a}e^{ikx}
\displaystyle  \displaystyle  = \displaystyle  Be^{ikx}

Thus in both cases, the wave function represents a wave travelling to the right, with no leftward component. That is, there is no reflected wave. The modulus of the wave for large {x} is

\displaystyle   \lim_{x\rightarrow\infty}\left|\psi_{k}(x)\right|^{2} \displaystyle  = \displaystyle  \left|B\right|^{2}\left|\frac{(ik-a)^{2}}{a^{2}+k^{2}}\right|^{2}
\displaystyle  \displaystyle  = \displaystyle  \left|B\right|^{2}
\displaystyle  \displaystyle  = \displaystyle  \lim_{x\rightarrow-\infty}\left|\psi_{k}(x)\right|^{2}

Thus the transmission coefficient is 1 for all positive energies, which means that any particle coming in from the left passes straight through with no reflection. There is, however, a change of phase due to the factor of {\frac{(ik-a)^{2}}{a^{2}+k^{2}}}.

Finite drop potential

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.35.

The problem of the finite step potential can be inverted to give a finite drop potential by replacing {V_{0}} by {-V_{0}}, so the potential is given by

\displaystyle  V(x)=\begin{cases} 0 & x\le0\\ -V_{0} & x>0 \end{cases}

If we assume particles coming in from the left, then we must have {E>0} (otherwise the wave function would decay exponentially inside the barrier and we couldn’t have particles coming in from infinity on the left). In this case the reflection coefficient is

\displaystyle  R=\left[\frac{E-\sqrt{E\left(E+V_{0}\right)}}{E+\sqrt{E\left(E+V_{0}\right)}}\right]^{2}

and the transmission coefficient is

\displaystyle  T=\frac{4E^{3/2}\sqrt{E+V_{0}}}{\left(E+\sqrt{E\left(E+V_{0}\right)}\right)^{2}}

For {V_{0}=0} the problem reduces to that of the free particle, and {R=0}, {T=1} as we’d expect. As {V_{0}} gets very large, {R\rightarrow1}, {T\rightarrow0}.

If we take {E=V_{0}/3}, then {R=1/9}.

Although the graph of the potential looks like a cliff, it doesn’t represent the behaviour of an object, such as a car, falling over a cliff. Classically, the energy of a car is kinetic + potential, which in the absence of other forces, remains a constant. If a car had a speed {v_{1}} in a region where {V=0}, then its total energy is kinetic: {E=K_{1}=\frac{1}{2}mv_{1}^{2}}. If it suddenly encounters a region where {V=-V_{0}}, then we’d have {E=K_{2}-V_{0}}, so {K_{2}} is larger than {K_{1}}, meaning that the car would instantaneously increase its speed, which of course doesn’t happen. In reality, a car driving off a cliff encounters a potential energy of {-mgy} where {y} is the distance it has fallen, so its kinetic energy increases gradually. Besides, a car falling off a cliff is essentially a two-dimensional problem, so trying to analyze it in one dimension won’t work.

A slightly more realistic case is that of a neutron which is fired at an atomic nucleus. The neutron experiences a sudden drop in potential from {V=0} outside the nucleus to {V=-V_{0}=-12} MeV inside. If we give the neutron an initial kinetic energy of {E=4} MeV, then the probability of transmission into the nucleus is

\displaystyle   T \displaystyle  = \displaystyle  \frac{4\times4^{3/2}\sqrt{4+12}}{\left(4+\sqrt{4(4+12)}\right)^{2}}
\displaystyle  \displaystyle  = \displaystyle  \frac{128}{144}
\displaystyle  \displaystyle  = \displaystyle  \frac{8}{9}

Finite step potential – scattering

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.34.

A variant of the finite square well is the finite step, which has the potential

\displaystyle  V(x)=\begin{cases} 0 & x\le0\\ V_{0} & x>0 \end{cases} \ \ \ \ \ (1)

where {V_{0}} is a positive constant energy.

There are two distinct cases here:

  1. Energy below the barrier: {0\le E\le V_{0}}
  2. Energy greater than the barrier: {E>V_{0}}

We’ll consider first the case where {0\le E\le V_{0}}.

In this case, the Schrödinger equation for {x>0} is

\displaystyle   -\frac{\hbar^{2}}{2m}\psi''+V_{0}\psi \displaystyle  = \displaystyle  E\psi\ \ \ \ \ (2)
\displaystyle  \psi'' \displaystyle  = \displaystyle  \mu^{2}\psi \ \ \ \ \ (3)

where

\displaystyle  \mu=\frac{\sqrt{2m(V_{0}-E)}}{\hbar} \ \ \ \ \ (4)

This has solution

\displaystyle  \psi(x)=Ce^{\mu x}+De^{-\mu x} \ \ \ \ \ (5)

To keep the solution finite as {x\rightarrow\infty} we must have {C=0} so the solution is an exponentially decaying wave function:

\displaystyle  \psi(x)=De^{-\mu x} \ \ \ \ \ (6)

To the left of the barrier, the Schrödinger equation is

\displaystyle  \psi''=E\psi \ \ \ \ \ (7)

Assuming particles coming in from the left, we have

\displaystyle  \psi(x)=Ae^{ikx}+Be^{-ikx} \ \ \ \ \ (8)

where

\displaystyle  k=\frac{\sqrt{2mE}}{\hbar} \ \ \ \ \ (9)

Since the potential is finite everywhere, both {\psi} and {\psi'} are continuous everywhere, which gives us two boundary conditions at {x=0}.

\displaystyle   A+B \displaystyle  = \displaystyle  D\ \ \ \ \ (10)
\displaystyle  ik\left(A-B\right) \displaystyle  = \displaystyle  -\mu D \ \ \ \ \ (11)

This has solution

\displaystyle   B \displaystyle  = \displaystyle  \frac{ik+\mu}{ik-\mu}A\ \ \ \ \ (12)
\displaystyle  D \displaystyle  = \displaystyle  \frac{2ik}{ik-\mu}A \ \ \ \ \ (13)

The reflection coefficient is

\displaystyle   R \displaystyle  = \displaystyle  \frac{\left|B\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  1 \ \ \ \ \ (15)

That is, the probability of an incoming particle being reflected is 1. This is because the wave function for {x>0} is exponentially decaying, so the probability of a particle reaching infinity is zero, thus no particles can be transmitted.

For {E>V_{0}} the Schrödinger equation for {x>0} is

\displaystyle   -\frac{\hbar^{2}}{2m}\psi''+V_{0}\psi \displaystyle  = \displaystyle  E\psi\ \ \ \ \ (16)
\displaystyle  \psi'' \displaystyle  = \displaystyle  -\kappa^{2}\psi \ \ \ \ \ (17)

where

\displaystyle  \kappa=\frac{\sqrt{2m(E-V_{0})}}{\hbar} \ \ \ \ \ (18)

We now get travelling wave solutions instead of exponentially decaying ones:

\displaystyle  \psi(x)=Ce^{i\kappa x}+De^{-i\kappa x} \ \ \ \ \ (19)

Assuming incoming particles arrive only from the left, we can set {D=0}. Applying the boundary conditions, we get

\displaystyle   A+B \displaystyle  = \displaystyle  C\ \ \ \ \ (20)
\displaystyle  ik\left(A-B\right) \displaystyle  = \displaystyle  i\kappa C \ \ \ \ \ (21)

with solutions

\displaystyle   B \displaystyle  = \displaystyle  \frac{k-\kappa}{k+\kappa}A\ \ \ \ \ (22)
\displaystyle  C \displaystyle  = \displaystyle  \frac{2k}{k+\kappa}A \ \ \ \ \ (23)

In this case, the reflection coefficient is

\displaystyle   R \displaystyle  = \displaystyle  \frac{\left|B\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{k-\kappa}{k+\kappa}\right)^{2} \ \ \ \ \ (25)

Substituting the expressions for {k} and {\kappa} we get

\displaystyle  R=\left[\frac{E-\sqrt{E\left(E-V_{0}\right)}}{E+\sqrt{E\left(E-V_{0}\right)}}\right]^{2} \ \ \ \ \ (26)

From this we can get the transmission coefficient

\displaystyle   T \displaystyle  = \displaystyle  1-R\ \ \ \ \ (27)
\displaystyle  \displaystyle  = \displaystyle  \frac{4E^{3/2}\sqrt{E-V_{0}}}{\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}} \ \ \ \ \ (28)

Note that this is not equal to {\left|C\right|^{2}/\left|A\right|^{2}=4E^{2}/\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}}. Have we done something wrong?

The answer lies in the fact that the wave for {x>0} is not the same as the wave for {x<0}, since the net energy on the right is {E-V_{0}} while on the left it is just {E}. One way of looking at it is in terms of the probability current for the free particle. The probability current must be conserved; this is just a way of saying that particles cannot vanish into, nor arise from, thin air. Since the probability current for a free particle with stationary state

\displaystyle  \Psi(x,t)=Ae^{ikx}e^{-i\hbar k^{2}t/2m} \ \ \ \ \ (29)

is

\displaystyle  J=\frac{\hbar k}{m}\left|A\right|^{2} \ \ \ \ \ (30)

the conservation law implies, for the case of the step potential

\displaystyle  \frac{\hbar k}{m}\left[\left|A\right|^{2}-\left|B\right|^{2}\right]=\frac{\hbar\kappa}{m}\left|C\right|^{2} \ \ \ \ \ (31)

That is, the influx of particles from the left minus the reflected beam must equal the transmitted beam. Dividing through by {\frac{\hbar k}{m}\left|A\right|^{2}} we get

\displaystyle  \frac{\left|B\right|^{2}}{\left|A\right|^{2}}+\frac{\kappa^{2}}{k^{2}}\frac{\left|C\right|^{2}}{\left|A\right|^{2}}=1 \ \ \ \ \ (32)

The first term is the reflection coefficient we calculated in 26. The second term is the transmission coefficient, which works out to

\displaystyle   T \displaystyle  = \displaystyle  \sqrt{\frac{E-V_{0}}{E}}\frac{\left|C\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (33)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{E-V_{0}}{E}}\frac{4E^{2}}{\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}}\ \ \ \ \ (34)
\displaystyle  \displaystyle  = \displaystyle  \frac{4E^{3/2}\sqrt{E-V_{0}}}{\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}} \ \ \ \ \ (35)

which is what we got earlier.

Finite square barrier – scattering

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.33.

We’ve analyzed the scattering problem in the finite square well, and we can use similar techniques to analyze a finite square barrier, which has the potential

\displaystyle  V(x)=\begin{cases} 0 & x<-a\\ V_{0} & -a\le x\le a\\ 0 & x>a \end{cases} \ \ \ \ \ (1)

where {V_{0}} is a positive constant energy, and {a} is a constant location on the {x} axis.

There are three distinct cases here:

  1. Energy below the barrier: {0\le E<V_{0}}
  2. Energy exactly equal to the barrier: {E=V_{0}}
  3. Energy greater than the barrier: {E>V_{0}}

In all three cases, the wave function away from the barrier on either side has the same form; it is only within the barrier that the three cases differ. We’ll consider first the case where {0\le E<V_{0}}.

In this case, the Schrödinger equation within the barrier is

\displaystyle   -\frac{\hbar^{2}}{2m}\psi''+V_{0}\psi \displaystyle  = \displaystyle  E\psi\ \ \ \ \ (2)
\displaystyle  \psi'' \displaystyle  = \displaystyle  \mu^{2}\psi \ \ \ \ \ (3)

where

\displaystyle  \mu=\frac{\sqrt{2m(V_{0}-E)}}{\hbar} \ \ \ \ \ (4)

This has solution

\displaystyle  \psi(x)=Ce^{\mu x}+De^{-\mu x} \ \ \ \ \ (5)

Outside the barrier, the Schrödinger equation is

\displaystyle  \psi''=E\psi \ \ \ \ \ (6)

Outside the barrier on the left, the solution is the sum of travelling waves (assuming particles are incident from the left only), while to the right we have right propagating waves only. Thus

\displaystyle  \psi(x)=\begin{cases} Ae^{ikx}+Be^{-ikx} & x<-a\\ Fe^{ikx} & x>a \end{cases} \ \ \ \ \ (7)

where

\displaystyle  k=\frac{\sqrt{2mE}}{\hbar} \ \ \ \ \ (8)

Since the potential is finite everywhere, both {\psi} and {\psi'} are continuous everywhere, which gives us four boundary conditions.

At {x=-a} we have

\displaystyle   Ae^{-ika}+Be^{ika} \displaystyle  = \displaystyle  Ce^{-\mu a}+De^{\mu a}\ \ \ \ \ (9)
\displaystyle  ik\left(Ae^{-ika}-Be^{ika}\right) \displaystyle  = \displaystyle  \mu\left(Ce^{-\mu a}-De^{\mu a}\right) \ \ \ \ \ (10)

At {x=a}:

\displaystyle   Ce^{\mu a}+De^{-\mu a} \displaystyle  = \displaystyle  Fe^{ika}\ \ \ \ \ (11)
\displaystyle  \mu\left(Ce^{\mu a}-De^{-\mu a}\right) \displaystyle  = \displaystyle  ikFe^{ika} \ \ \ \ \ (12)

We can solve these linear equations to get the other four constants in terms of {A}. The results are

\displaystyle   B \displaystyle  = \displaystyle  \frac{e^{-2ika}\left(k^{2}+\mu^{2}\right)\left(e^{2\mu a}-e^{-2\mu a}\right)}{2ik\mu\left(e^{2\mu a}+e^{-2\mu a}\right)+\left(k^{2}-\mu^{2}\right)\left(e^{2\mu a}-e^{-2\mu a}\right)}A\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \frac{e^{-2ika}\left(k^{2}+\mu^{2}\right)\sinh\left(2\mu a\right)}{2ik\mu\cosh\left(2\mu a\right)+\left(k^{2}-\mu^{2}\right)\sinh\left(2\mu a\right)}A\ \ \ \ \ (14)
\displaystyle  C \displaystyle  = \displaystyle  \frac{e^{-a\mu-iak}\left(-k^{2}+k\mu i\right)}{2ik\mu\cosh\left(2\mu a\right)+\left(k^{2}-\mu^{2}\right)\sinh\left(2\mu a\right)}A\ \ \ \ \ (15)
\displaystyle  D \displaystyle  = \displaystyle  \frac{e^{-a\mu-iak}\left(k^{2}+k\mu i\right)}{2ik\mu\cosh\left(2\mu a\right)+\left(k^{2}-\mu^{2}\right)\sinh\left(2\mu a\right)}A\ \ \ \ \ (16)
\displaystyle  F \displaystyle  = \displaystyle  \frac{2e^{-2ika}k\mu i}{2ik\mu\cosh\left(2\mu a\right)+\left(k^{2}-\mu^{2}\right)\sinh\left(2\mu a\right)}A \ \ \ \ \ (17)

From here we can get the transmission coefficient as

\displaystyle   T \displaystyle  = \displaystyle  \frac{\left|F\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  \frac{4\mu^{2}k^{2}}{\left[\mu^{4}+2\mu^{2}k^{2}+k^{4}\right]\sinh^{2}\left(2\mu a\right)+4\mu^{2}k^{2}}\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\left(\mu^{2}+k^{2}\right)^{2}\sinh^{2}\left(2\mu a\right)/4\mu^{2}k^{2}+1} \ \ \ \ \ (20)

The reciprocal of {T} is then, substituting to get the physical quantities back:

\displaystyle   T^{-1} \displaystyle  = \displaystyle  1+\frac{\left(\mu^{2}+k^{2}\right)^{2}\sinh^{2}\left(2\mu a\right)}{4\mu^{2}k^{2}}\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  1+\frac{V_{0}^{2}}{4E\left(V_{0}-E\right)}\sinh^{2}\left(\frac{2a}{\hbar}\sqrt{2m\left(V_{0}-E\right)}\right) \ \ \ \ \ (22)

The second case is where {E=V_{0}}. In this case, the outer two solutions are the same as before, but in the barrier region we have

\displaystyle  \psi''=0 \ \ \ \ \ (23)

which has the solution

\displaystyle  \psi=Cx+D \ \ \ \ \ (24)

Applying the boundary conditions we have, at {x=-a}

\displaystyle   Ae^{-ika}+Be^{ika} \displaystyle  = \displaystyle  -Ca+D\ \ \ \ \ (25)
\displaystyle  ik\left(Ae^{-ika}-Be^{ika}\right) \displaystyle  = \displaystyle  C \ \ \ \ \ (26)

At {x=a} we have

\displaystyle   Ca+D \displaystyle  = \displaystyle  Fe^{ika}\ \ \ \ \ (27)
\displaystyle  C \displaystyle  = \displaystyle  ikFe^{ika} \ \ \ \ \ (28)

Solving these equations we get

\displaystyle   B \displaystyle  = \displaystyle  \frac{kae^{-2ika}}{ka+i}A\ \ \ \ \ (29)
\displaystyle  C \displaystyle  = \displaystyle  \frac{ke^{-ika}}{-ka-i}A\ \ \ \ \ (30)
\displaystyle  D \displaystyle  = \displaystyle  e^{-ika}A\ \ \ \ \ (31)
\displaystyle  F \displaystyle  = \displaystyle  \frac{e^{-2ika}}{1-ika}A \ \ \ \ \ (32)

In this case the transmission coefficient is

\displaystyle   T \displaystyle  = \displaystyle  \frac{\left|F\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (33)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{1+k^{2}a^{2}}\ \ \ \ \ (34)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{1+2mEa^{2}/\hbar^{2}} \ \ \ \ \ (35)

Finally, for {E>V_{0}} the Schrödinger equation within the barrier is

\displaystyle   -\frac{\hbar^{2}}{2m}\psi'' \displaystyle  = \displaystyle  \left(E-V_{0}\right)\psi\ \ \ \ \ (36)
\displaystyle  \psi'' \displaystyle  = \displaystyle  -\frac{2m\left(E-V_{0}\right)}{\hbar^{2}}\psi\ \ \ \ \ (37)
\displaystyle  \displaystyle  = \displaystyle  -\lambda^{2}\psi \ \ \ \ \ (38)

where

\displaystyle  \lambda=\frac{\sqrt{2m\left(E-V_{0}\right)}}{\hbar} \ \ \ \ \ (39)

The solution within the barrier is thus

\displaystyle  \psi(x)=Ce^{i\lambda x}+De^{-i\lambda x} \ \ \ \ \ (40)

Boundary conditions give at {x=-a}

\displaystyle   Ae^{-ika}+Be^{ika} \displaystyle  = \displaystyle  Ce^{-i\lambda a}+De^{i\lambda a}\ \ \ \ \ (41)
\displaystyle  ik\left(Ae^{-ika}-Be^{ika}\right) \displaystyle  = \displaystyle  i\lambda\left(Ce^{-i\lambda a}-De^{i\lambda a}\right) \ \ \ \ \ (42)

At {x=a}:

\displaystyle   Ce^{i\lambda a}+De^{-i\lambda a} \displaystyle  = \displaystyle  Fe^{ika}\ \ \ \ \ (43)
\displaystyle  i\lambda\left(Ce^{i\lambda a}-De^{-i\lambda a}\right) \displaystyle  = \displaystyle  ikFe^{ika} \ \ \ \ \ (44)

Solving these four equations gives

\displaystyle   B \displaystyle  = \displaystyle  \frac{e^{-2ika}\left(k^{2}-\lambda^{2}\right)\sin\left(2\lambda a\right)}{\left(k^{2}+\lambda^{2}\right)\sin\left(2\lambda a\right)+2i\lambda k\cos\left(2\lambda a\right)}A\ \ \ \ \ (45)
\displaystyle  C \displaystyle  = \displaystyle  \frac{e^{-ia\left(\lambda+k\right)}\left(\lambda+k\right)k}{-i\left(k^{2}+\lambda^{2}\right)\sin\left(2\lambda a\right)+2\lambda k\cos\left(2\lambda a\right)}A\ \ \ \ \ (46)
\displaystyle  D \displaystyle  = \displaystyle  \frac{e^{-ia\left(k-\lambda\right)}\left(k-\lambda\right)k}{i\left(k^{2}+\lambda^{2}\right)\sin\left(2\lambda a\right)-2\lambda k\cos\left(2\lambda a\right)}A\ \ \ \ \ (47)
\displaystyle  F \displaystyle  = \displaystyle  \frac{2k\lambda e^{-2ika}}{-i\left(k^{2}+\lambda^{2}\right)\sin\left(2\lambda a\right)+2\lambda k\cos\left(2\lambda a\right)}A \ \ \ \ \ (48)

The transmission coefficient is

\displaystyle   T \displaystyle  = \displaystyle  \frac{\left|F\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (49)
\displaystyle  \displaystyle  = \displaystyle  \frac{4\lambda^{2}k^{2}}{\left(\lambda^{4}-2\lambda^{2}k^{2}+k^{4}\right)\sin^{2}\left(2\lambda a\right)+4\lambda^{2}k^{2}}\ \ \ \ \ (50)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{1+\left(\lambda^{2}-k^{2}\right)^{2}\sin^{2}\left(2\lambda a\right)/4\lambda^{2}k^{2}} \ \ \ \ \ (51)

The reciprocal is

\displaystyle   T^{-1} \displaystyle  = \displaystyle  1+\frac{\left(\lambda^{2}-k^{2}\right)^{2}\sin^{2}\left(2\lambda a\right)}{4\lambda^{2}k^{2}}\ \ \ \ \ (52)
\displaystyle  \displaystyle  = \displaystyle  1+\frac{V_{0}^{2}}{4E\left(E-V_{0}\right)}\sin^{2}\left(\frac{2a}{\hbar}\sqrt{2m\left(E-V_{0}\right)}\right) \ \ \ \ \ (53)

Delta function well as limit of finite square well

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.31.

The finite square well has the potential

\displaystyle  V(x)=\begin{cases} 0 & x<-a\\ -V_{0} & -a\le x\le a\\ 0 & x>a \end{cases}

where {V_{0}} is a positive constant energy, and {a} is a constant location on the {x} axis.

The delta function potential {V=-\alpha\delta(x)} can be thought of as the limit of the finite square well as {a\rightarrow0} and {V_{0}\rightarrow\infty} in such a way that the area of the rectangle in the well is a constant. That is, the integral of the potential is the same in both cases, so that

\displaystyle  2aV_{0}=\alpha

The energies of the bound states for the even solution of the finite square well are given by

\displaystyle  \tan z=\sqrt{\frac{z_{0}^{2}}{z^{2}}-1} \ \ \ \ \ (1)


where

\displaystyle   z_{0} \displaystyle  \equiv \displaystyle  \frac{a}{\hbar}\sqrt{2mV_{0}}
\displaystyle  z \displaystyle  \equiv \displaystyle  \frac{a}{\hbar}\sqrt{2m(E+V_{0})}

Substituting for {V_{0}} we get

\displaystyle   z_{0} \displaystyle  = \displaystyle  \frac{1}{\hbar}\sqrt{ma\alpha}
\displaystyle  z \displaystyle  = \displaystyle  \frac{1}{\hbar}\sqrt{2ma^{2}E+ma\alpha}

As {a\rightarrow0}, {z_{0}/z\rightarrow1}, so {\tan z} becomes very small. In this limit, {\tan z\approx z}, so we can approximate equation 1 by

\displaystyle   z \displaystyle  = \displaystyle  \sqrt{\frac{z_{0}^{2}}{z^{2}}-1}
\displaystyle  \frac{1}{\hbar}\sqrt{2ma^{2}E+ma\alpha} \displaystyle  = \displaystyle  \sqrt{\frac{\alpha}{2aE+\alpha}-1}
\displaystyle  \frac{1}{\hbar^{2}}\left(2ma^{2}E+ma\alpha\right) \displaystyle  = \displaystyle  \frac{\alpha}{2aE+\alpha}-1
\displaystyle  \left(2ma^{2}E+ma\alpha\right)\left(2aE+\alpha\right) \displaystyle  = \displaystyle  -2aE\hbar^{2}

If we retain only the term in {a} (discarding higher powers of {a}), we get

\displaystyle   ma\alpha^{2} \displaystyle  = \displaystyle  -2aE\hbar^{2}
\displaystyle  E \displaystyle  = \displaystyle  -\frac{m\alpha^{2}}{2\hbar^{2}}

This is the energy we found earlier when analyzing the delta function well.

We can do a similar analysis for the scattering states. The transmission coefficient for the finite square well is

\displaystyle  T^{-1}=1+\frac{V_{0}^{2}}{4E\left(E+V_{0}\right)}\sin^{2}\left(\frac{2a}{\hbar}\sqrt{2m\left(E+V_{0}\right)}\right)

For small {a} we can use the approximation {\sin x\approx x} and we get

\displaystyle  T^{-1}\approx1+\frac{2mV_{0}^{2}a^{2}}{\hbar^{2}E}

Substituting for {V_{0}} gives

\displaystyle   T^{-1} \displaystyle  \approx \displaystyle  1+\frac{m\alpha^{2}}{2\hbar^{2}E}
\displaystyle  T \displaystyle  = \displaystyle  \frac{1}{1+m\alpha^{2}/2\hbar^{2}E}

This is the same formula we obtained when analyzing the delta function potential directly.

Double delta function well – scattering states

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.28.

In the last post we had a look at the bound states of the double delta function potential

\displaystyle  V(x)=-\alpha\left[\delta(x+a)+\delta(x-a)\right]

where {\alpha} gives the strength of the well. In this post, we’ll look at the scattering states of this potential.

We will use a similar approach to that for the single delta function potential. At first glance, you might think the problem is a trivial extension of the single delta function case. If a stream of particles enters from the left, then a fraction will get reflected at the first delta function, with the remainder being transmitted. Of those that are transmitted, another fraction will get reflected at the second delta function and those left over from that reflection will be transmitted to travel on to infinity on the right.

The flaw in this argument is that those that get reflected at the second delta function will travel back to the left, and some of them will be reflected back to the right again when they reach the first delta function. This process continues ad infinitum, with part of the particle stream being bounced back and forth between the two delta functions. Thus we are faced with an infinite series of reflections and transmissions.

Probably the easiest way to analyze the problem is just to confront the mathematics head on and see where it leads. We therefore follow the procedure for the single delta function to obtain the solutions in each of the three regions. Since we are proposing a particle stream entering from the left, there is no left-travelling stream from the right, so the solution is asymmetric, meaning we can’t propose even and odd solutions to the problem.

The most general solution of this equation is (with {k\equiv\frac{\sqrt{2mE}}{\hbar}}; remember {E} is positive so {k} is real)

\displaystyle  \psi(x)=\begin{cases} Ae^{ikx}+Be^{-ikx} & x<-a\\ Ce^{ikx}+De^{-ikx} & -a<x<a\\ Fe^{ikx} & x>a \end{cases}

We can apply boundary conditions to eliminate some of the constants.

Continuity of the wave function at {x=-a} gives

\displaystyle  Ae^{-ika}+Be^{ika}=Ce^{-ika}+De^{ika}

The same condition at {x=a} gives

\displaystyle  Ce^{ika}+De^{-ika}=Fe^{ika}

The change in derivative of the wave function across the delta function boundary satisfies the following condition at {x=\pm a} (this is the same condition that we applied to the single delta function at {x=0}):

\displaystyle  \Delta\psi'=-\frac{2m\alpha}{\hbar^{2}}\psi(\pm a)

At {x=-a} we have

\displaystyle   \Delta\psi' \displaystyle  = \displaystyle  ik\left[Ce^{-ika}-De^{ika}-Ae^{-ika}+Be^{ika}\right]
\displaystyle  \displaystyle  = \displaystyle  -\frac{2m\alpha}{\hbar^{2}}\left(Ae^{-ika}+Be^{ika}\right)

Similarly at {x=a} we have

\displaystyle   \Delta\psi' \displaystyle  = \displaystyle  ik\left[Fe^{ika}-Ce^{ika}+De^{-ika}\right]
\displaystyle  \displaystyle  = \displaystyle  -\frac{2m\alpha}{\hbar^{2}}Fe^{ika}

We now have four equations in the five unknowns {A,B,C,D} and {F}. To get the transmission and reflection coefficients, however, we need only express the last four constants in terms of {A}. The four equations constitute a system of linear equations in the constants, so it is a straightforward matter of algebra to solve them. Doing this by hand is fairly laborious, but we can use software such as Maple’s ‘solve’ command to do it for us.

The results are

\displaystyle   B \displaystyle  = \displaystyle  \frac{iz\left[4k\cos\left(2ka\right)-2z\sin\left(2ka\right)\right]}{4k\left(k-iz\right)+z^{2}\left(e^{4ika}-1\right)}A
\displaystyle  C \displaystyle  = \displaystyle  -\frac{2ki(z+2ki)}{4k\left(k-iz\right)+z^{2}\left(e^{4ika}-1\right)}A
\displaystyle  D \displaystyle  = \displaystyle  \frac{2ikze^{2ika}}{4k\left(k-iz\right)+z^{2}\left(e^{4ika}-1\right)}A
\displaystyle  F \displaystyle  = \displaystyle  \frac{4k^{2}}{4k\left(k-iz\right)+z^{2}\left(e^{4ika}-1\right)}A

where

\displaystyle  z\equiv\frac{2m\alpha}{\hbar^{2}}

The transmission coefficient is then

\displaystyle   T \displaystyle  = \displaystyle  \frac{\left|F\right|^{2}}{\left|A\right|^{2}}
\displaystyle  \displaystyle  = \displaystyle  \frac{8k^{4}}{8k^{4}+4k^{2}z^{2}+z^{4}-4kz^{3}\sin\left(4ka\right)+z^{2}\cos\left(4ka\right)\left[4k^{2}-z^{2}\right]}

The reflection coefficient is

\displaystyle   R \displaystyle  = \displaystyle  \frac{\left|B\right|^{2}}{\left|A\right|^{2}}
\displaystyle  \displaystyle  = \displaystyle  \frac{2z^{2}\left(2k\cos\left(2ka\right)-z\sin\left(2ka\right)\right)^{2}}{8k^{4}+4k^{2}z^{2}+z^{4}-4kz^{3}\sin\left(4ka\right)+z^{2}\cos\left(4ka\right)\left[4k^{2}-z^{2}\right]}
\displaystyle  \displaystyle  = \displaystyle  \frac{4k^{2}z^{2}+z^{4}-4kz^{3}\sin\left(4ka\right)+z^{2}\cos\left(4ka\right)\left[4k^{2}-z^{2}\right]}{8k^{4}+4k^{2}z^{2}+z^{4}-4kz^{3}\sin\left(4ka\right)+z^{2}\cos\left(4ka\right)\left[4k^{2}-z^{2}\right]}

As a check, we note that {R+T=1}.

For reference, the two internal rates are

\displaystyle   T_{i}=\frac{\left|C\right|^{2}}{\left|A\right|^{2}} \displaystyle  = \displaystyle  \frac{2k^{2}z^{2}+8k^{4}}{8k^{4}+4k^{2}z^{2}+z^{4}-4kz^{3}\sin\left(4ka\right)+z^{2}\cos\left(4ka\right)\left[4k^{2}-z^{2}\right]}
\displaystyle  R_{i}=\frac{\left|D\right|^{2}}{\left|A\right|^{2}} \displaystyle  = \displaystyle  \frac{2k^{2}z^{2}}{8k^{4}+4k^{2}z^{2}+z^{4}-4kz^{3}\sin\left(4ka\right)+z^{2}\cos\left(4ka\right)\left[4k^{2}-z^{2}\right]}

The first quantity represents the flow to the right after the first delta function, and we observe that it is larger than the second quantity, which represents the flow to the left. This makes sense, since we would expect that as the main particle stream enters from the left, and of that which gets transmitted past the first well, some will get transmitted past the second well and escape, while some will get reflected back towards the first well. In fact, we have {T+R_{i}=T_{i}} which says that the probability of being transmitted past the first well is the sum of the probabilities of being reflected from the second well and being transmitted through the second well.

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