# Every attractive 1-dimensional potential has a bound state

Reference: Lecture by Barton Zwiebach in MIT course 8.05.1x, week 1, Deep Dive 2 (not in the PDF notes).

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.2, Exercise 5.2.2b.

An interesting application of the variational principle in quantum mechances is the following theorem:

Theorem Every 1-dimensional attractive potential has at least one bound state.

To prove this, we need first to define what we mean by an attractive potential ${V\left(x\right)}$. ${V\left(x\right)}$ must satisfy the following conditions:

• ${V\left(x\right)\rightarrow0}$ as ${x\rightarrow\pm\infty}$.
• ${V\left(x\right)<0}$ everywhere.
• ${V\left(x\right)}$ is piecewise continuous. This means that it may have a finite number of jump discontinuities.

One possible form for ${V\left(x\right)}$ is as shown:

This is a particularly simple potential that satisfies the above conditions. We could introduce a few step functions, multiple local maxima and minima, and so on, provided we don’t violate any of the 3 conditions above.

Since ${V\left(x\right)<0}$ everywhere, we can write it as

$\displaystyle V\left(x\right)=-\left|V\left(x\right)\right| \ \ \ \ \ (1)$

What we would like to prove is that for any hamiltonian of the form

$\displaystyle H=-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}-\left|V\left(x\right)\right| \ \ \ \ \ (2)$

the ground state ${E_{0}}$ is a bound state, that is

$\displaystyle E_{0}<0 \ \ \ \ \ (3)$

We can apply the variational principle, which states

If ${\psi}$ is any normalized function and ${H}$ is a hamiltonian, then the ground state energy ${E_{0}}$ of this hamiltonian has an upper bound given by

$\displaystyle E_{0}\le\left\langle \psi\left|H\right|\psi\right\rangle \equiv\left\langle H\right\rangle \ \ \ \ \ (4)$

The use of the variational principle to prove the above theorem involves a bit of a convoluted argument, but the mathematics involved is fairly simple. Our goal is to find some wave function ${\psi_{\alpha}}$ (where ${\alpha}$ is some parameter that we can vary) so that

$\displaystyle E_{0}\le\left\langle \psi_{\alpha}\left|H\right|\psi_{\alpha}\right\rangle =\left\langle H\right\rangle _{\psi_{\alpha}}<0 \ \ \ \ \ (5)$

From 2 we have

 $\displaystyle \left\langle \hat{H}\right\rangle _{\psi_{\alpha}}$ $\displaystyle =$ $\displaystyle \int dx\;\psi_{\alpha}\left(x\right)\hat{H}\psi_{\alpha}\left(x\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle T\right\rangle _{\psi_{\alpha}}-\left\langle \left|V\left(x\right)\right|\right\rangle _{\psi_{\alpha}} \ \ \ \ \ (7)$

where

 $\displaystyle \left\langle T\right\rangle _{\psi_{\alpha}}$ $\displaystyle =$ $\displaystyle -\int dx\;\psi_{\alpha}\left(x\right)\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi_{\alpha}\left(x\right)\ \ \ \ \ (8)$ $\displaystyle \left\langle \left|V\left(x\right)\right|\right\rangle _{\psi_{\alpha}}$ $\displaystyle =$ $\displaystyle \int dx\;\psi_{\alpha}\left(x\right)\left|V\left(x\right)\right|\psi_{\alpha}\left(x\right) \ \ \ \ \ (9)$

We can integrate 8 by parts once to get

 $\displaystyle \left\langle T\right\rangle _{\psi_{\alpha}}$ $\displaystyle =$ $\displaystyle -\int dx\;\psi_{\alpha}\left(x\right)\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi_{\alpha}\left(x\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left.\frac{\hbar^{2}}{2m}\psi_{\alpha}\left(x\right)\frac{d}{dx}\psi_{\alpha}\left(x\right)\right|_{-\infty}^{\infty}+\frac{\hbar^{2}}{2m}\int dx\;\left(\frac{d}{dx}\psi_{\alpha}\left(x\right)\right)^{2}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{2m}\int_{-\infty}^{\infty}dx\;\left(\frac{d}{dx}\psi_{\alpha}\left(x\right)\right)^{2} \ \ \ \ \ (12)$

where we invoke the usual requirement that ${\psi_{\alpha}}$ and its first derivative vanish at infinity.

We therefore see that since the integrand in the last line is always positive (we’re assuming that ${\psi_{\alpha}}$ is not zero everywhere), that ${\left\langle T\right\rangle _{\psi_{\alpha}}>0}$. Likewise, from 9, ${\left\langle \left|V\left(x\right)\right|\right\rangle _{\psi_{\alpha}}>0}$. Thus in order that ${\left\langle H\right\rangle _{\psi_{\alpha}}<0}$, we must have

$\displaystyle \left\langle T\right\rangle _{\psi_{\alpha}}<\left\langle \left|V\left(x\right)\right|\right\rangle _{\psi_{\alpha}} \ \ \ \ \ (13)$

To get any further, we need to choose a test function ${\psi_{\alpha}\left(x\right)}$. We’ll pick (because it works!)

$\displaystyle \psi_{\alpha}=\left(\frac{\alpha}{\pi}\right)^{1/4}e^{-\frac{1}{2}\alpha x^{2}} \ \ \ \ \ (14)$

The factor of ${\left(\frac{\alpha}{\pi}\right)^{1/4}}$ is required so that ${\psi_{\alpha}}$ is normalized. The integral in 12 can be done using standard methods; I’ll just use Maple, and we find

$\displaystyle \left\langle T\right\rangle _{\psi_{\alpha}}=\frac{\hbar^{2}\alpha}{4m} \ \ \ \ \ (15)$

The integral 9 of course can’t be done exactly if we don’t know what ${V}$ is, so we have just

$\displaystyle \left\langle \left|V\left(x\right)\right|\right\rangle _{\psi_{\alpha}}=\int dx\;\psi_{\alpha}^{2}\left(x\right)\left|V\left(x\right)\right| \ \ \ \ \ (16)$

(No need for modulus signs around ${\psi_{\alpha}}$ since the function 14 is real.) To progress further, we need to start invoking some inequalities to get where we want to go. The argument consists of several steps, so watch carefully as we go along.

From 13 through 15 we have to show that we can satisfy the condition

$\displaystyle \frac{\left\langle \left|V\left(x\right)\right|\right\rangle _{\psi_{\alpha}}}{\left\langle T\right\rangle _{\psi_{\alpha}}}=\frac{4m}{\hbar^{2}\sqrt{\pi}}\frac{1}{\sqrt{\alpha}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}\left|V\left(x\right)\right|dx>1 \ \ \ \ \ (17)$

Since ${V}$ is arbitrary subject to the 3 conditions above, the only thing we can legitimately fiddle with is the value of ${\alpha}$. We can see that if we choose ${\alpha}$ small enough, we should be able to satisfy this inequality, since for small ${\alpha}$, the ${1/\sqrt{\alpha}}$ term gets large, while the ${e^{-\alpha x^{2}}}$ term in the integrand is bounded between 0 and 1. We need to find some upper limit for ${\alpha}$.

In what follows, you’ll need to refer to the following diagram:

First, we choose some point ${x_{0}}$ at which ${V\left(x_{0}\right)}$ is continuous (that is, we ensure that ${x_{0}}$ isn’t at one of the points where ${V\left(x\right)}$ has a discontinuity, or jump). The value of ${V\left(x_{0}\right)}$ is defined as ${-2v_{0}}$ where ${v_{0}>0}$. Because ${V\rightarrow0}$ at ${x\rightarrow\pm\infty}$, there must be points ${x_{1}}$ and ${x_{2}}$ on either side of ${x_{0}}$ where ${V}$ has the value ${-v_{0}}$ (actually, I’m not sure this is strictly true, because, as ${V}$ is allowed a few jumps, it might jump over the point where it’s equal to ${-v_{0}}$. However, as the number of jumps is required to be finite, there must be some points ${x_{1}}$ and ${x_{2}}$ on either side of ${x_{0}}$ where ${V}$ attains a value that is between ${-2v_{0}}$ and 0, and I think the argument below still works if we choose those points instead.)

Now for the first inequality. We know that, because the integrand is positive

$\displaystyle \int_{-\infty}^{\infty}e^{-\alpha x^{2}}\left|V\left(x\right)\right|dx>\int_{x_{1}}^{x_{2}}e^{-\alpha x^{2}}\left|V\left(x\right)\right|dx \ \ \ \ \ (18)$

Second inequality: in the interval ${x_{1}}$ to ${x_{2}}$, ${\left|V\left(x\right)\right|>v_{0}}$ (see the diagram!), so we have

$\displaystyle \int_{x_{1}}^{x_{2}}e^{-\alpha x^{2}}\left|V\left(x\right)\right|dx>v_{0}\int_{x_{1}}^{x_{2}}e^{-\alpha x^{2}}dx \ \ \ \ \ (19)$

The last integral has no closed form solution, but we know that in the interval ${x_{1}}$ to ${x_{2}}$

$\displaystyle e^{-\alpha x^{2}}>e^{-\alpha\max\left(x_{1}^{2},x_{2}^{2}\right)} \ \ \ \ \ (20)$

Therefore

 $\displaystyle v_{0}\int_{x_{1}}^{x_{2}}e^{-\alpha x^{2}}dx$ $\displaystyle >$ $\displaystyle v_{0}\int_{x_{1}}^{x_{2}}e^{-\alpha\max\left(x_{1}^{2},x_{2}^{2}\right)}dx\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle v_{0}\left(x_{2}-x_{1}\right)e^{-\alpha\max\left(x_{1}^{2},x_{2}^{2}\right)} \ \ \ \ \ (22)$

Now suppose we choose ${\alpha}$ to be

$\displaystyle \alpha<\frac{1}{\max\left(x_{1}^{2},x_{2}^{2}\right)} \ \ \ \ \ (23)$

Then

$\displaystyle e^{-\alpha\max\left(x_{1}^{2},x_{2}^{2}\right)}>e^{-1} \ \ \ \ \ (24)$

We can now summarize as follows:

$\displaystyle \int_{-\infty}^{\infty}e^{-\alpha x^{2}}\left|V\left(x\right)\right|dx>v_{0}\left(x_{2}-x_{1}\right)e^{-1} \ \ \ \ \ (25)$

provided we choose ${\alpha}$ according to 23. Plugging this back into 17 we have

$\displaystyle \frac{\left\langle \left|V\left(x\right)\right|\right\rangle _{\psi_{\alpha}}}{\left\langle T\right\rangle _{\psi_{\alpha}}}>\frac{4m}{\hbar^{2}\sqrt{\pi}}\frac{v_{0}\left(x_{2}-x_{1}\right)}{e}\frac{1}{\sqrt{\alpha}} \ \ \ \ \ (26)$

This expression will now be greater than 1 provided that

 $\displaystyle \sqrt{\alpha}$ $\displaystyle <$ $\displaystyle \frac{4m}{\hbar^{2}\sqrt{\pi}}\frac{v_{0}\left(x_{2}-x_{1}\right)}{e}\ \ \ \ \ (27)$ $\displaystyle \alpha$ $\displaystyle <$ $\displaystyle \left[\frac{4m}{\hbar^{2}\sqrt{\pi}}\frac{v_{0}\left(x_{2}-x_{1}\right)}{e}\right]^{2} \ \ \ \ \ (28)$

Comparing 23 and 28, we see that we can satisfy both conditions if we take

$\displaystyle \alpha<\min\left\{ \frac{1}{\max\left(x_{1}^{2},x_{2}^{2}\right)},\left[\frac{4m}{\hbar^{2}\sqrt{\pi}}\frac{v_{0}\left(x_{2}-x_{1}\right)}{e}\right]^{2}\right\} \ \ \ \ \ (29)$

This condition depends on ${x_{1}}$ and ${x_{2}}$ but that doesn’t matter, since both quantities in the RHS of 29 are positive, so there is always some positive value of ${\alpha}$ that satisfies the condition. In other words, going right back to 17 and then to 7, we can always find a value of ${\alpha}$ so that ${\left\langle H\right\rangle <0}$ which means that the ground state of ${H}$ must be negative, which makes it a bound state.

# Stark effect in hydrogen for n = 1 and n = 2

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.36.

The Zeeman effect occurs when an atom is placed in an external magnetic field, resulting in the interaction between field and the magnetic dipole moments of the atom causing splitting of the energy levels. The electrical analogue of the Zeeman effect, when an atom is placed in an external electric field, is called the Stark effect. We can use perturbation theory to analyze the effect on the energy levels of the electron.

The perturbation hamiltonian is, assuming the electric field points in the ${z}$ direction:

$\displaystyle H_{S}^{\prime}=eE_{ext}z=eE_{ext}r\cos\theta \ \ \ \ \ (1)$

To use perturbation theory, we’ll need the wave functions for unperturbed hydrogen, which are given in Griffiths as equation 4.89. For the ground state ${n=1}$, we have

$\displaystyle \left|100\right\rangle =\frac{2}{a^{3/2}}\frac{1}{\sqrt{4\pi}}e^{-r/a} \ \ \ \ \ (2)$

Since the ground state is non-degenerate, we can use non-degenerate perturbation theory:

 $\displaystyle E_{100,1}$ $\displaystyle =$ $\displaystyle \left\langle 100\right|H_{S}^{\prime}\left|100\right\rangle \ \ \ \ \ (3)$

Rather than writing out the integral, we observe that ${\left\langle 100\right|H_{S}^{\prime}\left|100\right\rangle }$ contains the integral of ${\cos\theta\sin\theta=\frac{1}{2}\sin2\theta}$ over ${\theta=0,..,\pi}$ which is zero, so ${E_{100,1}=0}$.

To analyze ${n=2}$, we need the four wave functions:

 $\displaystyle \left|200\right\rangle$ $\displaystyle =$ $\displaystyle R_{20}\left(r\right)Y_{0}^{0}\left(\theta,\phi\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}a^{3/2}}\left(1-\frac{r}{2a}\right)\frac{1}{\sqrt{4\pi}}e^{-r/2a}\ \ \ \ \ (5)$ $\displaystyle \left|211\right\rangle$ $\displaystyle =$ $\displaystyle R_{21}\left(r\right)Y_{1}^{1}\left(\theta,\phi\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(\frac{3}{8\pi}\right)^{1/2}\frac{1}{\sqrt{24}a^{5/2}}re^{-r/2a}\sin\theta e^{i\phi}\ \ \ \ \ (7)$ $\displaystyle \left|210\right\rangle$ $\displaystyle =$ $\displaystyle R_{21}\left(r\right)Y_{1}^{0}\left(\theta,\phi\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{3}{4\pi}\right)^{1/2}\frac{1}{\sqrt{24}a^{5/2}}re^{-r/2a}\cos\theta\ \ \ \ \ (9)$ $\displaystyle \left|21-1\right\rangle$ $\displaystyle =$ $\displaystyle R_{21}\left(r\right)Y_{1}^{-1}\left(\theta,\phi\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{3}{8\pi}\right)^{1/2}\frac{1}{\sqrt{24}a^{5/2}}re^{-r/2a}\sin\theta e^{-i\phi} \ \ \ \ \ (11)$

Since all four of these states have the same unperturbed energy, we need to use degenerate perturbation theory, so we’ll need to find the matrix ${W}$ with elements

$\displaystyle W_{a,b}=\left\langle a\right|H_{S}^{\prime}\left|b\right\rangle \ \ \ \ \ (12)$

where ${a}$ and ${b}$ represent one of the four states above.

First, we’ll look at the ${\theta}$ integrals. All matrix elements involve integrals of the form (remember that ${H_{S}^{\prime}}$ always contributes a ${\cos\theta}$ and the spherical volume element always contributes a ${\sin\theta}$):

$\displaystyle I_{nm}=\int_{0}^{\pi}\sin^{n}\theta\cos^{m}\theta d\theta \ \ \ \ \ (13)$

For the possible values of ${n}$ and ${m}$ in this problem, the only non-zero integrals of this form are

 $\displaystyle I_{12}$ $\displaystyle =$ $\displaystyle \frac{2}{3}\ \ \ \ \ (14)$ $\displaystyle I_{22}$ $\displaystyle =$ $\displaystyle \frac{\pi}{8} \ \ \ \ \ (15)$

${I_{12}}$ arises in ${\left\langle 200\right|H_{S}^{\prime}\left|210\right\rangle }$ (and its transpose) and ${I_{22}}$ arises in ${\left\langle 211\right|H_{S}^{\prime}\left|210\right\rangle }$ and ${\left\langle 210\right|H_{S}^{\prime}\left|21-1\right\rangle }$ (and their transposes). Thus these are the only possible non-zero entries in ${W}$. However, ${\left\langle 211\right|H_{S}^{\prime}\left|210\right\rangle }$ and ${\left\langle 210\right|H_{S}^{\prime}\left|21-1\right\rangle }$ involve integrating ${e^{\pm i\phi}}$ over ${\phi=0..2\pi}$ which gives zero. Thus the only non-zero matrix elements are ${\left\langle 200\right|H_{S}^{\prime}\left|210\right\rangle }$ (and its transpose). This gives

 $\displaystyle \left\langle 200\right|H_{S}^{\prime}\left|210\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}a^{3/2}}\left(\frac{3}{4\pi}\right)^{1/2}\frac{1}{\sqrt{24}a^{5/2}}\frac{1}{\sqrt{4\pi}}eE_{ext}\times\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\left(1-\frac{r}{2a}\right)re^{-r/a}\cos^{2}\theta r^{2}\sin\theta d\phi d\theta dr\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -3aeE_{ext} \ \ \ \ \ (17)$

(The integral can be done with software, or by hand using integration by parts.) The matrix ${W}$ is therefore

$\displaystyle W=\left[\begin{array}{cccc} 0 & 0 & -3aeE_{ext} & 0\\ 0 & 0 & 0 & 0\\ -3aeE_{ext} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (18)$

The eigenvalues are 0, 0, ${\pm3aeE_{ext}}$ so the ${n=2}$ state splits into 3 states, one with energy ${E_{2,0}}$ (degeneracy 2) and two with energies ${E_{2,0}\pm3aeE_{ext}}$ (each with degeneracy 1). The eigenvectors are ${\left[0,1,0,0\right]}$ and ${\left[0,0,0,1\right]}$ for eigenvalue 0, ${\left[-1,0,1,0\right]}$ for ${3aeE_{ext}}$ and ${\left[1,0,1,0\right]}$ for ${-3aeE_{ext}}$. Thus the ‘good’ states are

$\displaystyle \left|211\right\rangle ,\left|21-1\right\rangle ,\frac{1}{\sqrt{2}}\left(-\left|200\right\rangle +\left|210\right\rangle \right),\frac{1}{\sqrt{2}}\left(\left|200\right\rangle +\left|210\right\rangle \right) \ \ \ \ \ (19)$

The electric dipole moment of hydrogen is (treating the proton and electron as point charges):

 $\displaystyle \mathbf{p}$ $\displaystyle =$ $\displaystyle -e\mathbf{r}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -er\left(\sin\theta\cos\phi\hat{\mathbf{x}}+\sin\theta\sin\phi\hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}}\right) \ \ \ \ \ (21)$

We can work out the expectation value of ${\mathbf{p}}$ in each of the ‘good’ states by straightforward integration: ${\left\langle \mathbf{p}\right\rangle =\left\langle a\right|\mathbf{p}\left|a\right\rangle }$ where ${a}$ stands for one of the ‘good’ states. Note that if ${a=\left|211\right\rangle }$ or ${a=\left|21-1\right\rangle }$, then ${\left\langle a\right|\mathbf{p}\left|a\right\rangle }$ has only a ${z}$ component that is non-zero, since the complex exponentials in ${\phi}$ cancel out and the integral of ${\sin\phi}$ or ${\cos\phi}$ in the ${x}$ or ${y}$ components is zero. Similarly, if ${a=\frac{1}{\sqrt{2}}\left(-\left|200\right\rangle +\left|210\right\rangle \right)}$ or ${a=\frac{1}{\sqrt{2}}\left(\left|200\right\rangle +\left|210\right\rangle \right)}$, the ${x}$ and ${y}$ components are again zero, since these wave functions are independent of ${\phi}$ so the integral of ${\sin\phi}$ or ${\cos\phi}$ in the ${x}$ or ${y}$ components gives zero again. Therefore, ${\left\langle \mathbf{p}\right\rangle }$ is always in the ${z}$ direction, and can be calculated from

$\displaystyle \left\langle \mathbf{p}\right\rangle =-e\left\langle a\right|r\cos\theta\left|a\right\rangle \hat{\mathbf{z}} \ \ \ \ \ (22)$

Doing the integrals results in

$\displaystyle \left\langle \mathbf{p}\right\rangle =0,0,3a\hat{\mathbf{z}},-3a\hat{\mathbf{z}} \ \ \ \ \ (23)$

respectively.

# Infinite square well with variable delta function barrier: location of the particle

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.8.

Continuing our example of the infinite square well with the growing delta function barrier, we can now find the probability of the particle being found in the right section of the well. The wave function is

$\displaystyle \psi\left(x\right)=\begin{cases} Ae^{ikx}+Be^{-ikx} & 0\le x<\frac{a}{2}+\epsilon\\ Ce^{ikx}+De^{-ikx} & \frac{a}{2}+\epsilon

With the boundary conditions we can solve for 3 of the coefficients in terms of the fourth and we got

 $\displaystyle B$ $\displaystyle =$ $\displaystyle -A\ \ \ \ \ (2)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle e^{-iz}D\frac{\sin\left[\frac{1}{2}z\left(1-\delta\right)\right]}{\sin\left[\frac{1}{2}z\left(1+\delta\right)\right]}\ \ \ \ \ (3)$ $\displaystyle C$ $\displaystyle =$ $\displaystyle -De^{-2iz} \ \ \ \ \ (4)$

where ${z\equiv ka}$.

We can call the left and right wave functions ${\psi_{l}}$ and ${\psi_{r}}$ and we get

 $\displaystyle \psi_{l}$ $\displaystyle =$ $\displaystyle 2ie^{-iz}D\frac{\sin\left[\frac{1}{2}z\left(1-\delta\right)\right]}{\sin\left[\frac{1}{2}z\left(1+\delta\right)\right]}\sin kx\ \ \ \ \ (5)$ $\displaystyle \psi_{r}$ $\displaystyle =$ $\displaystyle 2ie^{-iz}D\sin\left(z-kx\right) \ \ \ \ \ (6)$

The probability of being in the right well is

$\displaystyle P_{r}=\frac{\int_{\frac{a}{2}+\epsilon}^{a}\left|\psi_{r}\right|^{2}dx}{\int_{0}^{\frac{a}{2}+\epsilon}\left|\psi_{l}\right|^{2}dx+\int_{\frac{a}{2}+\epsilon}^{a}\left|\psi_{r}\right|^{2}dx} \ \ \ \ \ (7)$

With ${\delta\equiv2\epsilon/a}$ this is

$\displaystyle P_{r}=\frac{\int_{\frac{a}{2}\left(1+\delta\right)}^{a}\left|\psi_{r}\right|^{2}dx}{\int_{0}^{\frac{a}{2}\left(1+\delta\right)}\left|\psi_{l}\right|^{2}dx+\int_{\frac{a}{2}\left(1+\delta\right)}^{a}\left|\psi_{r}\right|^{2}dx} \ \ \ \ \ (8)$

We get

 $\displaystyle X_{-}\equiv\int_{\frac{a}{2}\left(1+\delta\right)}^{a}\left|\psi_{r}\right|^{2}dx$ $\displaystyle =$ $\displaystyle \frac{D^{2}}{4k}\left[z\left(1-\delta\right)-\sin\left(z\left(1-\delta\right)\right)\right]\ \ \ \ \ (9)$ $\displaystyle X_{+}\equiv\int_{0}^{\frac{a}{2}\left(1+\delta\right)}\left|\psi_{l}\right|^{2}dx$ $\displaystyle =$ $\displaystyle \frac{D^{2}}{4k}\frac{\sin^{2}\left[\frac{1}{2}z\left(1-\delta\right)\right]}{\sin^{2}\left[\frac{1}{2}z\left(1+\delta\right)\right]}\left[\left[z\left(1+\delta\right)-2\sin\left[\frac{z}{2}\left(1+\delta\right)\right]\cos\left[\frac{z}{2}\left(1+\delta\right)\right]\right]\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{D^{2}}{4k}\frac{\sin^{2}\left[\frac{1}{2}z\left(1-\delta\right)\right]}{\sin^{2}\left[\frac{1}{2}z\left(1+\delta\right)\right]}\left(z\left(1+\delta\right)-\sin\left(z\left(1+\delta\right)\right)\right) \ \ \ \ \ (11)$

So

 $\displaystyle P_{r}$ $\displaystyle =$ $\displaystyle \frac{X_{-}}{X_{+}+X_{-}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{1+X_{+}/X_{-}} \ \ \ \ \ (13)$

with

 $\displaystyle \frac{X_{+}}{X_{-}}$ $\displaystyle =$ $\displaystyle \frac{z\left(1+\delta\right)-\sin\left(z\left(1+\delta\right)\right)}{z\left(1-\delta\right)-\sin\left(z\left(1-\delta\right)\right)}\times\frac{\sin^{2}\left[\frac{1}{2}z\left(1-\delta\right)\right]}{\sin^{2}\left[\frac{1}{2}z\left(1+\delta\right)\right]}\equiv\frac{I_{+}}{I_{-}}\ \ \ \ \ (14)$ $\displaystyle I_{+}$ $\displaystyle \equiv$ $\displaystyle \left[z\left(1+\delta\right)-\sin\left(z\left(1+\delta\right)\right)\right]\sin^{2}\left[\frac{1}{2}z\left(1-\delta\right)\right]\ \ \ \ \ (15)$ $\displaystyle I_{-}$ $\displaystyle \equiv$ $\displaystyle \left[z\left(1-\delta\right)-\sin\left(z\left(1-\delta\right)\right)\right]\sin^{2}\left[\frac{1}{2}z\left(1+\delta\right)\right] \ \ \ \ \ (16)$

We can work out this probability for several values of ${T}$, with ${\delta=0.01}$:

 ${T}$ ${z}$ ${P_{r}}$ 1 3.673 0.487 5 4.760 0.471 20 5.720 0.401 100 6.135 0.147 1000 6.215 0.00248

The probability of the particle being to the right of the barrier drops to zero as the barrier strength becomes infinite, which is just what we saw above. Because the left well has a lower ground state energy, the particle favours that region.

We can see this by plotting the wave functions 5 and 6 for these same values of ${T}$. We’ve used ${\delta=0.01}$, ${a=1}$ and omitted the common factor of ${2ie^{-iz}D}$ in the plots, so the wave functions aren’t normalized, but all we’re interested in is the relative wave functions in the two regions of the well.

We can see that the wave function drops off as ${T}$ increases.

# Infinite square well with variable delta function barrier: ground state energy

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.8.

Here’s another example of the adiabatic theorem. This time, we have an infinite square well in which a delta function barrier is inserted slowly at a position that is slightly off centre, so that for ${0 we have the potential

$\displaystyle V\left(x\right)=f\left(t\right)\delta\left(x-\frac{a}{2}-\epsilon\right) \ \ \ \ \ (1)$

where ${f\left(t\right)}$ is a function that rises slowly from 0 to ${\infty}$. The adiabatic theorem says that the system will remain in the ground state of the time-varying hamiltonian.

First, we’ll look at what the state is when the barrier has attained infinite strength, so that ${t\rightarrow\infty}$. [OK, the delta function itself is always infinite at a single point, but it can have a constant ‘strength’ factor multiplying it. We’ve looked at the case of the infinite square well with a constant delta function barrier and we’ve seen that increasing the strength factor to ${\infty}$ effectively divides the well into two wells that are isolated from each other, while a finite strength barrier does allow the wave function to communicate across the barrier.]

For an infinitely strong delta function barrier then, we have one well of width ${\frac{a}{2}+\epsilon}$ and one well of width ${\frac{a}{2}-\epsilon}$. The wave functions in both wells must be zero at their boundaries, so we get for the ground state (${n=1}$):

$\displaystyle \psi\left(x\right)=\begin{cases} A\sin\frac{\pi}{\frac{a}{2}+\epsilon}x & 0\le x<\frac{a}{2}+\epsilon\\ A\sin\left[\frac{\pi}{\frac{a}{2}-\epsilon}\left(x-\frac{a}{2}-\epsilon\right)\right] & \frac{a}{2}+\epsilon

 $\displaystyle E_{l}$ $\displaystyle =$ $\displaystyle \frac{\pi^{2}\hbar^{2}}{2m\left(\frac{a}{2}+\epsilon\right)^{2}}\ \ \ \ \ (3)$ $\displaystyle E_{r}$ $\displaystyle =$ $\displaystyle \frac{\pi^{2}\hbar^{2}}{2m\left(\frac{a}{2}-\epsilon\right)^{2}} \ \ \ \ \ (4)$

Thus ${E_{l} so the ground state confines the particle to the left well. The wave function for the ground state is

$\displaystyle \psi\left(x\right)=\begin{cases} \sqrt{\frac{2}{\frac{a}{2}+\epsilon}}\sin\frac{\pi}{\frac{a}{2}+\epsilon}x & 0\le x<\frac{a}{2}+\epsilon\\ 0 & \frac{a}{2}+\epsilon

The plot looks like this:

Now for the general case where ${f\left(t\right)}$ is finite. In this case we can write the wave functions as

$\displaystyle \psi\left(x\right)=\begin{cases} Ae^{ikx}+Be^{-ikx} & 0\le x<\frac{a}{2}+\epsilon\\ Ce^{ikx}+De^{-ikx} & \frac{a}{2}+\epsilon

where

$\displaystyle k\equiv\frac{\sqrt{2mE}}{\hbar} \ \ \ \ \ (7)$

The barriers at ${x=0}$ and ${x=a}$ are still infinite so the wave function must be zero there, giving

 $\displaystyle A$ $\displaystyle =$ $\displaystyle -B\ \ \ \ \ (8)$ $\displaystyle C$ $\displaystyle =$ $\displaystyle -De^{-2ika} \ \ \ \ \ (9)$

The wave function must be continuous at the barrier, so we get

$\displaystyle A\left(e^{ik\left(\frac{a}{2}+\epsilon\right)}-e^{-ik\left(\frac{a}{2}+\epsilon\right)}\right)=D\left(-e^{-2ika}e^{ik\left(\frac{a}{2}+\epsilon\right)}+e^{-ik\left(\frac{a}{2}+\epsilon\right)}\right) \ \ \ \ \ (10)$

Finally, we can analyze the derivative at the barrier in the same way we did for the delta function well and we get

 $\displaystyle -\frac{\hbar^{2}}{2m}\int_{\frac{a}{2}+\epsilon-\beta}^{\frac{a}{2}+\epsilon+\beta}\frac{d^{2}\psi}{dx^{2}}dx+f\left(t\right)\int_{\frac{a}{2}+\epsilon-\beta}^{\frac{a}{2}+\epsilon+\beta}\delta(x)\psi dx$ $\displaystyle =$ $\displaystyle E\int_{\frac{a}{2}+\epsilon-\beta}^{\frac{a}{2}+\epsilon+\beta}\psi dx\ \ \ \ \ (11)$ $\displaystyle -\frac{\hbar^{2}}{2m}\left.\frac{d\psi}{dx}\right|_{\frac{a}{2}+\epsilon-\beta}^{\frac{a}{2}+\epsilon+\beta}+f\left(t\right)\psi\left(\frac{a}{2}+\epsilon\right)$ $\displaystyle =$ $\displaystyle E\int_{\frac{a}{2}+\epsilon-\beta}^{\frac{a}{2}+\epsilon+\beta}\psi dx \ \ \ \ \ (12)$

The integral on the RHS goes to zero as ${\beta\rightarrow0}$ since ${\psi}$ is finite, so

 $\displaystyle A\frac{2mf\left(t\right)}{\hbar^{2}}\left(e^{ik\left(\frac{a}{2}+\epsilon\right)}-e^{-ik\left(\frac{a}{2}+\epsilon\right)}\right)$ $\displaystyle =$ $\displaystyle -ikD\left(e^{-2ika}e^{ik\left(\frac{a}{2}+\epsilon\right)}+e^{-ik\left(\frac{a}{2}+\epsilon\right)}\right)-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle ikA\left(e^{ik\left(\frac{a}{2}+\epsilon\right)}+e^{-ik\left(\frac{a}{2}+\epsilon\right)}\right) \ \ \ \ \ (13)$

If we now define

 $\displaystyle z$ $\displaystyle \equiv$ $\displaystyle ka\ \ \ \ \ (14)$ $\displaystyle \delta$ $\displaystyle \equiv$ $\displaystyle \frac{2\epsilon}{a}\ \ \ \ \ (15)$ $\displaystyle k\left(\frac{a}{2}+\epsilon\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}z\left(1+\delta\right) \ \ \ \ \ (16)$

we get, transforming the complex exponentials to trig functions

$\displaystyle A\frac{4imf\left(t\right)}{\hbar^{2}}\sin\left[\frac{1}{2}z\left(1+\delta\right)\right]=-2ikDe^{-iz}\cos\left[\frac{1}{2}z\left(1-\delta\right)\right]-2ikA\cos\left[\frac{1}{2}z\left(1+\delta\right)\right] \ \ \ \ \ (17)$

Multiplying through by ${a}$ and defining

$\displaystyle T\equiv\frac{maf\left(t\right)}{\hbar^{2}} \ \ \ \ \ (18)$

we get

$\displaystyle 2AT\sin\left[\frac{1}{2}z\left(1+\delta\right)\right]=-zDe^{-iz}\cos\left[\frac{1}{2}z\left(1-\delta\right)\right]-zA\cos\left[\frac{1}{2}z\left(1+\delta\right)\right] \ \ \ \ \ (19)$

We can write 10 as

 $\displaystyle 2iA\sin\left[\frac{1}{2}z\left(1+\delta\right)\right]$ $\displaystyle =$ $\displaystyle 2ie^{-iz}D\sin\left[\frac{1}{2}z\left(1-\delta\right)\right]\ \ \ \ \ (20)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle e^{-iz}D\frac{\sin\left[\frac{1}{2}z\left(1-\delta\right)\right]}{\sin\left[\frac{1}{2}z\left(1+\delta\right)\right]} \ \ \ \ \ (21)$

Substituting this into 19, multiplying through by ${\sin\left[\frac{1}{2}z\left(1+\delta\right)\right]}$ and cancelling terms we get

 $\displaystyle 2T\sin\left[\frac{1}{2}z\left(1-\delta\right)\right]\sin\left[\frac{1}{2}z\left(1+\delta\right)\right]$ $\displaystyle =$ $\displaystyle -z\left[\cos\left[\frac{1}{2}z\left(1-\delta\right)\right]\sin\left[\frac{1}{2}z\left(1+\delta\right)\right]+\sin\left[\frac{1}{2}z\left(1-\delta\right)\right]\cos\left[\frac{1}{2}z\left(1+\delta\right)\right]\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -z\sin\left[\frac{1}{2}z\left(1-\delta\right)+\frac{1}{2}z\left(1+\delta\right)\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -z\sin z \ \ \ \ \ (24)$

The LHS can be transformed using

 $\displaystyle 2\sin\left[\frac{1}{2}z\left(1-\delta\right)\right]\sin\left[\frac{1}{2}z\left(1+\delta\right)\right]$ $\displaystyle =$ $\displaystyle \cos\left[\frac{1}{2}z\left(1-\delta\right)-\frac{1}{2}z\left(1+\delta\right)\right]-\cos\left[\frac{1}{2}z\left(1-\delta\right)+\frac{1}{2}z\left(1+\delta\right)\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos z\delta-\cos z \ \ \ \ \ (26)$

Putting it together we get the transcendental equation for the ground state energy

$\displaystyle z\sin z=T\left(\cos z-\cos z\delta\right) \ \ \ \ \ (27)$

The energy is

$\displaystyle E=\frac{\hbar^{2}k^{2}}{2m}=\frac{\hbar^{2}z^{2}}{2ma^{2}} \ \ \ \ \ (28)$

We can solve 27 graphically or numerically. For ${\delta=0}$, here are plots of ${z\sin z}$ (red) and ${T\left(\cos z-1\right)}$ (green) for 3 values of ${T}$:

Discounting the trivial solutions at ${z=0}$ and ${z=2\pi}$ which are valid for all ${T}$, we see that the intersection point moves from ${z=\pi}$ out to ${z=2\pi}$ as ${T}$ increases. This is equivalent to the energy moving from ${\frac{\hbar^{2}\pi^{2}}{2ma^{2}}}$ up to ${\frac{4\hbar^{2}\pi^{2}}{2ma^{2}}}$. The first energy is that of a well of width ${a}$ while the second energy is that of a well of width ${\frac{a}{2}}$, which is what we’d expect. As the barrier gets stronger, the ground state energy approaches that of a well of half the width of the original.

Using Maple’s fsolve command we can work out some other values of ${z}$ for ${\delta=0.01}$:

 ${T}$ ${z}$ 1 3.673 5 4.760 20 5.720 100 6.135 1000 6.215

# 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 eigenfunction 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

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

L&B are dealing with a 3-d particle in a box with periodic boundary conditions (rather than requiring the wave function to be zero outside the box), so the momentum eigenstate ${\left|\mathbf{p}\right\rangle }$ is just a plane wave so that its position space representation is

$\displaystyle \left\langle \mathbf{x}\left|\mathbf{p}\right.\right\rangle =\frac{1}{\sqrt{\mathcal{V}}}e^{i\mathbf{p}\cdot\mathbf{x}} \ \ \ \ \ (15)$

where ${\mathcal{V}}$ is the volume of the box.

They begin with a one-particle state

 $\displaystyle \left\langle \mathbf{p}\left|\mathbf{p}'\right.\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \hat{a}_{\mathbf{p}}^{\dagger}0\left|\hat{a}_{\mathbf{p}'}^{\dagger}\right.0\right\rangle \ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|\hat{a}_{\mathbf{p}}\hat{a}_{\mathbf{p}'}^{\dagger}\right|0\right\rangle \ \ \ \ \ (17)$

We can now use the relation 14 to get

$\displaystyle \hat{a}_{\mathbf{p}}\hat{a}_{\mathbf{p}'}^{\dagger}=\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right)+\hat{a}_{\mathbf{p}'}^{\dagger}\hat{a}_{\mathbf{p}} \ \ \ \ \ (18)$

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

 $\displaystyle \left\langle \mathbf{p}\left|\mathbf{p}'\right.\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right)\right|0\right\rangle \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right) \ \ \ \ \ (20)$

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

 $\displaystyle \phi_{\mathbf{p}}\left(\mathbf{x}\right)$ $\displaystyle \equiv$ $\displaystyle \left\langle \mathbf{x}\left|\mathbf{p}\right.\right\rangle \ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}'\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{x}'\right)\left|\mathbf{p}\right\rangle \ \ \ \ \ (22)$

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:

 $\displaystyle \left|\mathbf{x}\right\rangle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\left|\mathbf{q}\right\rangle \left\langle \mathbf{q}\left|\mathbf{x}\right.\right\rangle \ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\left|\mathbf{q}\right\rangle \left\langle \mathbf{x}\left|\mathbf{q}\right.\right\rangle ^*\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\phi_{\mathbf{q}}^*\left(\mathbf{x}\right)\left|\mathbf{q}\right\rangle \ \ \ \ \ (25)$

Therefore

 $\displaystyle \left\langle \mathbf{x}\right|$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\phi_{\mathbf{q}}\left(\mathbf{x}\right)\left\langle \mathbf{q}\right|\ \ \ \ \ (26)$ $\displaystyle \left\langle \mathbf{x}\left|\mathbf{p}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\phi_{\mathbf{q}}\left(\mathbf{x}\right)\left\langle \mathbf{q}\left|\mathbf{p}\right.\right\rangle \ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{q}\phi_{\mathbf{q}}\left(\mathbf{x}\right)\delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{p}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \phi_{\mathbf{p}}\left(\mathbf{x}\right) \ \ \ \ \ (29)$

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

$\displaystyle \left\langle \mathbf{p}'\mathbf{q'}\left|\mathbf{p}\mathbf{q}\right.\right\rangle =\left\langle 0\left|\hat{a}_{\mathbf{p}'}\hat{a}_{\mathbf{q}'}\hat{a}_{\mathbf{q}}^{\dagger}\hat{a}_{\mathbf{p}}^{\dagger}\right|0\right\rangle \ \ \ \ \ (30)$

From commutation relations 14 we get

 $\displaystyle \hat{a}_{\mathbf{p}'}\hat{a}_{\mathbf{q}'}\hat{a}_{\mathbf{q}}^{\dagger}\hat{a}_{\mathbf{p}}^{\dagger}$ $\displaystyle =$ $\displaystyle \hat{a}_{\mathbf{p}'}\left(\delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{q}'\right)+\hat{a}_{\mathbf{q}}^{\dagger}\hat{a}_{\mathbf{q}'}\right)\hat{a}_{\mathbf{p}}^{\dagger}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{q}'\right)\left(\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right)+\hat{a}_{\mathbf{p}}^{\dagger}\hat{a}_{\mathbf{p}'}\right)+\hat{a}_{\mathbf{p}'}\hat{a}_{\mathbf{q}}^{\dagger}\hat{a}_{\mathbf{q}'}\hat{a}_{\mathbf{p}}^{\dagger}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{q}'\right)\left(\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right)+\hat{a}_{\mathbf{p}}^{\dagger}\hat{a}_{\mathbf{p}'}\right)+\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(\delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{p}'\right)+\hat{a}_{\mathbf{q}}^{\dagger}\hat{a}_{\mathbf{p}'}\right)\left(\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}'\right)+\hat{a}_{\mathbf{p}}^{\dagger}\hat{a}_{\mathbf{q}'}\right) \ \ \ \ \ (34)$

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

$\displaystyle \left\langle \mathbf{p}'\mathbf{q'}\left|\mathbf{p}\mathbf{q}\right.\right\rangle =\delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{q}'\right)\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}'\right)+\delta^{\left(3\right)}\left(\mathbf{q}-\mathbf{p}'\right)\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}'\right) \ \ \ \ \ (35)$

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 ${\left|\mathbf{xy}\right\rangle }$. Since the particles are independent, we can represent the compound state as the product of two single-particle states:

$\displaystyle \left|\mathbf{p}'\mathbf{q}'\right\rangle =\left|\mathbf{p}'\right\rangle \left|\mathbf{q}'\right\rangle \ \ \ \ \ (36)$

where each single-particle state has its own coordinates, independent of the other state. The inner product of two such states is

$\displaystyle \left\langle \mathbf{rs}\left|\mathbf{p}\mathbf{q}\right.\right\rangle =\left\langle \mathbf{r}\left|\mathbf{p}\right.\right\rangle \left\langle \mathbf{s}\left|\mathbf{q}\right.\right\rangle \ \ \ \ \ (37)$

since each particle’s coordinates are independent of the other particle.

Using the more familiar wave function representation, we could have something like this:

 $\displaystyle \left\langle \mathbf{rs}\left|\mathbf{p}\mathbf{q}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}\int d^{3}\mathbf{y}\phi_{\mathbf{r}}^*\left(\mathbf{x}\right)\phi_{\mathbf{s}}^*\left(\mathbf{y}\right)\psi_{\mathbf{p}}\left(\mathbf{x}\right)\psi_{\mathbf{q}}\left(\mathbf{y}\right)\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}\phi_{\mathbf{r}}^*\left(\mathbf{x}\right)\psi_{\mathbf{p}}\left(\mathbf{x}\right)\int d^{3}\mathbf{y}\phi_{\mathbf{s}}^*\left(\mathbf{y}\right)\psi_{\mathbf{q}}\left(\mathbf{y}\right)\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \mathbf{r}\left|\mathbf{p}\right.\right\rangle \left\langle \mathbf{s}\left|\mathbf{q}\right.\right\rangle \ \ \ \ \ (40)$

In that case we can write

 $\displaystyle \left|\mathbf{xy}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2!}}\int d^{3}p'\int d^{3}q'\left|\mathbf{p}'\right\rangle \left|\mathbf{q}'\right\rangle \left\langle \mathbf{q'}\left|\mathbf{y}\right.\right\rangle \left\langle \mathbf{p'}\left|\mathbf{x}\right.\right\rangle \ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2!}}\int d^{3}p'\int d^{3}q'\phi_{\mathbf{p}'}^*\left(\mathbf{x}\right)\phi_{\mathbf{q}'}^*\left(\mathbf{y}\right)\left|\mathbf{p}'\right\rangle \left|\mathbf{q}'\right\rangle \ \ \ \ \ (42)$ $\displaystyle \left\langle \mathbf{xy}\right|$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2!}}\int d^{3}p'\int d^{3}q'\phi_{\mathbf{p}'}\left(\mathbf{x}\right)\phi_{\mathbf{q}'}\left(\mathbf{y}\right)\left\langle \mathbf{p}'\right|\left\langle \mathbf{q}'\right|\ \ \ \ \ (43)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2!}}\int d^{3}p'\int d^{3}q'\phi_{\mathbf{p}'}\left(\mathbf{x}\right)\phi_{\mathbf{q}'}\left(\mathbf{y}\right)\left\langle \mathbf{p}'\mathbf{q}'\right| \ \ \ \ \ (44)$

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 35

 $\displaystyle \left\langle \mathbf{xy}\left|\mathbf{p}\mathbf{q}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2!}}\int d^{3}p'\int d^{3}q'\phi_{\mathbf{p}'}\left(\mathbf{x}\right)\phi_{\mathbf{q}'}\left(\mathbf{y}\right)\left\langle \mathbf{p}'\mathbf{q}'\left|\mathbf{p}\mathbf{q}\right.\right\rangle \ \ \ \ \ (45)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[\phi_{\mathbf{p}}\left(\mathbf{x}\right)\phi_{\mathbf{q}}\left(\mathbf{y}\right)+\phi_{\mathbf{q}}\left(\mathbf{x}\right)\phi_{\mathbf{p}}\left(\mathbf{y}\right)\right] \ \ \ \ \ (46)$

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

$\displaystyle \left\langle \mathbf{xy}\left|\mathbf{p}\mathbf{q}\right.\right\rangle _{fermion}=\frac{1}{\sqrt{2}}\left[\phi_{\mathbf{p}}\left(\mathbf{x}\right)\phi_{\mathbf{q}}\left(\mathbf{y}\right)-\phi_{\mathbf{q}}\left(\mathbf{x}\right)\phi_{\mathbf{p}}\left(\mathbf{y}\right)\right] \ \ \ \ \ (47)$

# 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

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

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.

# 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

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

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:

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

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 – Chapter 10, Post 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 Schrödinger equation; rather they are the solutions of the time-independent Schrödinger 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

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

# 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 Schrödinger equation; rather they are the solutions of the time-independent Schrödinger 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

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