# Harmonic oscillator in 2 dimensions: comparison with rectangular coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.7 (8) – (10).

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

In this post, we’ll continue with the solution of the 2-d isotropic harmonic oscillator. In the previous post, we found that the radial equation ${R}$ can be written as

$\displaystyle R\left(y\right)=y^{\left|m\right|}e^{-y^{2}/2}U\left(y\right) \ \ \ \ \ (1)$

where the dimensionless variables are given by

 $\displaystyle y$ $\displaystyle \equiv$ $\displaystyle \sqrt{\frac{\mu\omega}{\hbar}}\rho\ \ \ \ \ (2)$ $\displaystyle \varepsilon$ $\displaystyle \equiv$ $\displaystyle \frac{E}{\hbar\omega} \ \ \ \ \ (3)$

and ${U}$ has a solution as a power series

$\displaystyle U\left(y\right)=\sum_{r=0}^{\infty}C_{r}y^{r} \ \ \ \ \ (4)$

The coefficients ${C_{r}}$ satisfy the recursion relation

$\displaystyle C_{r+2}=\frac{2\left(r+\left|m\right|+1-\varepsilon\right)}{\left(r+2\right)^{2}}C_{r} \ \ \ \ \ (5)$

Only ${C_{r}}$ for even ${r}$ are non-zero.

In order for ${U}$ to remain finite for large ${y}$, the series must terminate, which gives the allowable values for the energy as

$\displaystyle E=\hbar\omega\left(n+1\right) \ \ \ \ \ (6)$

with

$\displaystyle n\equiv2k+\left|m\right| \ \ \ \ \ (7)$

and ${k=0,1,2,\ldots}$.

We can now compare the solution obtained in polar coordinates with our earlier solution in terms of rectangular coordinates. First, what are the possible values for ${m}$ for a given energy ${E=\hbar\omega\left(n+1\right)}$? From the relation 7, we can look at even and odd ${n}$ separately. For even ${n}$, ${k}$ can take values ${0,1,\ldots,\frac{n}{2}-1,\frac{n}{2}}$. The first ${\frac{n}{2}}$ of these values for ${k}$ (that is, for ${k=0,1,\ldots,\frac{n}{2}-1}$) each allow two values of ${m}$ such that ${\left|m\right|=n-2k}$, namely ${m=\pm\left(n-2k\right)}$. If ${k=\frac{n}{2}}$, then we must have ${m=0}$. Thus for even ${n}$ the total number of combinations is ${2\times\frac{n}{2}+1=n+1}$.

For odd ${n}$, ${k}$ can take on values ${0,1,\ldots,\frac{n-1}{2}}$, giving a total of ${\frac{n+1}{2}}$ possible values. (If this isn’t obvious, write it out for the first few values of odd ${n}$ to see the pattern.) For each of these values of ${k}$, ${m}$ can take on the two values ${m=\pm\left(n-2k\right)}$, thus there are again ${2\times\frac{n+1}{2}=n+1}$ different combinations. Thus a state with energy ${E=\hbar\omega\left(n+1\right)}$ has a degeneracy ${n+1}$.

We can construct the actual eigenfunctions for a couple of values of ${n}$ by plugging in the appropriate formulas. For ${n=0}$ there is only one function, which we find by setting ${k=m=0}$. From 4, we have

$\displaystyle U_{0}\left(y\right)=C_{0} \ \ \ \ \ (8)$

and from 1 we have

$\displaystyle R_{0}\left(y\right)=C_{0}e^{-y^{2}/2} \ \ \ \ \ (9)$

or, in terms of the original variables

$\displaystyle R_{0}\left(\rho\right)=C_{0}e^{-\mu\omega\rho^{2}/2\hbar} \ \ \ \ \ (10)$

The complete solution is given by

 $\displaystyle \psi_{m}\left(\rho,\phi\right)$ $\displaystyle =$ $\displaystyle R\left(\rho\right)\Phi_{m}\left(\phi\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}R\left(\rho\right)e^{im\phi} \ \ \ \ \ (12)$

so for ${m=0}$ we have

$\displaystyle \psi_{0}\left(\rho,\phi\right)=\frac{C_{0}}{\sqrt{2\pi}}e^{-\mu\omega\rho^{2}/2\hbar} \ \ \ \ \ (13)$

The constant ${C_{0}}$ can be found by normalizing:

 $\displaystyle 1$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}\int_{0}^{2\pi}\left|\psi_{0}\right|^{2}\rho d\phi d\rho\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|C_{0}\right|^{2}\int_{0}^{\infty}e^{-\mu\omega\rho^{2}/\hbar}\rho d\rho\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|C_{0}\right|^{2}\frac{\hbar}{2\mu\omega}\ \ \ \ \ (16)$ $\displaystyle C_{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\mu\omega}{\hbar}}\ \ \ \ \ (17)$ $\displaystyle \psi_{0}\left(\rho,\phi\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\mu\omega}{\pi\hbar}}e^{-\mu\omega\rho^{2}/2\hbar} \ \ \ \ \ (18)$

This agrees with the earlier result in rectangular coordinates (eqn 26 in this post). This must be the case, since the ${n=0}$ state is non-degenerate.

For ${n=1}$, we have ${k=0}$ and ${m=\pm1}$ so we have two solutions:

 $\displaystyle \psi_{1}$ $\displaystyle =$ $\displaystyle \frac{C_{0}}{\sqrt{2\pi}}\sqrt{\frac{\mu\omega}{\hbar}}\rho e^{-\mu\omega\rho^{2}/2\hbar}e^{i\phi}\ \ \ \ \ (19)$ $\displaystyle \psi_{-1}$ $\displaystyle =$ $\displaystyle \frac{C_{0}}{\sqrt{2\pi}}\sqrt{\frac{\mu\omega}{\hbar}}\rho e^{-\mu\omega\rho^{2}/2\hbar}e^{-i\phi} \ \ \ \ \ (20)$

Again, we normalize

 $\displaystyle 1$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}\int_{0}^{2\pi}\left|\psi_{\pm1}\right|^{2}\rho d\phi d\rho\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu\omega}{\hbar}\left|C_{0}\right|^{2}\int_{0}^{\infty}e^{-\mu\omega\rho^{2}/\hbar}\rho^{3}d\rho\ \ \ \ \ (22)$ $\displaystyle C_{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\mu\omega}{\hbar}}\ \ \ \ \ (23)$ $\displaystyle \psi_{\pm1}$ $\displaystyle =$ $\displaystyle \frac{\mu\omega}{\hbar\sqrt{\pi}}\rho e^{-\mu\omega\rho^{2}/2\hbar}e^{\pm i\phi} \ \ \ \ \ (24)$

These solutions are linear combinations of the corresponding solutions in rectangular coordinates (converted to polar for comparison):

 $\displaystyle \psi_{10}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2}\mu\omega}{\hbar\sqrt{\pi}}e^{-\mu\omega\rho^{2}/2\hbar}\rho\cos\phi\ \ \ \ \ (25)$ $\displaystyle \psi_{01}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2}\mu\omega}{\hbar\sqrt{\pi}}e^{-\mu\omega\rho^{2}/2\hbar}\rho\sin\phi \ \ \ \ \ (26)$

The combinations are

 $\displaystyle \psi_{+1}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\psi_{10}+i\psi_{01}\right)\ \ \ \ \ (27)$ $\displaystyle \psi_{-1}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\psi_{10}-i\psi_{01}\right) \ \ \ \ \ (28)$

The parity of the states is found from their behaviour under the transformation (in rectangular coordinates) ${x\rightarrow-x}$ and ${y\rightarrow-y}$. In polar coordinates this is equivalent to the transformation ${\phi\rightarrow\phi+\pi}$ and from 18 and 24 we see that

 $\displaystyle \psi_{0}\left(\rho,\phi+\pi\right)$ $\displaystyle =$ $\displaystyle \psi_{0}\left(\rho,\phi\right)\ \ \ \ \ (29)$ $\displaystyle \psi_{\pm1}\left(\rho,\phi+\pi\right)$ $\displaystyle =$ $\displaystyle \psi_{\pm1}\left(\rho,\phi\right)e^{\pm\pi}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\psi_{\pm1}\left(\rho,\phi\right) \ \ \ \ \ (31)$

Thus the parity of ${n=0}$ is even, and that of ${n=1}$ is odd. In general, since the ${\phi}$ dependence enters only through the term ${e^{im\phi}=e^{in\phi}e^{-2ik\phi}}$, we see that adding ${\pi}$ to ${\phi}$ leaves the ${e^{-2ik\phi}}$ term unchanged and multiplies the ${e^{in\phi}}$ term by ${e^{in\pi\phi}=\left(-1\right)^{n}}$, so the parity of state ${n}$ is ${\left(-1\right)^{n}}$.

# Two-dimensional harmonic oscillator – Part 2: Series solution

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.7 (6) – (7).

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In this post, we’ll continue with the solution of the 2-d isotropic harmonic oscillator. In the last post, we started with the ODE for the radial function in the form

$\displaystyle -\frac{\hbar^{2}}{2\mu}\left(\frac{d^{2}}{d\rho^{2}}+\frac{1}{\rho}\frac{d}{d\rho}-\frac{m^{2}}{\rho^{2}}\right)R+\frac{1}{2}\mu\omega^{2}\rho^{2}R=ER \ \ \ \ \ (1)$

We introduced dimensionless variables

 $\displaystyle y$ $\displaystyle \equiv$ $\displaystyle \sqrt{\frac{\mu\omega}{\hbar}}\rho\ \ \ \ \ (2)$ $\displaystyle \varepsilon$ $\displaystyle \equiv$ $\displaystyle \frac{E}{\hbar\omega} \ \ \ \ \ (3)$

and found that ${R}$ could be written as

$\displaystyle R\left(y\right)=y^{\left|m\right|}e^{-y^{2}/2}U\left(y\right) \ \ \ \ \ (4)$

with ${U}$ given by the solution of the ODE

$\displaystyle U^{\prime\prime}+\left(\frac{2\left|m\right|+1}{y}-2y\right)U^{\prime}+\left(2\varepsilon-2\left|m\right|-2\right)U=0 \ \ \ \ \ (5)$

We can solve this by using a power series of the form

$\displaystyle U\left(y\right)=\sum_{r=0}^{\infty}C_{r}y^{r} \ \ \ \ \ (6)$

where the coefficients ${C_{r}}$ are constants.

The derivatives are

 $\displaystyle U^{\prime}$ $\displaystyle =$ $\displaystyle \sum_{r=0}^{\infty}C_{r}ry^{r-1}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0+C_{1}+2C_{2}y+3C_{3}+y^{2}+\ldots\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{r=0}^{\infty}C_{r+1}\left(r+1\right)y^{r}\ \ \ \ \ (9)$ $\displaystyle U^{\prime\prime}$ $\displaystyle =$ $\displaystyle \sum_{r=0}^{\infty}C_{r+1}r\left(r+1\right)y^{r-1}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0+\left(1\right)\left(2\right)C_{2}+\left(2\right)\left(3\right)C_{3}y+\ldots\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{r=0}^{\infty}C_{r+2}\left(r+1\right)\left(r+2\right)y^{r} \ \ \ \ \ (12)$

Plugging these into 5 we have (we’ll drop the absolute value signs on ${\left|m\right|}$ to make the notation simpler; we can restore them at the end):

 $\displaystyle \sum_{r=0}^{\infty}C_{r+2}\left(r+1\right)\left(r+2\right)y^{r}+\left(2m+1\right)\sum_{r=0}^{\infty}C_{r}ry^{r-2}-$ $\displaystyle$ $\displaystyle \ldots\ \ \ \ \ (13)$ $\displaystyle 2\sum_{r=0}^{\infty}C_{r}ry^{r}+2\left(\varepsilon-m-1\right)\sum_{r=0}^{\infty}C_{r}y^{r}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (14)$

The second sum in the first line is

 $\displaystyle \sum_{r=0}^{\infty}C_{r}ry^{r-2}$ $\displaystyle =$ $\displaystyle 0+C_{1}y^{-1}+2C_{2}+3C_{3}y+\ldots\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{r=-1}^{\infty}C_{r+2}\left(r+2\right)y^{r} \ \ \ \ \ (16)$

The sum thus becomes

 $\displaystyle \left(2m+1\right)C_{1}y^{-1}+\sum_{r=0}^{\infty}y^{r}C_{r+2}\left(r+2\right)^{2}+2\sum_{r=0}^{\infty}y^{r}C_{r}\left[-r+\varepsilon-m-1\right]$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (17)$

A basic theorem about power series is that if the sum of the series equals zero for all ${y}$, then the coefficient of each power must be zero. This shows that ${C_{1}=0}$ since otherwise the series would blow up as ${y\rightarrow0}$. This results in a recursion relation for the ${C_{r}}$:

$\displaystyle C_{r+2}=\frac{2\left(r+m+1-\varepsilon\right)}{\left(r+2\right)^{2}}C_{r} \ \ \ \ \ (18)$

Since ${C_{1}=0}$, all ${C_{r}=0}$ for odd ${r}$. For large ${r}$ we have

$\displaystyle \frac{C_{r+2}}{C_{r}}\rightarrow\frac{2}{r} \ \ \ \ \ (19)$

If the series is allowed to be infinite, this leads to a divergent series as we can see from the following (based on Shankar’s section 7.3). Suppose we look at ${y^{m}e^{y^{2}}}$, which clearly goes to infinity at large ${y}$ (remember, ${m}$ is positive). In series form this is

$\displaystyle y^{m}e^{y^{2}}=\sum_{k=0}^{\infty}\frac{y^{2k+m}}{k!} \ \ \ \ \ (20)$

The coefficient ${C_{n}}$ of ${y^{n}}$ , with ${n=2k+m}$ in this series is

$\displaystyle C_{n}=\frac{1}{\left[\left(n-m\right)/2\right]!} \ \ \ \ \ (21)$

Similarly,

$\displaystyle C_{n+2}=\frac{1}{\left[\left(n+2-m\right)/2\right]!} \ \ \ \ \ (22)$

The ratio is

 $\displaystyle \frac{C_{n+2}}{C_{n}}$ $\displaystyle =$ $\displaystyle \frac{\left[\left(n-m\right)/2\right]!}{\left[\left(n+2-m\right)/2\right]!}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(n-m\right)/2+1}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle \rightarrow$ $\displaystyle \frac{2}{n} \ \ \ \ \ (25)$

In other words, the coefficients of our series solution have the same behaviour 19 for large ${r}$ as those in the series for ${y^{m}e^{y^{2}}}$. Referring back to 4, we see that this gives an overall behaviour for the radial function ${R}$ of

$\displaystyle R\rightarrow y^{m}e^{-y^{2}/2}y^{m}e^{y^{2}}=y^{2m}e^{y^{2}/2} \ \ \ \ \ (26)$

Thus if we allow the series for ${U}$ to be infinite, the overall solution diverges, which is not acceptable. We therefore require that the series terminates at some finite value of ${r}$, and from 18 we see that this happens if

$\displaystyle \varepsilon=r+m+1 \ \ \ \ \ (27)$

for some ${r}$. From the definition 3 this gives us the allowed values for the energy:

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \hbar\omega\left(r+\left|m\right|+1\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\omega\left(2k+\left|m\right|+1\right) \ \ \ \ \ (29)$

where the last line follows because ${r}$ must be even. If

$\displaystyle n\equiv2k+\left|m\right| \ \ \ \ \ (30)$

then the allowed energies are

$\displaystyle E=\hbar\omega\left(n+1\right) \ \ \ \ \ (31)$

# Two-dimensional harmonic oscillator – Part 1

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.7 (1) – (5).

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

In this problem, we’ll look at solving the 2-dimensional isotropic harmonic oscillator. The solution is fairly lengthy, so we’ll split it into two posts, with this being the first. The method of solution is similar to that used in the one-dimensional harmonic oscillator, so you may wish to refer back to that before proceeding.

The Hamiltonian is, in rectangular coordinates:

$\displaystyle H=\frac{P_{x}^{2}+P_{y}^{2}}{2\mu}+\frac{1}{2}\mu\omega^{2}\left(X^{2}+Y^{2}\right) \ \ \ \ \ (1)$

The potential term is radially symmetric (it doesn’t depend on the polar angle ${\phi}$) so we have a problem of the form considered earlier. We saw there that for such potentials ${\left[H,L_{z}\right]=0}$. [If you don’t believe this, you can grind through the calculations using the commutation relations for ${L_{z}}$ with the rectangular momenta and coordinates, but I won’t go through that here.]

As a result, ${L_{z}}$ and ${H}$ have simultaneous eigenfunctions of form

$\displaystyle \psi\left(\rho,\phi\right)=R\left(\rho\right)\Phi_{m}\left(\phi\right) \ \ \ \ \ (2)$

where

$\displaystyle \Phi_{m}\left(\phi\right)=\frac{1}{\sqrt{2\pi}}e^{im\phi} \ \ \ \ \ (3)$

The radial function satisfies the ODE

$\displaystyle -\frac{\hbar^{2}}{2\mu}\left(\frac{d^{2}R}{d\rho^{2}}+\frac{1}{\rho}\frac{dR}{d\rho}-\frac{m^{2}}{\rho^{2}}R\right)+V\left(\rho\right)R=ER \ \ \ \ \ (4)$

where in this case

$\displaystyle V\left(\rho\right)=\frac{1}{2}\mu\omega^{2}\rho^{2} \ \ \ \ \ (5)$

Thus the equation we must solve is

$\displaystyle -\frac{\hbar^{2}}{2\mu}\left(\frac{d^{2}}{d\rho^{2}}+\frac{1}{\rho}\frac{d}{d\rho}-\frac{m^{2}}{\rho^{2}}\right)R+\frac{1}{2}\mu\omega^{2}\rho^{2}R=ER \ \ \ \ \ (6)$

To get a feel for the solution, we examine the behaviour in two limiting cases: ${\rho\rightarrow0}$ and ${\rho\rightarrow\infty}$. It’s actually easier if we introduce dimensionless variables now, rather than in Shankar’s step 4, so we define

 $\displaystyle y$ $\displaystyle \equiv$ $\displaystyle \sqrt{\frac{\mu\omega}{\hbar}}\rho\ \ \ \ \ (7)$ $\displaystyle \varepsilon$ $\displaystyle \equiv$ $\displaystyle \frac{E}{\hbar\omega} \ \ \ \ \ (8)$

This transforms 6 to

 $\displaystyle -\frac{\hbar^{2}}{2\mu}\left(\frac{\mu\omega}{\hbar}\right)\left(\frac{d^{2}}{dy^{2}}+\frac{1}{y}\frac{d}{dy}-\frac{m^{2}}{y^{2}}\right)R+\frac{1}{2}\hbar\omega y^{2}R$ $\displaystyle =$ $\displaystyle ER\ \ \ \ \ (9)$ $\displaystyle -\left(\frac{d^{2}}{dy^{2}}+\frac{1}{y}\frac{d}{dy}-\frac{m^{2}}{y^{2}}+2\varepsilon\right)R+y^{2}R$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (10)$

We can now look at ${y\rightarrow0}$, and we neglect the terms ${2\varepsilon R}$ and ${y^{2}R}$ to get

$\displaystyle \left(\frac{d^{2}}{dy^{2}}+\frac{1}{y}\frac{d}{dy}-\frac{m^{2}}{y^{2}}\right)R=0 \ \ \ \ \ (11)$

If we try a solution of form

$\displaystyle R=y^{\left|m\right|} \ \ \ \ \ (12)$

we have

$\displaystyle \left|m\right|\left(\left|m\right|-1\right)y^{\left|m\right|-2}+\left|m\right|y^{\left|m\right|-2}-m^{2}y^{\left|m\right|-2}=0 \ \ \ \ \ (13)$

Thus 12 is indeed a solution in this limiting case.

For ${y\rightarrow\infty}$, we can ignore the terms ${\frac{1}{y}\frac{d}{dy}}$, ${\frac{m^{2}}{y^{2}}R}$ and ${2\varepsilon R}$ to get

$\displaystyle -\frac{d^{2}}{dy^{2}}R+y^{2}R=0 \ \ \ \ \ (14)$

or

$\displaystyle R^{\prime\prime}=y^{2}R \ \ \ \ \ (15)$

We try a solution of form

$\displaystyle R=y^{a}e^{-y^{2}/2} \ \ \ \ \ (16)$

where ${a}$ is some constant. We find

 $\displaystyle R^{\prime}$ $\displaystyle =$ $\displaystyle \left(ay^{a-1}-y^{a+1}\right)e^{-y^{2}/2}\ \ \ \ \ (17)$ $\displaystyle R^{\prime\prime}$ $\displaystyle =$ $\displaystyle \left(a\left(a-1\right)y^{a-2}-\left(a+1\right)y^{a}-ay^{a}+y^{a+2}\right)e^{-y^{2}/2}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle y^{a+2}\left(\frac{a\left(a-1\right)}{y^{4}}-\frac{2a+1}{y^{2}}+1\right)e^{-y^{2}/2} \ \ \ \ \ (19)$

As ${y\rightarrow\infty}$, the last line tends to

$\displaystyle R^{\prime\prime}\rightarrow y^{a+2}e^{-y^{2}/2}=y^{2}R \ \ \ \ \ (20)$

so in this limit 16 is a solution. We can therefore propose that ${R}$ has the general form

$\displaystyle R\left(y\right)=y^{\left|m\right|}e^{-y^{2}/2}U\left(y\right) \ \ \ \ \ (21)$

where ${U}$ is a function to be determined by solving the exact ODE 10. We can get an ODE for ${U}$ by substituting 21 into 10, although the calculation gets somewhat messy. As Shankar suggests, we can do this in two stages. First, we substitute

$\displaystyle R=y^{\left|m\right|}f\left(y\right) \ \ \ \ \ (22)$

where

$\displaystyle f\left(y\right)=e^{-y^{2}/2}U\left(y\right) \ \ \ \ \ (23)$

The required derivatives are (To make the notation simpler, I’ll drop the absolute value signs around ${m}$; you should assume that wherever ${m}$ occurs, it should really be ${\left|m\right|}$. We can replace the absolute value sign at the end.)

 $\displaystyle R^{\prime}$ $\displaystyle =$ $\displaystyle my^{m-1}f+y^{m}f^{\prime}\ \ \ \ \ (24)$ $\displaystyle R^{\prime\prime}$ $\displaystyle =$ $\displaystyle m\left(m-1\right)y^{m-2}f+2my^{m-1}f^{\prime}+y^{m}f^{\prime\prime} \ \ \ \ \ (25)$

Plugging these into 10 we have

 $\displaystyle -\left(m\left(m-1\right)y^{m-2}f+2my^{m-1}f^{\prime}+y^{m}f^{\prime\prime}\right)-\left(my^{m-2}f+y^{m-1}f^{\prime}\right)+$ $\displaystyle$ $\displaystyle \ldots\ \ \ \ \ (26)$ $\displaystyle m^{2}y^{m-2}f-2\varepsilon y^{m}f+y^{m+2}f$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (27)$

Collecting terms and dividing through by ${-y^{m}}$, we get

$\displaystyle f^{\prime\prime}+f^{\prime}\left(\frac{2m+1}{y}\right)+f\left(2\varepsilon-y^{2}\right)=0 \ \ \ \ \ (28)$

We now get the derivatives of ${f}$:

 $\displaystyle f^{\prime}$ $\displaystyle =$ $\displaystyle -ye^{-y^{2}/2}U+e^{-y^{2}/2}U^{\prime}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-y^{2}/2}\left(U^{\prime}-yU\right)\ \ \ \ \ (30)$ $\displaystyle f^{\prime\prime}$ $\displaystyle =$ $\displaystyle \left[-y\left(U^{\prime}-yU\right)+U^{\prime\prime}-U-yU^{\prime}\right]e^{-y^{2}/2}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(U^{\prime\prime}-2yU^{\prime}+\left(y^{2}-1\right)U\right)e^{-y^{2}/2} \ \ \ \ \ (32)$

When we plug these into 28, the exponential factor cancels out, so we get

$\displaystyle U^{\prime\prime}-2yU^{\prime}+\left(y^{2}-1\right)U+\frac{2m+1}{y}\left(U^{\prime}-yU\right)+U\left(2\varepsilon-y^{2}\right)=0 \ \ \ \ \ (33)$

Collecting terms, we get, upon restoring the absolute values:

$\displaystyle U^{\prime\prime}+\left(\frac{2\left|m\right|+1}{y}-2y\right)U^{\prime}+\left(2\varepsilon-2\left|m\right|-2\right)U=0 \ \ \ \ \ (34)$

We can solve this ODE using a power series in ${y}$, but we’ll leave that till the next post.

# Radially symmetric potentials, angular momentum and centrifugal force

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.5.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

We’ve seen that the eigenfunctions of two-dimensional angular momentum have the form

$\displaystyle \psi\left(\rho,\phi\right)=R\left(\rho\right)\Phi_{m}\left(\phi\right) \ \ \ \ \ (1)$

where

$\displaystyle \Phi_{m}\left(\phi\right)=\frac{1}{\sqrt{2\pi}}e^{im\phi} \ \ \ \ \ (2)$

In 2 dimensions and polar coordinates, the hamiltonian can be written as

$\displaystyle H=-\frac{\hbar^{2}}{2\mu}\left(\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}\right)+V\left(\rho,\phi\right) \ \ \ \ \ (3)$

If the potential is radially symmetric, that is, it doesn’t depend on ${\phi}$, then

$\displaystyle H=-\frac{\hbar^{2}}{2\mu}\left(\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}\right)+V\left(\rho\right) \ \ \ \ \ (4)$

In polar coordinates, the angular momentum operator has the form

$\displaystyle L_{z}=-i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (5)$

Thus ${L_{z}}$ commutes with every term in the hamiltonian 4, so for ${V=V\left(\rho\right)}$, we find

$\displaystyle \left[H,L_{z}\right]=0 \ \ \ \ \ (6)$

meaning that we can find a set of functions that are simultaneously eigenfunctions of both ${H}$ and ${L_{z}}$. Since we already know what the most general eigenfunctions of ${L_{z}}$ are (eqn 1), the problem is then to find the radial function ${R\left(\rho\right)}$ so that

$\displaystyle H\left[R\left(\rho\right)\Phi_{m}\left(\phi\right)\right]=ER\left(\rho\right)\Phi_{m}\left(\phi\right) \ \ \ \ \ (7)$

If we use 4 for ${H}$ and 2 for ${\Phi}$ we find that we must solve the differential equation

$\displaystyle -\frac{\hbar^{2}}{2\mu}\left(\frac{d^{2}R}{d\rho^{2}}+\frac{1}{\rho}\frac{dR}{d\rho}-\frac{m^{2}}{\rho^{2}}R\right)+V\left(\rho\right)R=ER \ \ \ \ \ (8)$

We’ve replaced the partial derivatives in 4 by ordinary derivatives, since we now have an ODE in one independent variable, namely ${\rho}$.

The term arising from the ${\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}}$ term in 4 is similar to a potential term, since it doesn’t involve any derivatives of ${R}$. The potential term is

$\displaystyle V_{c}=\frac{\hbar^{2}}{2\mu}\frac{m^{2}}{\rho^{2}} \ \ \ \ \ (9)$

We can find the force corresponding to ${V_{c}}$ by taking the gradient, which in this case amounts to

$\displaystyle F_{c}=\frac{\partial V_{c}}{\partial\rho}=-\frac{\hbar^{2}m^{2}}{\mu\rho^{3}} \ \ \ \ \ (10)$

Since the quantum angular momentum is ${\ell_{z}=m\hbar}$, this can be written as

$\displaystyle F_{c}=-\frac{\ell_{z}^{2}}{\mu\rho^{3}} \ \ \ \ \ (11)$

If the particle is in a circular orbit, then ${\ell_{z}=\rho p}$ where ${p}$ is its momentum, so this becomes

$\displaystyle F_{c}=-\frac{p^{2}}{\mu\rho} \ \ \ \ \ (12)$

Classically, ${p=\mu v^{2}}$ so this is equivalent to

$\displaystyle F_{c}=-\frac{\mu v^{2}}{\rho} \ \ \ \ \ (13)$

which is the formula for centripetal force in Newtonian physics. (Shankar calls it the centrifugal force, but the minus sign indicates it acts towards the centre of rotation rather than outwards, and of course as we well know, the centrifugal force is a fictitious force anyway.)

# Angular momentum: probabilities of eigenvalues in two dimensions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercises 12.3.3 – 12.3.4.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

We’ve seen that the eigenfunctions of two-dimensional angular momentum have the form

$\displaystyle \psi\left(\rho,\phi\right)=R\left(\rho\right)e^{i\ell_{z}\phi/\hbar} \ \ \ \ \ (1)$

where ${\ell_{z}}$ (the eigenvalue) is an integral multiple of ${\hbar}$ and ${R\left(\rho\right)}$ is some function of the radial coordinate ${\rho}$ which depends on the particular potential function in the hamiltonian. It’s more convenient to write the angular function as

$\displaystyle \Phi_{m}\left(\phi\right)=\frac{1}{\sqrt{2\pi}}e^{im\phi} \ \ \ \ \ (2)$

This set of functions is orthonormal over the interval ${\phi\in\left[0,2\pi\right]}$, that is

$\displaystyle \int_{0}^{2\pi}\Phi_{m}^*\left(\phi\right)\Phi_{m^{\prime}}\left(\phi\right)d\phi=\delta_{mm^{\prime}} \ \ \ \ \ (3)$

This set of functions forms the angular part of the eigenfunctions of ${L_{z}}$, which in some cases allows us to determine the probabilities of a system being in a particular eigenstate of ${L_{z}}$. Here are a couple of examples.

Example 1 A particle is described by the wave function

$\displaystyle \psi\left(\rho,\phi\right)=Ae^{-\rho^{2}/2\Delta^{2}}\cos^{2}\phi \ \ \ \ \ (4)$

where ${A}$ is a normalization constant, and ${\Delta}$ is another constant.

We can use the trig identity

$\displaystyle \cos^{2}\phi=\frac{1}{2}\left(1+\cos2\phi\right) \ \ \ \ \ (5)$

to write this wave function as

 $\displaystyle \psi\left(\rho,\phi\right)$ $\displaystyle =$ $\displaystyle \frac{A}{2}e^{-\rho^{2}/2\Delta^{2}}\left[1+\cos2\phi\right]\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{A}{2}e^{-\rho^{2}/2\Delta^{2}}\left(1+\frac{e^{2i\phi}+e^{-2i\phi}}{2}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{A\sqrt{2\pi}}{2}e^{-\rho^{2}/2\Delta^{2}}\left(\Phi_{0}+\frac{1}{2}\left(\Phi_{2}+\Phi_{-2}\right)\right) \ \ \ \ \ (8)$

Thus the wave function has the form

$\displaystyle \psi\left(\rho,\phi\right)=c_{0}\Phi_{0}+c_{2}\Phi_{2}+c_{-2}\Phi_{-2} \ \ \ \ \ (9)$

where the coefficients ${c_{m}}$ can be found by comparison with 8. Since the ${\Phi_{m}}$ are orthonormal functions, the probability of the particle being in state ${i}$ is

$\displaystyle P\left(\ell_{z}=m\hbar\right)=\frac{\left|c_{m}\right|^{2}}{\sum_{j}\left|c_{j}\right|^{2}} \ \ \ \ \ (10)$

We can see from this formula that the factor of ${\frac{A\sqrt{2\pi}}{2}e^{-\rho^{2}/2\Delta^{2}}}$ cancels out of the probability formula, so we have

 $\displaystyle P\left(\ell_{z}=0\right)$ $\displaystyle =$ $\displaystyle \frac{\left|c_{0}\right|^{2}}{\sum_{j}\left|c_{j}\right|^{2}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{1+\frac{1}{4}+\frac{1}{4}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{3}\ \ \ \ \ (13)$ $\displaystyle P\left(\ell_{z}=2\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|c_{2}\right|^{2}}{\sum_{j}\left|c_{j}\right|^{2}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\frac{1}{4}}{1+\frac{1}{4}+\frac{1}{4}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{6}\ \ \ \ \ (16)$ $\displaystyle P\left(\ell_{z}=-2\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|c_{-2}\right|^{2}}{\sum_{j}\left|c_{j}\right|^{2}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\frac{1}{4}}{1+\frac{1}{4}+\frac{1}{4}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{6} \ \ \ \ \ (19)$

Example 2 Now we have the wave function

$\displaystyle \psi\left(\rho,\phi\right)=Ae^{-\rho^{2}/2\Delta^{2}}\left(\frac{\rho}{\Delta}\cos\phi+\sin\phi\right) \ \ \ \ \ (20)$

Again, we write the trig functions in terms of ${\Phi_{m}}$ to get

 $\displaystyle \psi\left(\rho,\phi\right)$ $\displaystyle =$ $\displaystyle Ae^{-\rho^{2}/2\Delta^{2}}\left(\frac{\rho}{\Delta}\frac{e^{i\phi}+e^{-i\phi}}{2}+\frac{e^{i\phi}-e^{-i\phi}}{2i}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\sqrt{2\pi}e^{-\rho^{2}/2\Delta^{2}}\left[\left(\frac{\rho}{2\Delta}+\frac{1}{2i}\right)\Phi_{1}+\left(\frac{\rho}{2\Delta}-\frac{1}{2i}\right)\Phi_{-1}\right] \ \ \ \ \ (22)$

As above, the factor of ${A\sqrt{2\pi}e^{-\rho^{2}/2\Delta^{2}}}$ cancels out when calculating probabilities, so we have

 $\displaystyle P\left(\ell_{z}=\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|c_{1}\right|^{2}}{\left|c_{1}\right|^{2}+\left|c_{-1}\right|^{2}}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|\frac{\rho}{2\Delta}+\frac{1}{2i}\right|^{2}}{\left|\frac{\rho}{2\Delta}+\frac{1}{2i}\right|^{2}+\left|\frac{\rho}{2\Delta}-\frac{1}{2i}\right|^{2}}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(\frac{\rho}{2\Delta}\right)^{2}+\frac{1}{4}}{2\left[\left(\frac{\rho}{2\Delta}\right)^{2}+\frac{1}{4}\right]}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\ \ \ \ \ (26)$ $\displaystyle P\left(\ell_{z}=-\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|c_{-1}\right|^{2}}{\left|c_{1}\right|^{2}+\left|c_{-1}\right|^{2}}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|\frac{\rho}{2\Delta}-\frac{1}{2i}\right|^{2}}{\left|\frac{\rho}{2\Delta}+\frac{1}{2i}\right|^{2}+\left|\frac{\rho}{2\Delta}-\frac{1}{2i}\right|^{2}}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(\frac{\rho}{2\Delta}\right)^{2}+\frac{1}{4}}{2\left[\left(\frac{\rho}{2\Delta}\right)^{2}+\frac{1}{4}\right]}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2} \ \ \ \ \ (30)$

Thus in this case, the ${\rho}$ dependence cancels out when calculating the probabilities, although we can’t expect this to be true in general.

# Eigenvalues of angular momentum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.2.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

One consequence of requiring the angular momentum operator ${L_{z}}$ to be hermitian is that the eigenvalues must be integral multiples of ${\hbar}$, so that ${\ell_{z}=m\hbar}$ for ${m=0,\pm1,\pm2,\ldots}$. Shankar proposes another method by which we might try to obtain this restriction on ${\ell_{z}}$. We start with a superposition of two eigenstates of ${L_{z}}$, so that

 $\displaystyle \psi\left(\rho,\phi\right)$ $\displaystyle =$ $\displaystyle A\left(\rho\right)e^{i\phi\ell_{z}/\hbar}+B\left(\rho\right)e^{i\phi\ell_{z}^{\prime}/\hbar}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{i\phi\ell_{z}^{\prime}/\hbar}\left[A\left(\rho\right)e^{i\phi\left(\ell_{z}-\ell_{z}^{\prime}\right)/\hbar}+B\left(\rho\right)\right] \ \ \ \ \ (2)$

where ${A}$ and ${B}$ are two unknown functions of the radial coordinate ${\rho}$, and ${\ell_{z}}$ and ${\ell_{z}^{\prime}}$ are two eigenvalues of ${L_{z}}$. If we rotate the system by a complete circle, so that ${\phi\rightarrow\phi+2\pi}$, the physical state should remain unchanged. This means that

$\displaystyle \left|\psi\left(\rho,\phi+2\pi\right)\right|=\left|\psi\left(\rho,\phi\right)\right| \ \ \ \ \ (3)$

so that ${\psi\left(\rho,\phi+2\pi\right)}$ may differ from ${\psi\left(\rho,\phi\right)}$ by a phase factor. From 2

$\displaystyle \psi\left(\rho,\phi+2\pi\right)=e^{i\left(\phi+2\pi\right)\ell_{z}^{\prime}/\hbar}\left[A\left(\rho\right)e^{i\left(\phi+2\pi\right)\left(\ell_{z}-\ell_{z}^{\prime}\right)/\hbar}+B\left(\rho\right)\right] \ \ \ \ \ (4)$

The phase factor of ${e^{i\left(\phi+2\pi\right)\ell_{z}^{\prime}/\hbar}}$ on the RHS can be anything (provided the exponent is purely imaginary), but the quantity in the square brackets must be numerically the same as the corresponding quantity in 2. This means that

$\displaystyle \frac{\left(\phi+2\pi\right)\left(\ell_{z}-\ell_{z}^{\prime}\right)}{\hbar}=\frac{\phi\left(\ell_{z}-\ell_{z}^{\prime}\right)}{\hbar}+2m\pi \ \ \ \ \ (5)$

where ${m}$ is an integer. This gives the condition

$\displaystyle \ell_{z}-\ell_{z}^{\prime}=m\hbar \ \ \ \ \ (6)$

To proceed further, we need to argue that ${\ell_{z}}$ is symmetric about zero, that is, if ${\ell_{z}}$ is an eigenvalue, then so is ${-\ell_{z}}$. I’m not sure if Shankar expects us to prove this rigorously, but it seems plausible, since the only difference between ${+\ell_{z}}$ and ${-\ell_{z}}$ is (classically, anyway) that the direction of rotation is reversed. Given this condition, ${\ell_{z}}$ must be a multiple of ${\frac{1}{2}\hbar}$, since any other value doesn’t satisfy both the conditions of symmetry about zero, and 6. (For example, if we try ${\ell_{z}=\frac{1}{4}\hbar}$, then the symmetry requirement means we must also allow ${\ell_{z}=-\frac{1}{4}\hbar}$, but this violates 6.) If ${\ell_{z}}$ is an odd multiple of ${\frac{1}{2}\hbar}$, then we get the sequence ${\ldots,-\frac{3}{2}\hbar}$,-${\frac{1}{2}\hbar,+\frac{1}{2}\hbar,+\frac{3}{2}\hbar,\ldots}$ while if ${\ell_{z}}$ is an even multiple of ${\frac{1}{2}\hbar}$ we get the sequence ${\ldots,-2\hbar,-\hbar,0,+\hbar,+2\hbar,\ldots}$. In reality, only the latter sequence is correct, but we can’t show that from this argument.

# Eigenvalues of two-dimensional angular momentum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.1.

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The angular momentum operator ${L_{z}}$ for rotations in two dimensions has the form, in polar coordinates, of

$\displaystyle L_{z}=-i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (1)$

To find the eigenvalues and eigenfunctions, we need to solve

$\displaystyle L_{z}\left|\ell_{z}\right\rangle =\ell_{z}\left|\ell_{z}\right\rangle \ \ \ \ \ (2)$

where ${\left|\ell_{z}\right\rangle }$ is the eigenfunction and ${\ell_{z}}$ is the corresponding eigenvalue. Using polar coordinates, we must solve

$\displaystyle -i\hbar\frac{\partial}{\partial\phi}\psi_{\ell_{z}}\left(\rho,\phi\right)=\ell_{z}\psi_{\ell_{z}}\left(\rho,\phi\right) \ \ \ \ \ (3)$

where ${\rho}$ is the radial coordinate. As the only derivative here is with respect to ${\phi}$, we can solve this using separation of variables by proposing a solution of form

$\displaystyle \psi_{\ell_{z}}\left(\rho,\phi\right)=R\left(\rho\right)\Phi\left(\phi\right) \ \ \ \ \ (4)$

Substituting this and cancelling off ${R\left(\rho\right)}$ we get

$\displaystyle -i\hbar\frac{\partial}{\partial\phi}\Phi\left(\phi\right)=\ell_{z}\Phi\left(\phi\right) \ \ \ \ \ (5)$

which has the solution

$\displaystyle \Phi\left(\phi\right)=Ae^{i\ell_{z}\phi/\hbar} \ \ \ \ \ (6)$

for some constant ${A}$, which we can absorb into ${R\left(\rho\right)}$ to give the general solution

$\displaystyle \psi_{\ell_{z}}\left(\rho,\phi\right)=R\left(\rho\right)e^{i\ell_{z}\phi/\hbar} \ \ \ \ \ (7)$

[This is actually the two-dimensional version of the more general 3-d case, in which the solution involved a radial function multiplied by a spherical harmonic.]

At this stage, the eigenvalue ${\ell_{z}}$ could be any number, real or complex, since they all satisfy 3. However, since ${L_{z}}$ is an observable, it must be hermitian, which implies that ${L_{z}^{\dagger}=L_{z}}$, so that

$\displaystyle \left\langle \psi_{1}\left|L_{z}\right|\psi_{2}\right\rangle =\left\langle \psi_{2}\left|L_{z}\right|\psi_{1}\right\rangle ^* \ \ \ \ \ (8)$

In the coordinate basis, we have

$\displaystyle \int_{0}^{\infty}\int_{0}^{2\pi}\psi_{1}^*\left(-i\hbar\frac{\partial}{\partial\phi}\right)\psi_{2}d\phi\;d\rho=\left[\int_{0}^{\infty}\int_{0}^{2\pi}\psi_{2}^*\left(-i\hbar\frac{\partial}{\partial\phi}\right)\psi_{1}d\phi\;d\rho\right]^* \ \ \ \ \ (9)$

Integrating the LHS by parts, we have

$\displaystyle \int_{0}^{\infty}\int_{0}^{2\pi}\psi_{1}^*\left(-i\hbar\frac{\partial}{\partial\phi}\right)\psi_{2}d\phi\;d\rho=-i\hbar\int_{0}^{\infty}\left.\psi_{1}^*\psi_{2}\right|_{0}^{2\pi}d\rho+i\hbar\int_{0}^{\infty}\int_{0}^{2\pi}\frac{\partial\psi_{1}^*}{\partial\phi}\psi_{2}d\phi\;d\rho \ \ \ \ \ (10)$

The second term on the RHS is seen to be equal to the RHS of 9, so in order for 9 to be true, we must have

$\displaystyle \int_{0}^{\infty}\left.\psi_{1}^*\psi_{2}\right|_{0}^{2\pi}d\rho=0 \ \ \ \ \ (11)$

Although two different eigenfunctions ${\psi_{1}}$ and ${\psi_{2}}$ are orthogonal and thus would satisfy this condition automatically, the condition must also be true when ${\psi_{1}=\psi_{2}}$. This gives us the condition that

$\displaystyle \psi_{\ell_{z}}\left(2\pi\right)=\psi_{\ell_{z}}\left(0\right) \ \ \ \ \ (12)$

That is, the eigenfunctions must be periodic with period ${2\pi}$. Looking back at 7, we see that this forces the eigenvalues ${\ell_{z}}$ to be integral multiples of ${\hbar}$:

 $\displaystyle \ell_{z}$ $\displaystyle =$ $\displaystyle m\hbar\ \ \ \ \ (13)$ $\displaystyle m$ $\displaystyle =$ $\displaystyle 0,\pm1,\pm2,\ldots \ \ \ \ \ (14)$

Here ${m}$ is the magnetic quantum number, not the mass.

# Combining translations and rotations

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.2.4.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

When it comes to symmetries in quantum mechanics, we’ve looked at translations and rotations in two dimensions, and found that the generators are the momenta ${P_{x}}$ and ${P_{y}}$ for translations, and the angular momentum ${L_{z}}$ for rotations.

From the fact that ${L_{z}}$ does not commute with either momentum or position operators, you might guess that if we performed some sequence of translations and rotations on a system that the order in which these operations are done matters. In fact, you can see this by considering simple two-dimensional geometry, without reference to quantum mechanics. Consider the ${x}$ and ${y}$ axes on a sheet of graph paper. First, translate these axes by adding the vector ${\mathbf{r}}$ to all points, so that the new origin of coordinates lies at position ${\mathbf{r}}$ as referenced in the original coordinates. Next, do a rotation about the original origin by some angle ${\phi}$. This will move the new origin around the original ${z}$ axis. Now, do the inverse of the original translation by adding ${-\mathbf{r}}$ to all points. Finally, do the inverse of the rotation by rotating the system by ${-\phi}$ around the original ${z}$ axis. You’ll find that the ${xy}$ axes that have undergone this sequence of transformations does not coincide with the original ${xy}$ axes. However, if you did the same set of four transformations in the order: translate by ${\mathbf{r}}$, translate by ${-\mathbf{r}}$, rotate by ${\phi}$, rotate by ${-\phi}$, the transformed axes would coincide with the original axes.

To see how this works in quantum mechanics, we can again consider infinitesimal translations and rotations. If we start with a point at location ${\left[x,y\right]}$ and apply the four transformations described above, but now for an infinitesimal translation ${\boldsymbol{\varepsilon}=\varepsilon_{x}\hat{\mathbf{x}}+\varepsilon_{y}\hat{\mathbf{y}}}$ and rotation ${\varepsilon_{z}\hat{\mathbf{z}}}$, then the successive transformations work as follows. In each case, we’ll retain terms up to order ${\varepsilon_{x}\varepsilon_{z}}$ and ${\varepsilon_{y}\varepsilon_{z}}$ but discard terms of order ${\varepsilon_{x}^{2}}$, ${\varepsilon_{y}^{2}}$, ${\varepsilon_{z}^{2}}$ and higher. [I’m not quite sure of the rationale that allows us to do this, apart from the fact that it gives the right answer.]

 $\displaystyle \left[\begin{array}{c} x\\ y \end{array}\right]$ $\displaystyle {\longrightarrow\atop T\left(\boldsymbol{\varepsilon}\right)}$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}\\ y+\varepsilon_{y} \end{array}\right]\ \ \ \ \ (1)$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}\\ y+\varepsilon_{y} \end{array}\right]$ $\displaystyle {\longrightarrow\atop R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)}$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}-\left(y+\varepsilon_{y}\right)\varepsilon_{z}\\ y+\varepsilon_{y}+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]\ \ \ \ \ (2)$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}-\left(y+\varepsilon_{y}\right)\varepsilon_{z}\\ y+\varepsilon_{y}+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]$ $\displaystyle {\longrightarrow\atop T\left(-\boldsymbol{\varepsilon}\right)}$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}-\left(y+\varepsilon_{y}\right)\varepsilon_{z}-\varepsilon_{x}\\ y+\varepsilon_{y}+\left(x+\varepsilon_{x}\right)\varepsilon_{z}-\varepsilon_{y} \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} x-\left(y+\varepsilon_{y}\right)\varepsilon_{z}\\ y+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle \left[\begin{array}{c} x-\left(y+\varepsilon_{y}\right)\varepsilon_{z}\\ y+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]$ $\displaystyle {\longrightarrow\atop R\left(-\varepsilon_{z}\hat{\mathbf{z}}\right)}$ $\displaystyle \left[\begin{array}{c} x-\left(y+\varepsilon_{y}\right)\varepsilon_{z}+\left[y+\left(x+\varepsilon_{x}\right)\varepsilon_{z}\right]\varepsilon_{z}\\ y+\left(x+\varepsilon_{x}\right)\varepsilon_{z}-\left[x-\left(y+\varepsilon_{y}\right)\varepsilon_{z}\right]\varepsilon_{z} \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} x-\varepsilon_{y}\varepsilon_{z}\\ y+\varepsilon_{x}\varepsilon_{z} \end{array}\right] \ \ \ \ \ (6)$

Thus, to this order in the infinitesimals, the combination of translation-rotation-translation-rotation is equivalent to a single translation by a distance ${\left[-\varepsilon_{y}\varepsilon_{z},\varepsilon_{x}\varepsilon_{z}\right]}$. We can write this in terms of the unitary quantum operators for translations and rotations as

$\displaystyle U\left[R\left(-\varepsilon_{z}\hat{\mathbf{z}}\right)\right]T\left(-\boldsymbol{\varepsilon}\right)U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]T\left(\boldsymbol{\varepsilon}\right)=T\left(-\varepsilon_{y}\varepsilon_{z}\hat{\mathbf{x}}+\varepsilon_{x}\varepsilon_{z}\hat{\mathbf{y}}\right) \ \ \ \ \ (7)$

Using the forms of these operators for infinitesimal transformations, we can expand both sides to give

 $\displaystyle \left(I+\frac{i\varepsilon_{z}}{\hbar}L_{z}\right)\left[I+\frac{i}{\hbar}\left(\varepsilon_{x}P_{x}+\varepsilon_{y}P_{y}\right)\right]\times$ $\displaystyle \left(I-\frac{i\varepsilon_{z}}{\hbar}L_{z}\right)\left[I-\frac{i}{\hbar}\left(\varepsilon_{x}P_{x}+\varepsilon_{y}P_{y}\right)\right]$ $\displaystyle = \ \ \ \ \ (8)$ $\displaystyle I-\frac{i}{\hbar}\left(-\varepsilon_{y}\varepsilon_{z}P_{x}+\varepsilon_{x}\varepsilon_{z}P_{y}\right)$

Since the infinitesimal displacements are arbitrary, this equation can be valid only if the coefficients of each combination of ${\varepsilon_{x},\varepsilon_{y}}$ and ${\varepsilon_{z}}$ are equal on both sides. As above, we’ll discard any terms of order ${\varepsilon_{x}^{2}}$, ${\varepsilon_{y}^{2}}$, ${\varepsilon_{z}^{2}}$ and higher. The algebra is straightforward although a bit tedious, so I’ll just give a couple of examples here.

The coefficient of ${\varepsilon_{z}}$ on its own is, on the LHS

$\displaystyle \frac{i\varepsilon_{z}}{\hbar}L_{z}-\frac{i\varepsilon_{z}}{\hbar}L_{z}=0 \ \ \ \ \ (9)$

On the RHS, there is no term in ${\varepsilon_{z}}$, so we get 0 on the RHS. In this case, we see the equation is consistent.

For the ${\varepsilon_{x}\varepsilon_{z}}$ term, we get on the LHS:

$\displaystyle \varepsilon_{x}\varepsilon_{z}\frac{i^{2}}{\hbar^{2}}\left(L_{z}P_{x}-L_{z}P_{x}-P_{x}L_{z}+L_{z}P_{x}\right)=-\varepsilon_{x}\varepsilon_{z}\frac{i^{2}}{\hbar^{2}}\left[P_{x},L_{z}\right] \ \ \ \ \ (10)$

On the RHS, the term is

$\displaystyle -\frac{i}{\hbar}\varepsilon_{x}\varepsilon_{z}P_{y} \ \ \ \ \ (11)$

Thus the condition here becomes

$\displaystyle \left[P_{x},L_{z}\right]=-i\hbar P_{y} \ \ \ \ \ (12)$

which agrees with the commutation relation we found earlier. By considering the coefficient of ${\varepsilon_{y}\varepsilon_{z}}$, we arrive at the other condition, which is

$\displaystyle \left[P_{y},L_{z}\right]=i\hbar P_{x} \ \ \ \ \ (13)$

The result of this calculation doesn’t tell us anything new about the translation or rotation operators, but it does show that the condition 7 is consistent with what we already know about the commutators of position, momentum and angular momentum.

As Shankar points out, we might think that we need to verify the conditions for an infinite number of combinations of rotations and translations, since each such combination gives rise to a different overall transformation. He says that it has actually been shown that the example above is sufficient to guarantee that all such combinations do in fact give valid results, although he doesn’t give the details. We are, however, given the exercise of verifying this claim for one special case, which we’ll consider now.

In this example, we’ll consider the same four transformations, in the same order, as above except that we’ll take the translation to be entirely in the ${x}$ direction so that ${\varepsilon_{y}=0}$. This time, we’ll retain terms up to ${\varepsilon_{x}\varepsilon_{z}^{2}}$ and see what we get. We start by repeating the calculations in 1 through 6. However, because we’re saving higher order terms, we need to represent the infinitesimal rotations by

$\displaystyle R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)=\left[\begin{array}{cc} 1-\frac{\varepsilon_{z}^{2}}{2} & -\varepsilon_{z}\\ \varepsilon_{z} & 1-\frac{\varepsilon_{z}^{2}}{2} \end{array}\right] \ \ \ \ \ (14)$

That is, we’re approximating ${\cos\varepsilon_{z}}$ by the first two terms in its expansion. Using this, we have

 $\displaystyle \left[\begin{array}{c} x\\ y \end{array}\right]$ $\displaystyle {\longrightarrow\atop T\left(\boldsymbol{\varepsilon}\right)}$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}\\ y \end{array}\right]\ \ \ \ \ (15)$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}\\ y \end{array}\right]$ $\displaystyle {\longrightarrow\atop R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)}$ $\displaystyle \left[\begin{array}{c} \left(x+\varepsilon_{x}\right)\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-y\varepsilon_{z}\\ y\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]\ \ \ \ \ (16)$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}-y\varepsilon_{z}\\ y+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]$ $\displaystyle {\longrightarrow\atop T\left(-\boldsymbol{\varepsilon}\right)}$ $\displaystyle \left[\begin{array}{c} \left(x+\varepsilon_{x}\right)\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-y\varepsilon_{z}-\varepsilon_{x}\\ y\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} x\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-y\varepsilon_{z}-\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\\ y\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\varepsilon_{z}x+\varepsilon_{x}\varepsilon_{z} \end{array}\right]\ \ \ \ \ (18)$ $\displaystyle \left[\begin{array}{c} x-y\varepsilon_{z}\\ y+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]$ $\displaystyle {\longrightarrow\atop R\left(-\varepsilon_{z}\hat{\mathbf{z}}\right)}$ $\displaystyle \left[\begin{array}{c} \left[x\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-y\varepsilon_{z}-\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\right]\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\left[y\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\varepsilon_{z}x+\varepsilon_{x}\varepsilon_{z}\right]\varepsilon_{z}\\ \left[y\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\varepsilon_{z}x+\varepsilon_{x}\varepsilon_{z}\right]\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-\left[x\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-y\varepsilon_{z}-\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\right]\varepsilon_{z} \end{array}\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} x\left(1+\frac{\varepsilon_{z}^{4}}{4}\right)+\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}+\frac{1}{4}\varepsilon_{x}\varepsilon_{z}^{4}\\ y\left(1+\frac{\varepsilon_{z}^{4}}{4}\right)+\varepsilon_{x}\varepsilon_{z} \end{array}\right] \ \ \ \ \ (20)$

To get the last line, I used Maple to do the algebra in multiplying out the terms. At this point, we can neglect the terms in ${\varepsilon_{z}^{4}}$, leaving us with the overall transformation:

$\displaystyle \left[\begin{array}{c} x\\ y \end{array}\right]\longrightarrow\left[\begin{array}{c} x+\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\\ y+\varepsilon_{x}\varepsilon_{z} \end{array}\right] \ \ \ \ \ (21)$

This is equivalent to a translation by ${\boldsymbol{\varepsilon}=\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\hat{\mathbf{x}}+\varepsilon_{x}\varepsilon_{z}\hat{\mathbf{y}}}$, so by analogy with 7, we have the condition

$\displaystyle U\left[R\left(-\varepsilon_{z}\hat{\mathbf{z}}\right)\right]T\left(-\boldsymbol{\varepsilon}\right)U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]T\left(\boldsymbol{\varepsilon}\right)=T\left(\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\hat{\mathbf{x}}+\varepsilon_{x}\varepsilon_{z}\hat{\mathbf{y}}\right) \ \ \ \ \ (22)$

To expand the operators on the LHS and retain terms up to ${\varepsilon_{x}\varepsilon_{z}^{2}}$, we need to expand the rotation operators up to order ${\varepsilon_{z}^{2}}$. Treating the rotation operator as an exponential, this expansion is

$\displaystyle R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)=I-\frac{i\varepsilon_{z}}{\hbar}L_{z}+\frac{i^{2}\varepsilon_{z}^{2}}{2\hbar^{2}}L_{z}^{2}+\ldots \ \ \ \ \ (23)$

Using this approximation gives us

 $\displaystyle \left(I+\frac{i\varepsilon_{z}}{\hbar}L_{z}+\frac{i^{2}\varepsilon_{z}^{2}}{2\hbar^{2}}L_{z}^{2}\right)\left[I+\frac{i}{\hbar}\varepsilon_{x}P_{x}\right]\left(I-\frac{i\varepsilon_{z}}{\hbar}L_{z}+\frac{i^{2}\varepsilon_{z}^{2}}{2\hbar^{2}}L_{z}^{2}\right)\left[I-\frac{i}{\hbar}\varepsilon_{x}P_{x}\right]$ $\displaystyle =$ $\displaystyle I-\frac{i}{\hbar}\left(\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}P_{x}+\varepsilon_{x}\varepsilon_{z}P_{y}\right) \ \ \ \ \ (24)$

By equating the coefficients of ${\varepsilon_{x}\varepsilon_{z}}$ we regain 12, so that condition checks out.

Extracting the coefficient of ${\varepsilon_{x}\varepsilon_{z}^{2}}$ on the LHS gives

 $\displaystyle \frac{i^{3}}{\hbar^{3}}\varepsilon_{x}\varepsilon_{z}^{2}\left(-L_{z}P_{x}L_{z}+\frac{L_{z}^{2}P_{x}}{2}-\frac{L_{z}^{2}P_{x}}{2}+\frac{P_{x}L_{z}^{2}}{2}-\frac{L_{z}^{2}P_{x}}{2}+L_{z}^{2}P_{x}\right)$ $\displaystyle =$ $\displaystyle \frac{i^{3}}{\hbar^{3}}\varepsilon_{x}\varepsilon_{z}^{2}\left(-L_{z}P_{x}L_{z}+\frac{L_{z}^{2}P_{x}}{2}+\frac{P_{x}L_{z}^{2}}{2}\right) \ \ \ \ \ (25)$

Matching this to the ${\varepsilon_{x}\varepsilon_{z}^{2}}$ term on the RHS of 24, we get the condition specified in Shankar’s problem:

$\displaystyle -2L_{z}P_{x}L_{z}+L_{z}^{2}P_{x}+P_{x}L_{z}^{2}=\hbar^{2}P_{x} \ \ \ \ \ (26)$

We can show that this condition reduces to the already-known commutators by using the identity

 $\displaystyle \left[\Lambda,\left[\Lambda,\Omega\right]\right]$ $\displaystyle =$ $\displaystyle \Lambda\left(\Lambda\Omega-\Omega\Lambda\right)-\left(\Lambda\Omega-\Omega\Lambda\right)\Lambda\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2\Lambda\Omega\Lambda+\Lambda^{2}\Omega+\Omega\Lambda^{2} \ \ \ \ \ (28)$

Applying this to 26 we have

 $\displaystyle -2L_{z}P_{x}L_{z}+L_{z}^{2}P_{x}+P_{x}L_{z}^{2}$ $\displaystyle =$ $\displaystyle \left[L_{z},\left[L_{z},P_{x}\right]\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[L_{z},P_{y}\right]\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(-i\hbar P_{x}\right)\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar^{2}P_{x} \ \ \ \ \ (32)$

Thus the more complicated condition 26 actually reduces to existing commutators.

# Rotations through a finite angle; use of polar coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.2.3.

The angluar momentum operator ${L_{z}}$ is the generator of rotations in the ${xy}$ plane. We did the derivation for infinitesimal rotations, but we can generalize this to finite rotations in a similar manner to that used for translations. The unitary transformation for an infinitesimal rotation is

$\displaystyle U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]=I-\frac{i\varepsilon_{z}L_{z}}{\hbar} \ \ \ \ \ (1)$

For rotation through a finite angle ${\phi_{0}}$, we divide up the angle into ${N}$ small angles, so ${\varepsilon_{z}=\phi_{0}/N}$. Rotation through the full angle ${\phi_{0}}$ is then given by

$\displaystyle U\left[R\left(\phi_{0}\hat{\mathbf{z}}\right)\right]=\lim_{N\rightarrow\infty}\left(I-\frac{i\phi_{0}L_{z}}{N\hbar}\right)^{N}=e^{-i\phi_{0}L_{z}/\hbar} \ \ \ \ \ (2)$

The limit follows because the only non-trivial operator involved is ${L_{z}}$, so no commutation problems arise.

In rectangular coordinates, ${L_{z}}$ has the relatively non-obvious form

 $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle XP_{y}-YP_{x}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right) \ \ \ \ \ (4)$

so it’s not immediately clear that 2 does in fact lead to the desired rotation. Trying to calculate the exponential with ${L_{z}}$ expressed this way is not easy, given that the two terms ${x\frac{\partial}{\partial y}}$ and ${y\frac{\partial}{\partial x}}$ don’t commute.

It turns out that ${L_{z}}$ has a much simpler form in polar coordinates, and there are two ways of converting it to polar form. First, we recall the transformation equations.

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \rho\cos\phi\ \ \ \ \ (5)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle \rho\sin\phi\ \ \ \ \ (6)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{x^{2}+y^{2}}\ \ \ \ \ (7)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \tan^{-1}\frac{y}{x} \ \ \ \ \ (8)$

From the chain rule, we can convert the derivatives:

 $\displaystyle \frac{\partial}{\partial x}$ $\displaystyle =$ $\displaystyle \frac{\partial\rho}{\partial x}\frac{\partial}{\partial\rho}+\frac{\partial\cos\phi}{\partial x}\frac{\partial}{\partial\left(\cos\phi\right)}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial\rho}{\partial x}\frac{\partial}{\partial\rho}-\sin\phi\frac{\partial\phi}{\partial x}\frac{\partial}{\left(-\sin\phi\right)\partial\phi}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{x}{\rho}\frac{\partial}{\partial\rho}-\sin\phi\frac{-y/x^{2}}{1+y^{2}/x^{2}}\left(\frac{-1}{\sin\phi}\right)\frac{\partial}{\partial\phi}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{x}{\rho}\frac{\partial}{\partial\rho}-\frac{y}{\rho^{2}}\frac{\partial}{\partial\phi} \ \ \ \ \ (12)$

Using similar methods, we get for the other derivative

 $\displaystyle \frac{\partial}{\partial y}$ $\displaystyle =$ $\displaystyle \frac{\partial\rho}{\partial y}\frac{\partial}{\partial\rho}+\frac{\partial\sin\phi}{\partial x}\frac{\partial}{\partial\left(\sin\phi\right)}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{y}{\rho}\frac{\partial}{\partial\rho}+\frac{x}{\rho^{2}}\frac{\partial}{\partial\phi} \ \ \ \ \ (14)$

Plugging these into 4 we have

 $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle -i\hbar\left[x\left(\frac{y}{\rho}\frac{\partial}{\partial\rho}+\frac{x}{\rho^{2}}\frac{\partial}{\partial\phi}\right)-y\left(\frac{x}{\rho}\frac{\partial}{\partial\rho}-\frac{y}{\rho^{2}}\frac{\partial}{\partial\phi}\right)\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\frac{x^{2}+y^{2}}{\rho^{2}}\frac{\partial}{\partial\phi}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (17)$

Another method of converting ${L_{z}}$ to polar coordinates is to consider the effect of ${U\left[R\right]}$ for an infinitesimal rotation ${\varepsilon_{z}}$ on a state vector expressed in polar coordinates ${\psi\left(\rho,\phi\right)}$. Shankar states that

$\displaystyle \left\langle \rho,\phi\left|U\left[R\right]\right|\psi\left(\rho,\phi\right)\right\rangle =\psi\left(\rho,\phi-\varepsilon_{z}\right) \ \ \ \ \ (18)$

If you don’t believe this, it can be shown using a method similar to that for the one-dimensional translation. In this case, we’re dealing with position eigenkets in polar coordinates, so we have

$\displaystyle U\left[R\right]\left|\rho,\phi\right\rangle =\left|\rho,\phi+\varepsilon_{z}\right\rangle \ \ \ \ \ (19)$

Applying this, we get

 $\displaystyle \left|\psi_{\varepsilon_{z}}\right\rangle$ $\displaystyle =$ $\displaystyle U\left[R\right]\left|\psi\right\rangle \ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U\left[R\right]\int_{0}^{2\pi}\int_{0}^{\infty}\left|\rho,\phi\right\rangle \left\langle \rho,\phi\left|\psi\right.\right\rangle \rho d\rho\;d\phi\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\infty}\left|\rho,\phi+\varepsilon_{z}\right\rangle \left\langle \rho,\phi\left|\psi\right.\right\rangle \rho d\rho\;d\phi\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\infty}\left|\rho^{\prime},\phi^{\prime}\right\rangle \left\langle \rho^{\prime},\phi^{\prime}-\varepsilon_{z}\left|\psi\right.\right\rangle \rho^{\prime}d\rho^{\prime}\;d\phi^{\prime} \ \ \ \ \ (23)$

where in the last line, we used the substitution ${\phi^{\prime}=\phi+\varepsilon_{z}}$. (The substitution ${\rho^{\prime}=\rho}$ is used just to give the radial variable a different name in the integrand.) We can use the same limits of integration for ${\phi}$ and ${\phi^{\prime}}$, since we just need to ensure that the integral covers the total range of angles. It then follows that

 $\displaystyle \left\langle \rho,\phi\left|\psi_{\varepsilon_{z}}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\infty}\left\langle \rho,\phi\left|\rho^{\prime},\phi^{\prime}\right.\right\rangle \left\langle \rho^{\prime},\phi^{\prime}-\varepsilon_{z}\left|\psi\right.\right\rangle \rho^{\prime}d\rho^{\prime}\;d\phi^{\prime}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\infty}\delta\left(\rho-\rho^{\prime}\right)\delta\left(\phi-\phi^{\prime}\right)\left\langle \rho^{\prime},\phi^{\prime}-\varepsilon_{z}\left|\psi\right.\right\rangle \rho^{\prime}d\rho^{\prime}\;d\phi^{\prime}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(\rho,\phi-\varepsilon_{z}\right) \ \ \ \ \ (26)$

Combining this with 1 we have

$\displaystyle \left\langle \rho,\phi\left|I-\frac{i\varepsilon_{z}L_{z}}{\hbar}\right|\psi\right\rangle =\psi\left(\rho,\phi-\varepsilon_{z}\right) \ \ \ \ \ (27)$

Expanding the RHS to order ${\varepsilon_{z}}$ we have

$\displaystyle \left\langle \rho,\phi\left|I-\frac{i\varepsilon_{z}L_{z}}{\hbar}\right|\psi\right\rangle =\psi\left(\rho,\phi\right)-\varepsilon_{z}\frac{\partial\psi}{\partial\phi} \ \ \ \ \ (28)$

from which 17 follows again.

Once we have ${L_{z}}$ in this form, the exponential form of a finite rotation is easier to interpret, for we have, from 2

 $\displaystyle e^{-i\phi_{0}L_{z}/\hbar}$ $\displaystyle =$ $\displaystyle \exp\left[-\phi_{0}\frac{\partial}{\partial\phi}\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1-\phi_{0}\frac{\partial}{\partial\phi}+\frac{\phi_{0}^{2}}{2!}\frac{\partial^{2}}{\partial\phi^{2}}+\ldots \ \ \ \ \ (30)$

Applying this to a state function ${\psi\left(\rho,\phi\right)}$, we see that we get the Taylor series for ${\psi\left(\rho,\phi-\phi_{0}\right)}$, so the exponential does indeed represent a rotation through a finite angle.

# Rotational transformations using passive transformations

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.2.2.

We can also derive the generator of rotations ${L_{z}}$ by considering passive transformations of the position and momentum operators, in a way similar to that used for deriving the generator of translations. In a passive transformation, the operators are modified while the state vectors remain the same. For an infinitesimal rotation ${\varepsilon_{z}\hat{\mathbf{z}}}$ about the ${z}$ axis in two dimensions, the unitary operator has the form

$\displaystyle U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]=I-\frac{i\varepsilon_{z}L_{z}}{\hbar} \ \ \ \ \ (1)$

For a finite rotation by ${\phi_{0}\hat{\mathbf{z}}}$ the transformations are given by

 $\displaystyle \left\langle X\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle X\right\rangle \cos\phi_{0}-\left\langle Y\right\rangle \sin\phi_{0}\ \ \ \ \ (2)$ $\displaystyle \left\langle Y\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle X\right\rangle \sin\phi_{0}+\left\langle Y\right\rangle \cos\phi_{0}\ \ \ \ \ (3)$ $\displaystyle \left\langle P_{x}\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle P_{x}\right\rangle \cos\phi_{0}-\left\langle P_{y}\right\rangle \sin\phi_{0}\ \ \ \ \ (4)$ $\displaystyle \left\langle P_{y}\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle P_{x}\right\rangle \sin\phi_{0}+\left\langle P_{y}\right\rangle \cos\phi_{0} \ \ \ \ \ (5)$

For the infinitesimal transformation, ${\phi_{0}=\varepsilon_{z}}$ and these equations reduce to

 $\displaystyle \left\langle X\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle X\right\rangle -\left\langle Y\right\rangle \varepsilon_{z}\ \ \ \ \ (6)$ $\displaystyle \left\langle Y\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle X\right\rangle \varepsilon_{z}+\left\langle Y\right\rangle \ \ \ \ \ (7)$ $\displaystyle \left\langle P_{x}\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle P_{x}\right\rangle -\left\langle P_{y}\right\rangle \varepsilon_{z}\ \ \ \ \ (8)$ $\displaystyle \left\langle P_{y}\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle P_{x}\right\rangle \varepsilon_{z}+\left\langle P_{y}\right\rangle \ \ \ \ \ (9)$

In the passive transformation scheme, we move the transformation to the operators to get

 $\displaystyle U^{\dagger}\left[R\right]XU\left[R\right]$ $\displaystyle =$ $\displaystyle X-Y\varepsilon_{z}\ \ \ \ \ (10)$ $\displaystyle U^{\dagger}\left[R\right]YU\left[R\right]$ $\displaystyle =$ $\displaystyle X\varepsilon_{z}+Y\ \ \ \ \ (11)$ $\displaystyle U^{\dagger}\left[R\right]P_{x}U\left[R\right]$ $\displaystyle =$ $\displaystyle P_{x}-P_{y}\varepsilon_{z}\ \ \ \ \ (12)$ $\displaystyle U^{\dagger}\left[R\right]P_{y}U\left[R\right]$ $\displaystyle =$ $\displaystyle P_{x}\varepsilon_{z}+P_{y} \ \ \ \ \ (13)$

Substituting 1 into these equations gives us the commutation relations satisfied by ${L_{z}}$. For example, in the first equation we have

 $\displaystyle U^{\dagger}\left[R\right]XU\left[R\right]$ $\displaystyle =$ $\displaystyle \left(I+\frac{i\varepsilon_{z}L_{z}}{\hbar}\right)X\left(I-\frac{i\varepsilon_{z}L_{z}}{\hbar}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle X+\frac{i\varepsilon_{z}}{\hbar}\left(L_{z}X-XL_{z}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle X-Y\varepsilon_{z} \ \ \ \ \ (16)$

Equating the last two lines, we get

$\displaystyle \left[X,L_{z}\right]=-i\hbar Y \ \ \ \ \ (17)$

Similarly, for the other three equations we get

 $\displaystyle \left[Y,L_{z}\right]$ $\displaystyle =$ $\displaystyle i\hbar X\ \ \ \ \ (18)$ $\displaystyle \left[P_{x},L_{z}\right]$ $\displaystyle =$ $\displaystyle -i\hbar P_{y}\ \ \ \ \ (19)$ $\displaystyle \left[P_{y},L_{z}\right]$ $\displaystyle =$ $\displaystyle i\hbar P_{x} \ \ \ \ \ (20)$

We can use these commutation relations to derive the form of ${L_{z}}$ by using the commutation relations for coordinates and momenta:

$\displaystyle \left[X,P_{x}\right]=\left[Y,P_{y}\right]=i\hbar \ \ \ \ \ (21)$

with all other commutators involving ${X,Y,P_{x}}$ and ${P_{y}}$ being zero. Starting with 17, we see that

$\displaystyle \left[X,L_{z}\right]=-\left[X,P_{x}\right]Y \ \ \ \ \ (22)$

We can therefore deduce that

$\displaystyle L_{z}=-P_{x}Y+f\left(X,Y,P_{y}\right) \ \ \ \ \ (23)$

where ${f}$ is some unknown function. We must include ${f}$ since the commutators of ${X}$ with ${X,Y}$ and ${P_{y}}$ are all zero, so adding on ${f}$ still satisfies 17. (You can think of it as similar to adding on the constant in an indefinite integral.)

Now from 18, we have

$\displaystyle \left[Y,L_{z}\right]=\left[Y,P_{y}\right]X \ \ \ \ \ (24)$

so combining this with 23 we have

$\displaystyle L_{z}=-P_{x}Y+P_{y}X+g\left(X,Y\right) \ \ \ \ \ (25)$

The undetermined function is now a function only of ${X}$ and ${Y}$, since the dependence of ${L_{z}}$ on ${P_{x}}$ and ${P_{y}}$ has been determined uniquely by the commutators 17 and 18.

From 19 we have

$\displaystyle \left[P_{x},L_{z}\right]=\left[P_{x},X\right]P_{y} \ \ \ \ \ (26)$

We can see that this is satisfied already by 25, except that we now know that the function ${g}$ cannot depend on ${X}$, since then ${\left[P_{x},g\right]\ne0}$. Thus we have narrowed down ${L_{z}}$ to

$\displaystyle L_{z}=-P_{x}Y+P_{y}X+h\left(Y\right) \ \ \ \ \ (27)$

Finally, from 20 we have

$\displaystyle \left[P_{y},L_{z}\right]=-\left[P_{y},Y\right]P_{x} \ \ \ \ \ (28)$

This is satisfied by 27 if we take ${h=0}$ (well, technically, we could take ${h}$ to be some constant, but we might as well take the constant to be zero), giving us the final form for ${L_{z}}$:

$\displaystyle L_{z}=-P_{x}Y+P_{y}X \ \ \ \ \ (29)$