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# Hydrogen atom – radial function at large r

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.1.3.

[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 solving the 3-d Schrödinger equation for a spherically symmetric potential, the radial function has the asymptotic form for large ${r}$, and for the energy ${E<0}$:

$\displaystyle U_{El}\left(r\right)\underset{r\rightarrow\infty}{\longrightarrow}Ar^{\pm me^{2}/\kappa\hbar^{2}}e^{-\kappa r} \ \ \ \ \ (1)$

where

$\displaystyle \kappa\equiv\sqrt{\frac{2m\left|E\right|}{\hbar^{2}}} \ \ \ \ \ (2)$

For the hydrogen atom, the function ${U_{El}}$ is obtained from a series solution of the differential equation with the result

 $\displaystyle U_{El}$ $\displaystyle =$ $\displaystyle e^{-\rho}v_{El}\ \ \ \ \ (3)$ $\displaystyle v_{El}$ $\displaystyle =$ $\displaystyle \rho^{l+1}\sum_{k=0}^{\infty}C_{k}\rho^{k}\ \ \ \ \ (4)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{\frac{-2mE}{\hbar^{2}}}r\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \kappa r \ \ \ \ \ (6)$

To keep the wave function finite at large ${r}$, we require the series to terminate, which leads to the quantized energy levels, given by

$\displaystyle E_{n}=-\frac{me^{4}}{2\hbar^{2}n^{2}} \ \ \ \ \ (7)$

The series in 4 terminates at a value of ${k=n-l-1}$, so the function ${v_{El}}$ is a polynomial in ${\rho}$, and thus in ${r}$, of degree ${n}$. Since the actual radial function is

$\displaystyle R_{nl}=\frac{U_{El}}{r} \ \ \ \ \ (8)$

we have that ${R_{nl}}$ is a polynomial of degree ${n-1}$ in ${r}$ multiplied by the exponential ${e^{-\rho}=e^{-\kappa r}}$. That is, for large ${r}$

$\displaystyle R_{nl}\sim r^{n-1}e^{-\kappa r} \ \ \ \ \ (9)$

To show that this is consistent with 1, we use 7 and 2.

 $\displaystyle n$ $\displaystyle =$ $\displaystyle \sqrt{\frac{me^{4}}{2\hbar^{2}\left|E_{n}\right|}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{me^{4}}{2\hbar^{2}}}\sqrt{\frac{2m}{\hbar^{2}}}\frac{1}{\kappa}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{me^{2}}{\kappa\hbar^{2}} \ \ \ \ \ (12)$

Comparing this with 1, we see that

$\displaystyle r^{n}=r^{me^{2}/\kappa\hbar^{2}} \ \ \ \ \ (13)$

so the condition is satisfied.

# Radial function for large r

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Section 12.6.

[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 solving the Schrödinger equation for spherically symmetric potentials, we found that we could reduce the problem to the equation

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (1)$

where ${U_{El}\left(r\right)}$ is related to the radial function by

$\displaystyle R_{El}\left(r\right)=\frac{U_{El}\left(r\right)}{r} \ \ \ \ \ (2)$

We’ve looked at some properties of ${U_{El}}$ (which Griffiths calls ${u}$) for the hydrogen atom, but we can also try to extract some information about ${U_{El}}$ in the more general case where we don’t need to specify the potential ${V}$ precisely. Here we’ll examine what happens as ${r\rightarrow\infty}$.

By looking at 1, we can see that the centrifugal barrier term (the last term in the square brackets) disappears for large ${r}$, so the behaviour is determined by the nature of the potential ${V}$. We might think that, provided ${V\underset{r\rightarrow\infty}{\longrightarrow}0}$, we can just ignore the potential and solve the reduced equation

$\displaystyle \frac{d^{2}U_{E}}{dr^{2}}=-\frac{2\mu E}{\hbar^{2}}U_{E} \ \ \ \ \ (3)$

where we’ve dropped the subscript ${l}$ since we’re ignoring the centrifugal barrier, which is the only term in which ${l}$ appears. In fact, this assumption proves to be faulty, in that the analysis is valid only if ${V\rightarrow\frac{1}{r^{a}}}$ where ${a>1}$, or in other words, if ${rV\left(r\right)\rightarrow0}$. To see why, we need to consider two cases: ${E>0}$ (so that the particle can escape to infinity, since we’re assuming ${V\le0}$ everywhere, and thus that ${E}$ can take on any positive value); ${E<0}$, so that the particle is bound, and there are discrete energy levels. Shakar treats the ${E>0}$ case so we’ll look at the ${E<0}$ case. In this case, 3 has the general solution

$\displaystyle U_{E}=Ae^{-\kappa r}+Be^{\kappa r} \ \ \ \ \ (4)$

where

$\displaystyle \kappa=\sqrt{-\frac{2\mu E}{\hbar^{2}}}=\sqrt{\frac{2\mu\left|E\right|}{\hbar^{2}}} \ \ \ \ \ (5)$

In the most general case, the constants ${A}$ and ${B}$ can be anything, subject to the usual constraint that the overall wave function is normalized. However, in order for this normalization to occur, we can’t have the ${e^{\kappa r}}$ term, since that terms blows up as ${r\rightarrow\infty}$. As we’ve seen in the specific example of the hydrogen atom, when we express the radial function as a series in powers of ${r}$, the series must terminate after a finite number of terms in order to keep the wave function finite, and it is this that results in the quantized energy levels. Although a direct link between the series solution and the form 4 isn’t obvious, the net effect is that, when the energy has one of the allowed discrete values, the term ${Be^{\kappa r}}$ disappears from the asymptotic solution.

The form 4 is valid only under the restriction that ${rV\left(r\right)\rightarrow0}$ for large ${r}$. To see why, suppose we write

$\displaystyle U_{E}=f\left(r\right)e^{\pm\kappa r} \ \ \ \ \ (6)$

for some function ${f}$. If 4 is valid, then ${f}$ should tend to a constant for large ${r}$. We can plug 6 into 1 and ignore the centrifugal term since we’re looking only at large ${r}$. This gives

$\displaystyle \frac{d^{2}U_{E}}{dr^{2}}-\frac{2\mu}{\hbar^{2}}VU_{E}-\kappa^{2}U_{E}=0 \ \ \ \ \ (7)$

Calculating the derivative, we have

 $\displaystyle \frac{dU_{E}}{dr}$ $\displaystyle =$ $\displaystyle \left(f^{\prime}\pm\kappa f\right)e^{\pm\kappa r}\ \ \ \ \ (8)$ $\displaystyle \frac{d^{2}U_{E}}{dr^{2}}$ $\displaystyle =$ $\displaystyle \left(f^{\prime\prime}\pm\kappa f^{\prime}\pm\kappa f^{\prime}+\kappa^{2}f\right)e^{\pm\kappa r}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(f^{\prime\prime}\pm2\kappa f^{\prime}+\kappa^{2}f\right)e^{\pm\kappa r} \ \ \ \ \ (10)$

Plugging this into 7 we get

$\displaystyle f^{\prime\prime}\pm2\kappa f^{\prime}-\frac{2\mu}{\hbar^{2}}Vf=0 \ \ \ \ \ (11)$

At this point, Shankar assumes that ${f}$ is slowly varying for large ${r}$, which seems reasonable, so we can disregard the second derivative, to get

$\displaystyle f^{\prime}=\mp\frac{\mu}{\kappa\hbar^{2}}Vf \ \ \ \ \ (12)$

or

$\displaystyle \frac{df}{f}=\mp\frac{\mu}{\kappa\hbar^{2}}V\left(r\right)\;dr \ \ \ \ \ (13)$

If we integrate this from some constant lower value ${r_{0}}$ up to an arbitrary large value ${r}$, we have

$\displaystyle f\left(r\right)=f\left(r_{0}\right)\exp\left[\mp\frac{\mu}{\kappa\hbar^{2}}\int_{r_{0}}^{r}V\left(r^{\prime}\right)dr^{\prime}\right] \ \ \ \ \ (14)$

The point now is that if ${V\left(r\right)\rightarrow\frac{1}{r^{a}}}$ with ${a>1}$, then the integral of ${V}$ will be an inverse power of ${r}$, and thus will go to zero as ${r\rightarrow\infty}$. In that case, the RHS of 14 does indeed tend to a constant as ${r\rightarrow\infty}$, and the asymptotic solution 4 is valid. However, if ${V=-\frac{e^{2}}{r}}$ (the Coulomb potential, as found in the hydrogen atom), then the integral of ${V}$ is a logarithm and does not tend to zero for large ${r}$. In this case, we get

 $\displaystyle f\left(r\right)$ $\displaystyle =$ $\displaystyle f\left(r_{0}\right)\exp\left[\pm\frac{\mu e^{2}}{\kappa\hbar^{2}}\ln\frac{r}{r_{0}}\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[r_{0}^{\mp\mu e^{2}/\kappa\hbar^{2}}f\left(r_{0}\right)\right]r^{\pm\mu e^{2}/\kappa\hbar^{2}} \ \ \ \ \ (16)$

The quantity in square brackets is a constant, but the last factor is a power of ${r}$ which, for the positive exponent, continues to grow as ${r\rightarrow\infty}$. Thus the asymptotic solution 4 is valid only for potentials that fall off faster than ${\frac{1}{r}}$ for large ${r}$.

# Radial function for small r

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Section 12.6.

[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 solving the Schrödinger equation for spherically symmetric potentials, we found that we could reduce the problem to the equation

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (1)$

where ${U_{El}\left(r\right)}$ is related to the radial function by

$\displaystyle R_{El}\left(r\right)=\frac{U_{El}\left(r\right)}{r} \ \ \ \ \ (2)$

We’ve looked at some properties of ${U_{El}}$ (which Griffiths calls ${u}$) for the hydrogen atom, but we can also try to extract some information about ${U_{El}}$ in the more general case where we don’t need to specify the potential ${V}$ precisely. Here we’ll examine what happens as ${r\rightarrow0}$.

The quantity in the square brackets in 1 is an operator which will call ${D_{l}\left(r\right)}$:

$\displaystyle D_{l}\left(r\right)\equiv-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}} \ \ \ \ \ (3)$

If we require ${D_{L}}$ to be hermitian, this results in the condition that, for two functions ${U_{1}}$ and ${U_{2}}$,

$\displaystyle \left.U_{1}^*\frac{dU_{2}}{dr}\right|_{0}^{\infty}-\left.U_{2}\frac{dU_{1}^*}{dr}\right|_{0}^{\infty}=0 \ \ \ \ \ (4)$

If we require ${U_{El}}$ to be normalizable then it must satisfy either

$\displaystyle U_{El}\underset{r\rightarrow\infty}{\longrightarrow}0 \ \ \ \ \ (5)$

which is valid for bound states where ${E>V}$ as ${r\rightarrow\infty}$, or

$\displaystyle U_{El}\underset{r\rightarrow\infty}{\longrightarrow}e^{ikr} \ \ \ \ \ (6)$

where

$\displaystyle k=\sqrt{\frac{2\mu E}{\hbar^{2}}} \ \ \ \ \ (7)$

if ${E>V}$ as ${r\rightarrow\infty}$. In the latter case, we’re using the definition of normalization for an oscillating function. If ${U_{El}\underset{r\rightarrow\infty}{\longrightarrow}0}$ then 4 is 0 at the upper limit. Using the normalization condition for oscillating functions, if ${U_{El}\underset{r\rightarrow\infty}{\longrightarrow}e^{ikr}}$ then 4 is also zero (on average) at the upper limit. Thus in order for ${D_{l}}$ to be Hermitian, we must have

$\displaystyle \left.U_{1}^*\frac{dU_{2}}{dr}\right|_{0}-\left.U_{2}\frac{dU_{1}^*}{dr}\right|_{0}=0 \ \ \ \ \ (8)$

at the lower limit as well.

One way of satisfying this condition is if

$\displaystyle U_{El}\underset{r\rightarrow0}{\longrightarrow}c \ \ \ \ \ (9)$

Because the actual radial function is given by 2, a value of ${c\ne0}$ would give

$\displaystyle R\sim\frac{U}{r}\sim\frac{c}{r} \ \ \ \ \ (10)$

Such a function is still square integrable because an integral over all space introduces a factor of ${r^{2}}$ in the volume element:

$\displaystyle \int R^{2}r^{2}\sin\theta dr\;d\theta\;d\phi \ \ \ \ \ (11)$

Thus the integrand is still finite at ${r=0}$ so the integral itself can be finite.

The problem with ${c\ne0}$ is that the Laplacian of ${\frac{1}{r}}$ gives a delta function:

$\displaystyle \nabla^{2}\frac{1}{r}=-4\pi\delta^{3}\left(\mathbf{r}\right) \ \ \ \ \ (12)$

Unless the potential ${V}$ has a delta function at the origin (which would be quite unusual), the ${\nabla^{2}}$ in the Schrödinger equation can’t be allowed to generate a delta function there, so we must have ${c=0}$.

So far, everything is true for any potential. If we now assume that ${V}$ is less singular than ${\frac{1}{r^{2}}}$ (that is, ${V\underset{r\rightarrow0}{\longrightarrow}\frac{1}{r^{a}}}$ where ${a<2}$), the centrifugal barrier term in 1 will dominate as ${r\rightarrow0}$, so for small ${r}$, 1 reduces to the differential equation

$\displaystyle U_{l}^{\prime\prime}\simeq\frac{l\left(l+1\right)}{r^{2}}U_{l} \ \ \ \ \ (13)$

The ${E}$ in the suffix of ${U}$ has been dropped because the term involving ${E}$ in 1 is negligible compared to the centrifugal barrier for ${r\rightarrow0}$. This equation has solutions

$\displaystyle U_{l}\sim r^{\alpha} \ \ \ \ \ (14)$

Plugging this into 13 we get

$\displaystyle \alpha\left(\alpha-1\right)=l\left(l+1\right) \ \ \ \ \ (15)$

This is a quadratic equation in ${\alpha}$ which has the two solutions

$\displaystyle \alpha=-l,l+1 \ \ \ \ \ (16)$

If we are to have ${U_{El}\rightarrow0}$ as ${r\rightarrow0}$, then we must discard the solution ${U_{l}\sim r^{-l}}$, so we have that

$\displaystyle U_{El}\sim r^{l+1} \ \ \ \ \ (17)$

All of this works only if ${l\ne0}$ since in the case where ${l=0}$ (zero angular momentum), there is no centrifugal barrier and we must look at the form of the potential. Shankar notes that the problems he considers in his book are such that 17 is also valid for ${l=0}$.

# Hydrogen atom – a sample wave function

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.1.3.

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

The wave function for the hydrogen atom can be obtained by a series solution of the differential equation, leading to the result (which I’ve rewritten in Shankar’s notation, although my original post used Griffiths’s notation):

$\displaystyle \psi_{nlm}\left(r,\theta,\phi\right)=\frac{U_{El}\left(r\right)}{r}Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (1)$

Here, we have

 $\displaystyle U_{El}$ $\displaystyle =$ $\displaystyle e^{-\rho}v_{El}\ \ \ \ \ (2)$ $\displaystyle v_{El}$ $\displaystyle =$ $\displaystyle \rho^{l+1}\sum_{k=0}^{\infty}C_{k}\rho^{k}\ \ \ \ \ (3)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{\frac{-2mE}{\hbar^{2}}}r \ \ \ \ \ (4)$

The energy levels of the hydrogen atom are

$\displaystyle E=-\frac{me^{4}}{2\hbar^{2}n^{2}} \ \ \ \ \ (5)$

where ${n=1,2,3,\ldots}$. The coefficients ${C_{k}}$ in 3 are given by a recursion relation

 $\displaystyle C_{k+1}$ $\displaystyle =$ $\displaystyle \frac{-e^{2}\lambda+2\left(k+l+1\right)}{\left(k+l+2\right)\left(k+l+1\right)-l\left(l+1\right)}C_{k}\ \ \ \ \ (6)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \sqrt{-\frac{2m}{\hbar^{2}E}} \ \ \ \ \ (7)$

Combining ${\lambda}$ and ${E}$, the formula becomes, for a given ${n}$

$\displaystyle C_{k+1}=\frac{2\left(k+l+1\right)-2n}{\left(k+l+2\right)\left(k+l+1\right)-l\left(l+1\right)}C_{k}$

The coefficient ${C_{0}}$ which starts everything off is determined by normalization.

As an example, we can find the wave function ${\psi_{210}}$. In this case ${n=2}$ and ${l=1}$ so the first term in the recursion, with ${k=0}$ gives ${k+l+1=2}$ and ${C_{1}=0}$. The full wave function is then

 $\displaystyle \psi_{210}$ $\displaystyle =$ $\displaystyle \frac{1}{r}\rho^{2}e^{-\rho}C_{0}Y_{1}^{0} \ \ \ \ \ (8)$

To evaluate ${\rho}$ we use the energy for ${n=2}$:

$\displaystyle E_{2}=-\frac{me^{4}}{8\hbar^{2}} \ \ \ \ \ (9)$

This gives

$\displaystyle \rho=\sqrt{\frac{2m^{2}e^{4}}{8\hbar^{4}}}r=\frac{me^{2}}{2\hbar^{2}}r=\frac{r}{2a_{0}} \ \ \ \ \ (10)$

where ${a_{0}}$ is the Bohr radius

$\displaystyle a_{0}\equiv\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (11)$

Plugging everything into 8, using ${Y_{1}^{0}=\sqrt{\frac{3}{4\pi}}\cos\theta}$, we have

$\displaystyle \psi_{210}=\sqrt{\frac{3}{4\pi}}\frac{C_{0}}{4a_{0}^{2}}re^{-r/2a_{0}}\cos\theta \ \ \ \ \ (12)$

Normalizing gives the condition

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\psi_{210}^{2}r^{2}\sin\theta dr\;d\theta\;d\phi=1 \ \ \ \ \ (13)$

Working out the integral (using software or tables) gives

 $\displaystyle \frac{3}{2}a_{0}C_{0}^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (14)$ $\displaystyle C_{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{3a_{0}}} \ \ \ \ \ (15)$

So the final wave function is

$\displaystyle \psi_{210}=\frac{1}{\sqrt{32\pi a_{0}^{3}}}\frac{r}{a_{0}}e^{-r/2a_{0}}\cos\theta \ \ \ \ \ (16)$

which agrees with Shankar’s equation 13.1.27.

# Isotropic harmonic oscillator in 3-d – use of spherical harmonics

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.11.

[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 solved the 3-d isotropic harmonic oscillator before, so we’ve already solved most of Shankar’s exercise 12.6.11. We can quote the results here. The solution has the form

$\displaystyle \psi_{Elm}=\frac{U_{El}\left(r\right)}{r}Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (1)$

The earlier solution uses notation from Griffiths’s book, but as the end result is the same, it’s not worth going through the derivation again using Shankar’s notation.

The potential is

$\displaystyle V(r)=\frac{1}{2}m\omega^{2}r^{2} \ \ \ \ \ (2)$

The radial equation to be solved is

$\displaystyle \frac{d^{2}u}{d\rho^{2}}=\left(-1+\frac{l(l+1)}{\rho^{2}}+\rho_{0}^{2}\rho^{2}\right)u \ \ \ \ \ (3)$

If we define

 $\displaystyle \kappa^{2}$ $\displaystyle \equiv$ $\displaystyle \frac{2\mu E}{\hbar^{2}}\ \ \ \ \ (4)$ $\displaystyle \rho$ $\displaystyle \equiv$ $\displaystyle \kappa r\ \ \ \ \ (5)$ $\displaystyle \rho_{0}$ $\displaystyle \equiv$ $\displaystyle \frac{\mu\omega}{\hbar\kappa^{2}}=\frac{\hbar\omega}{2E} \ \ \ \ \ (6)$

Taking the asymptotic behaviour of the radial function for small and large ${r}$ into account leads us to a solution of form

$\displaystyle u(\rho)=\rho^{l+1}e^{-\rho_{0}\rho^{2}/2}v(\rho) \ \ \ \ \ (7)$

Note that Griffiths’s ${v}$ is not the same as Shankar’s ${v}$, the latter of which is defined by Shankar’s equation 12.6.49.

This gives a differential equation for Griffiths’s ${v}$

$\displaystyle \rho\frac{d^{2}v}{d\rho^{2}}+2\left(l+1-\rho_{0}\rho^{2}\right)\frac{dv}{d\rho}+\rho(1-\rho_{0}(2l+3))v=0 \ \ \ \ \ (8)$

The function ${v}$ can be solved as a power series, giving

$\displaystyle v(\rho)=\sum c_{j}\rho^{j} \ \ \ \ \ (9)$

Substituting into 8 leads to the recursion relation

$\displaystyle c_{q+2}=\frac{\rho_{0}(2q+2l+3)-1}{(q+2)(q+2l+3)}c_{q} \ \ \ \ \ (10)$

with ${c_{1}=0}$, so that ${c_{q}=0}$ for all odd ${q}$. The requirement that the series terminates at some finite value of ${j}$ leads to the quantization condition on ${E}$:

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \hbar\omega\left(q_{max}+l+\frac{3}{2}\right) \ \ \ \ \ (11)$

or, defining ${n=q_{max}+l}$,

$\displaystyle E_{n}=\hbar\omega\left(n+\frac{3}{2}\right) \ \ \ \ \ (12)$

We worked out the degeneracies in the earlier post as well, so that the degeneracy of ${E_{n}}$ is

$\displaystyle d\left(n\right)=\frac{1}{2}(n+1)(n+2) \ \ \ \ \ (13)$

To complete Shankar’s exercise, we need to work out the eigenfunctions for ${n=0}$ and ${n=1}$. For ${n=0}$, ${q_{max}=l=0}$, so only ${c_{0}\ne0}$ and we have

 $\displaystyle v\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\ \ \ \ \ (14)$ $\displaystyle u\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\rho e^{-\rho_{0}\rho^{2}/2}\ \ \ \ \ (15)$ $\displaystyle \psi_{000}$ $\displaystyle =$ $\displaystyle \frac{u}{r}Y_{0}^{0}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\kappa e^{-\rho_{0}\rho^{2}/2}Y_{0}^{0}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\sqrt{\frac{2\mu3\omega}{4\pi\hbar}}e^{-\mu\omega r^{2}/2\hbar} \ \ \ \ \ (18)$

where in the fourth line we used

 $\displaystyle \kappa$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2\mu E}}{\hbar}=\frac{\sqrt{2\mu\frac{3}{2}\hbar\omega}}{\hbar}=\sqrt{\frac{3\mu\omega}{\hbar}}\ \ \ \ \ (19)$ $\displaystyle Y_{0}^{0}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{4\pi}} \ \ \ \ \ (20)$

Normalizing this requires that

 $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\psi_{000}^{2}r^{2}\sin\theta dr\;d\theta\;d\phi$ $\displaystyle =$ $\displaystyle c_{0}^{2}\frac{6\mu\omega}{\hbar}\int_{0}^{\infty}e^{-\mu\omega r^{2}/\hbar}r^{2}dr\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (22)$

This is a standard Gaussian integral and can be done using software or tables so we get

$\displaystyle c_{0}=\frac{\sqrt{6}}{3}\left(\frac{\mu\omega}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (23)$

This gives a wave function of

$\displaystyle \psi_{000}=\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}e^{-\mu\omega r^{2}/2\hbar} \ \ \ \ \ (24)$

which agrees with the earlier result.

For ${n=1}$, the degeneracy is, from 13

$\displaystyle d\left(1\right)=3 \ \ \ \ \ (25)$

The three possibilities are ${m=0,\pm1}$ which are reflected in the three spherical harmonics ${Y_{1}^{0,\pm1}}$. The radial function is the same in all cases, and is obtained from ${q_{max}=0}$, ${l=1}$. From 7, this gives

 $\displaystyle v\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\ \ \ \ \ (26)$ $\displaystyle u\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\rho^{2}e^{-\rho_{0}\rho^{2}/2}\ \ \ \ \ (27)$ $\displaystyle \psi_{11m}$ $\displaystyle =$ $\displaystyle \frac{u}{r}Y_{1}^{m}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\kappa^{2}re^{-\rho_{0}\rho^{2}/2}Y_{1}^{m}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\frac{5\mu\omega}{\hbar}re^{-\mu\omega r^{2}/2\hbar}Y_{1}^{m} \ \ \ \ \ (30)$

Again, we work out ${c_{0}}$ by imposing normalization. For example

 $\displaystyle \psi_{111}$ $\displaystyle =$ $\displaystyle c_{0}\frac{5\mu\omega}{\hbar}re^{-\mu\omega r^{2}/2\hbar}Y_{1}^{1}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -c_{0}\frac{5\mu\omega}{\hbar}re^{-\mu\omega r^{2}/2\hbar}\sqrt{\frac{3}{8\pi}}\sin\theta e^{i\phi} \ \ \ \ \ (32)$

The normalization integral is

 $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\psi_{111}^{2}r^{2}\sin\theta dr\;d\theta\;d\phi$ $\displaystyle =$ $\displaystyle c_{0}^{2}\left(\frac{5\mu\omega}{\hbar}\right)^{2}\frac{3}{8\pi}2\pi\int_{0}^{\pi}\int_{0}^{\infty}e^{-\mu\omega r^{2}/\hbar}r^{4}\sin^{3}\theta dr\;d\theta\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}^{2}\frac{75}{8}\sqrt{\frac{\pi\hbar}{\mu\omega}}=1\ \ \ \ \ (34)$ $\displaystyle c_{0}$ $\displaystyle =$ $\displaystyle \frac{2\sqrt{6}}{15}\left(\frac{\mu\omega}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (35)$

I used Maple to do the integrals. This gives a wave function of

$\displaystyle \psi_{111}=-\sqrt{\frac{\mu\omega}{\hbar}}\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}re^{-\mu\omega r^{2}/2\hbar}\sin\theta e^{i\phi} \ \ \ \ \ (36)$

We can work out the other two wave functions the same way (I used Maple, so I won’t go into the details):

 $\displaystyle \psi_{11-1}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\mu\omega}{\hbar}}\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}re^{-\mu\omega r^{2}/2\hbar}\sin\theta e^{-i\phi}\ \ \ \ \ (37)$ $\displaystyle \psi_{110}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\mu\omega}{\hbar}}\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}re^{-\mu\omega r^{2}/2\hbar}\cos\theta \ \ \ \ \ (38)$

The ${\psi_{110}}$ here is the same as ${\psi_{001}}$ in our rectangular solution set. The other two are linear combinations of ${\psi_{100}}$ and ${\psi_{010}}$ from our rectangular set, which were (the suffixes in these 2 equations stand for ${x}$, ${y}$ and ${z}$, and not ${n}$, ${l}$ and ${m}$):

 $\displaystyle \psi_{100}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\cos\phi\ \ \ \ \ (39)$ $\displaystyle \psi_{010}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\sin\phi \ \ \ \ \ (40)$

We have

 $\displaystyle \psi_{111}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\psi_{100}+i\psi_{010}\right)\ \ \ \ \ (41)$ $\displaystyle \psi_{11-1}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\psi_{100}-i\psi_{010}\right) \ \ \ \ \ (42)$

# Free particle moving in the z direction

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.10.

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The radial function for a free particle can be either a spherical Bessel function ${j_{l}}$ or a spherical Neumann function ${n_{l}}$. If the solution space includes the origin, then only ${j_{l}}$ is acceptable since the ${n_{l}}$ functions diverge as ${r\rightarrow0}$.

In rectangular coordinates, a free particle wave function has the form

$\displaystyle \psi_{E}\left(x,y,z\right)=\frac{1}{\left(2\pi\hbar\right)^{3/2}}e^{i\mathbf{p}\cdot\mathbf{r}/\hbar} \ \ \ \ \ (1)$

where the energy ${E}$ is

$\displaystyle E=\frac{p^{2}}{2\mu}=\frac{\hbar^{2}k^{2}}{2\mu} \ \ \ \ \ (2)$

For a free particle travelling in the ${z}$ direction, this becomes

$\displaystyle \psi_{E}\left(r,\theta,\phi\right)=\frac{1}{\left(2\pi\hbar\right)^{3/2}}e^{ikr\cos\theta} \ \ \ \ \ (3)$

since ${z=r\cos\theta}$.

Since the solutions of the free-particle Schrödinger equation in spherical coordinations form a complete set, we must be able to express this wave function as a linear combination of these solutions, so that

$\displaystyle e^{ikr\cos\theta}=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}C_{l}^{m}j_{l}\left(kr\right)Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (4)$

where the ${C_{l}^{m}}$ are constants. Because we’re looking at motion in the ${z}$ direction, there is no angular momentum about the ${z}$ axis, which is reflected in the fact that ${\psi_{E}}$ does not depend on ${\phi}$. Thus ${L_{z}=m\hbar=0}$ and ${m=0}$. We therefore have

 $\displaystyle e^{ikr\cos\theta}$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}C_{l}^{0}j_{l}\left(kr\right)Y_{l}^{0}\left(\theta,\phi\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}\sqrt{\frac{2l+1}{4\pi}}C_{l}^{0}j_{l}\left(kr\right)P_{l}\left(\cos\theta\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}C_{l}j_{l}\left(kr\right)P_{l}\left(\cos\theta\right) \ \ \ \ \ (7)$

where

$\displaystyle C_{l}\equiv\sqrt{\frac{2l+1}{4\pi}}C_{l}^{0} \ \ \ \ \ (8)$

The problem, of course, is to find these constants. We can do this using the identities given by Shankar in his problem 12.6.10, which are

 $\displaystyle \int_{-1}^{1}P_{l}\left(x\right)P_{l^{\prime}}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{2\delta_{ll^{\prime}}}{2l+1}\ \ \ \ \ (9)$ $\displaystyle P_{l}\left(x\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2^{l}l!}\frac{d^{l}\left(x^{2}-1\right)^{l}}{dx^{l}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(-1\right)^{l}}{2^{l}l!}\frac{d^{l}\left(1-x^{2}\right)^{l}}{dx^{l}}\ \ \ \ \ (11)$ $\displaystyle \int_{0}^{1}\left(1-x^{2}\right)^{m}dx$ $\displaystyle =$ $\displaystyle \frac{\left(2m\right)!!}{\left(2m+1\right)!!}\ \ \ \ \ (12)$ $\displaystyle \int_{-1}^{1}\left(1-x^{2}\right)^{m}dx$ $\displaystyle =$ $\displaystyle \frac{2\left(2m\right)!!}{\left(2m+1\right)!!} \ \ \ \ \ (13)$

The last line follows because ${\left(1-x^{2}\right)^{m}}$ is an even function and is therefore symmetric about ${x=0}$.

We can use the standard procedure for isolating ${C_{l}}$ by multiplying both sides by ${C_{a}}$ and using 9.

 $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)e^{ikrx}dx$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}C_{l}j_{l}\left(kr\right)\int_{-1}^{1}P_{a}\left(x\right)P_{l}\left(x\right)dx\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{2a+1}C_{a}j_{a}\left(kr\right) \ \ \ \ \ (15)$

This relation must be true for all values of ${r}$, so we can look at the limit of small (but not zero, since both sides are then zero) ${r}$. We have the asymptotic relation for the spherical Bessel functions

 $\displaystyle j_{l}$ $\displaystyle \underset{\rho\rightarrow0}{\longrightarrow}$ $\displaystyle \frac{\rho^{l}}{\left(2l+1\right)!!} \ \ \ \ \ (16)$

We thus have

$\displaystyle \int_{-1}^{1}P_{a}\left(x\right)e^{ikrx}dx=\underset{r\rightarrow0}{\longrightarrow}\frac{2}{2a+1}\frac{k^{a}r^{a}}{\left(2a+1\right)!!}C_{a} \ \ \ \ \ (17)$

We can then look at the integral on the LHS and hope that, when we expand the exponential, that the terms in ${\left(kr\right)^{n}}$ for ${n vanish. We can then match the coefficients of ${\left(kr\right)^{a}}$ on both sides to find ${C_{a}}$.

We can see that this will work because the Legendre polynomials ${P_{l}}$ are a complete set of functions, and the polynomial ${P_{l}}$ has degree ${l}$. This means that any polynomial of degree ${a-1}$ can be written as a linear combination of the ${P_{l}}$, where ${l=0,\ldots,a-1}$. Because of 9, this means that

$\displaystyle \int_{-1}^{1}x^{l}P_{a}\left(x\right)dx=0\;\;\mbox{if \ensuremath{l

Therefore, when we expand ${e^{ikrx}}$ in a power series, we have

 $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)e^{ikrx}dx$ $\displaystyle =$ $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)\left(1+ikrx+\frac{\left(ikrx\right)^{2}}{2!}+\ldots\right)dx\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)\left(\frac{\left(ikrx\right)^{a}}{a!}+\ldots\right)dx \ \ \ \ \ (20)$

In the limit of small ${r}$, higher order terms in the sum on the RHS can be ignored, so we get

 $\displaystyle \frac{\left(ikr\right)^{a}}{a!}\int_{-1}^{1}x^{a}P_{a}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{2}{2a+1}\frac{k^{a}r^{a}}{\left(2a+1\right)!!}C_{a}\ \ \ \ \ (21)$ $\displaystyle C_{a}$ $\displaystyle =$ $\displaystyle \frac{i^{a}\left(2a+1\right)\left(2a+1\right)!!}{2a!}\int_{-1}^{1}x^{a}P_{a}\left(x\right)dx \ \ \ \ \ (22)$

Now consider the integral in the last line. Using 11 we have

$\displaystyle \int_{-1}^{1}x^{a}P_{a}\left(x\right)dx=\frac{\left(-1\right)^{a}}{2^{a}a!}\int_{-1}^{1}x^{a}\frac{d^{a}\left(1-x^{2}\right)^{a}}{dx^{a}}dx \ \ \ \ \ (23)$

We can integrate by parts repeatedly until the derivative in the integrand disappears. Note that the ${n}$th derivative of ${\left(1-x^{2}\right)^{a}}$ will always contain a factor of ${\left(1-x^{2}\right)}$ to some power for any ${n, and thus is zero at both limits of integration. Since the integrated term in the integration by parts always contains such a derivative, all integrated terms are zero at both limits. We therefore integrate ${\frac{d^{a}\left(1-x^{2}\right)^{a}}{dx^{a}}}$ (${a}$ times) and differentiate ${x^{a}}$ (${a}$ times) and keep only the residual integral after each iteration. The differentiation of ${x^{a}}$ (${a}$ times) introduces a factor of ${a!}$. Since the sign of the residual integral alternates as we perform each integration by parts, the final result is

 $\displaystyle \int_{-1}^{1}x^{a}P_{a}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{\left(-1\right)^{2a}}{2^{a}a!}a!\int_{-1}^{1}\left(1-x^{2}\right)^{a}dx\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2^{a}}\frac{2\left(2a\right)!!}{\left(2a+1\right)!!} \ \ \ \ \ (25)$

where we used 13 in the last line. The double factorial in the numerator can be written as

 $\displaystyle \left(2a\right)!!$ $\displaystyle =$ $\displaystyle \left(2a\right)\left(2a-2\right)\ldots\left(4\right)\left(2\right)\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2^{a}a\left(a-1\right)\ldots\left(2\right)\left(1\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2^{a}a! \ \ \ \ \ (28)$

We therefore have

 $\displaystyle \int_{-1}^{1}x^{a}P_{a}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{1}{2^{a}}\frac{2\times2^{a}a!}{\left(2a+1\right)!!}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2a!}{\left(2a+1\right)!!} \ \ \ \ \ (30)$

Plugging this back into 22 we have

$\displaystyle C_{a}=i^{a}\left(2a+1\right) \ \ \ \ \ (31)$

The wave function for a free particle moving in the ${z}$ direction is therefore

$\displaystyle \psi_{E}\left(r,\theta,\phi\right)=\frac{1}{\left(2\pi\hbar\right)^{3/2}}\sum_{l=0}^{\infty}i^{a}\left(2a+1\right)j_{l}\left(kr\right)P_{l}\left(\cos\theta\right) \ \ \ \ \ (32)$

# Spherical Bessel functions – behaviour for small arguments

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.7.

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

The general solution for a free particle in spherical coordinates involves the radial function, which turns out to be

$\displaystyle \frac{R_{l+1}}{\rho^{l+1}}=\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)^{l+1}\frac{R_{0}}{\rho^{0}} \ \ \ \ \ (1)$

where ${l}$ is the total angular momentum quantum number and

 $\displaystyle k^{2}$ $\displaystyle \equiv$ $\displaystyle \frac{2\mu E}{\hbar^{2}}\ \ \ \ \ (2)$ $\displaystyle \rho$ $\displaystyle \equiv$ $\displaystyle kr \ \ \ \ \ (3)$

We can rewrite this as

$\displaystyle R_{l}=\left(-\rho\right)^{l}\left(\frac{1}{\rho}\frac{d}{d\rho}\right)^{l}R_{0} \ \ \ \ \ (4)$

We saw earlier that the solutions for ${l=0}$ are, with ${U_{l}=\rho R_{l}}$

 $\displaystyle U_{0}^{A}\left(\rho\right)$ $\displaystyle =$ $\displaystyle \sin\rho\ \ \ \ \ (5)$ $\displaystyle U_{0}^{B}\left(\rho\right)$ $\displaystyle =$ $\displaystyle -\cos\rho \ \ \ \ \ (6)$

Thus the two solutions for ${l=0}$ are

 $\displaystyle R_{0}^{A}$ $\displaystyle =$ $\displaystyle \frac{\sin\rho}{\rho}\ \ \ \ \ (7)$ $\displaystyle R_{0}^{B}$ $\displaystyle =$ $\displaystyle -\frac{\cos\rho}{\rho} \ \ \ \ \ (8)$

From these starting points, we can generate all the solutions for higher values of ${l}$ using 4. These functions are

 $\displaystyle j_{l}\left(\rho\right)$ $\displaystyle =$ $\displaystyle \left(-\rho\right)^{l}\left(\frac{1}{\rho}\frac{d}{d\rho}\right)^{l}\frac{\sin\rho}{\rho}\ \ \ \ \ (9)$ $\displaystyle n_{l}\left(\rho\right)$ $\displaystyle =$ $\displaystyle -\left(-\rho\right)^{l}\left(\frac{1}{\rho}\frac{d}{d\rho}\right)^{l}\frac{\cos\rho}{\rho} \ \ \ \ \ (10)$

and are known as spherical Bessel functions ${j_{l}}$ and spherical Neumann functions ${n_{l}}$.

The asymptotic behaviour is given by

 $\displaystyle j_{l}$ $\displaystyle \underset{\rho\rightarrow\infty}{\longrightarrow}$ $\displaystyle \frac{1}{\rho}\sin\left(\rho-\frac{l\pi}{2}\right)\ \ \ \ \ (11)$ $\displaystyle n_{l}$ $\displaystyle \underset{\rho\rightarrow\infty}{\longrightarrow}$ $\displaystyle -\frac{1}{\rho}\cos\left(\rho-\frac{l\pi}{2}\right) \ \ \ \ \ (12)$

For ${\rho\rightarrow0}$, we have

 $\displaystyle j_{l}$ $\displaystyle \underset{\rho\rightarrow0}{\longrightarrow}$ $\displaystyle \frac{\rho^{l}}{\left(2l+1\right)!!}\ \ \ \ \ (13)$ $\displaystyle n_{l}$ $\displaystyle \underset{\rho\rightarrow0}{\longrightarrow}$ $\displaystyle -\frac{\left(2l-1\right)!!}{\rho^{l+1}} \ \ \ \ \ (14)$

We can verify the latter equation for ${j_{l}}$ for a couple of cases with small ${l}$. From 9, we can generate a couple of ${j_{l}}$s:

 $\displaystyle j_{0}$ $\displaystyle =$ $\displaystyle \frac{\sin\rho}{\rho}\ \ \ \ \ (15)$ $\displaystyle j_{1}$ $\displaystyle =$ $\displaystyle -\rho\frac{1}{\rho}\frac{d}{dr}\left(\frac{\sin\rho}{\rho}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\rho\frac{1}{\rho}\left(\frac{\cos\rho}{\rho}-\frac{\sin\rho}{\rho^{2}}\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sin\rho}{\rho^{2}}-\frac{\cos\rho}{\rho}\ \ \ \ \ (18)$ $\displaystyle j_{2}$ $\displaystyle =$ $\displaystyle \left(-\rho\right)^{2}\frac{1}{\rho}\frac{d}{d\rho}\left[\frac{1}{\rho}\frac{d}{dr}\left(\frac{\sin\rho}{\rho}\right)\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(-\rho\right)^{2}\frac{1}{\rho}\frac{d}{d\rho}\left[\frac{1}{\rho}\left(\frac{\cos\rho}{\rho}-\frac{\sin\rho}{\rho^{2}}\right)\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{3}{\rho^{3}}-\frac{1}{\rho}\right)\sin\rho-\frac{3\cos\rho}{\rho^{2}} \ \ \ \ \ (21)$

We can get the limits for ${\rho\rightarrow0}$ by expanding the sine and cosine. That is, we use the limiting forms

 $\displaystyle \sin\rho$ $\displaystyle \rightarrow$ $\displaystyle \rho-\frac{\rho^{3}}{3!}+\ldots\ \ \ \ \ (22)$ $\displaystyle \cos\rho$ $\displaystyle \rightarrow$ $\displaystyle 1-\frac{1}{2}\rho^{2}+\ldots \ \ \ \ \ (23)$

We need to retain enough terms for ${j_{l}}$ so that we get all the terms up to the first power of ${\rho}$ that doesn’t cancel out when we do the algebra. We get

 $\displaystyle j_{0}$ $\displaystyle \rightarrow$ $\displaystyle 1=\frac{\rho^{0}}{1!!}\ \ \ \ \ (24)$ $\displaystyle j_{1}$ $\displaystyle \rightarrow$ $\displaystyle \frac{1}{\rho}-\frac{\rho}{6}-\frac{1}{\rho}\left(1-\frac{1}{2}\rho^{2}\right)\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\rho}{3}=\frac{\rho^{1}}{3!!}\ \ \ \ \ (26)$ $\displaystyle j_{2}$ $\displaystyle \rightarrow$ $\displaystyle \left(\frac{3}{\rho^{3}}-\frac{1}{\rho}\right)\left(\rho-\frac{\rho^{3}}{6}+\frac{\rho^{5}}{120}\right)-\frac{3}{\rho^{2}}\left(1-\frac{1}{2}\rho^{2}+\frac{1}{24}\rho^{4}\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle \rightarrow$ $\displaystyle \left(\frac{1}{6}+\frac{1}{40}-\frac{1}{8}\right)\rho^{2}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{20+3-15}{120}\rho^{2}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\rho^{2}}{15}=\frac{\rho^{2}}{5!!} \ \ \ \ \ (30)$

# Free particle in spherical coordinates – finding the solutions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.6.

[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 solving the Schrödinger equation for spherically symmetric potentials, we found that we could reduce the problem to the equation

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (1)$

where ${U_{El}\left(r\right)}$ is related to the radial function by

$\displaystyle R_{El}\left(r\right)=\frac{U_{El}\left(r\right)}{r} \ \ \ \ \ (2)$

For a free particle, ${V=0}$ and ${E>0}$, so we have

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (3)$

Defining

 $\displaystyle k^{2}$ $\displaystyle \equiv$ $\displaystyle \frac{2\mu E}{\hbar^{2}}\ \ \ \ \ (4)$ $\displaystyle \rho$ $\displaystyle \equiv$ $\displaystyle kr \ \ \ \ \ (5)$

we convert the equation to

$\displaystyle \left(-\frac{d^{2}}{d\rho^{2}}+\frac{l\left(l+1\right)}{\rho^{2}}\right)U_{l}=U_{l} \ \ \ \ \ (6)$

This equation can be solved by a method similar to that for the harmonic oscillator and its raising and lowering operators. The entire solution is fairly involved, so we’ll start out here by showing how the new raising and lowering operators are defined.

We define

$\displaystyle d_{l}\equiv\frac{d}{d\rho}+\frac{l+1}{\rho} \ \ \ \ \ (7)$

$\displaystyle d_{l}^{\dagger}=-\frac{d}{d\rho}+\frac{l+1}{\rho} \ \ \ \ \ (8)$

To see where the minus sign comes from on the RHS, we need to recall that the momentum operator is defined in one dimension as

$\displaystyle P=-i\hbar\frac{\partial}{\partial x} \ \ \ \ \ (9)$

Since ${P}$ is an observable, it is hermitian, so that ${P^{\dagger}=P}$. Under the hermitian operation ${i\rightarrow-i}$, so we must also have ${\frac{\partial}{\partial x}\rightarrow-\frac{\partial}{\partial x}}$. Thus the first derivative with respect to a position variable is anti-hermitian. If this doesn’t convince you, you can also work out the integral:

$\displaystyle \int_{0}^{\infty}\psi_{2}^*\frac{d}{d\rho}\psi_{1}d\rho=\left.\psi_{2}^*\psi_{1}\right|_{0}^{\infty}-\int_{0}^{\infty}\psi_{1}\frac{d}{d\rho}\psi_{2}^*d\rho \ \ \ \ \ (10)$

Under the usual assumption that ${\psi\rightarrow0}$ at the limits, the integrated term is zero and we have

 $\displaystyle \int_{0}^{\infty}\psi_{2}^*\frac{d}{d\rho}\psi_{1}d\rho$ $\displaystyle =$ $\displaystyle -\int_{0}^{\infty}\psi_{1}\frac{d}{d\rho}\psi_{2}^*d\rho\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left[\int_{0}^{\infty}\psi_{1}^*\frac{d}{d\rho}\psi_{2}d\rho\right]^* \ \ \ \ \ (12)$

In bracket notation, this is

$\displaystyle \left\langle \psi_{2}\left|\frac{d}{d\rho}\psi_{1}\right.\right\rangle =-\left\langle \frac{d}{d\rho}\psi_{2}\left|\psi_{1}\right.\right\rangle \ \ \ \ \ (13)$

which shows that ${\frac{d}{d\rho}}$ is an anti-hermitian operator.

Returning to 7 and 8, we have

 $\displaystyle d_{l}d_{l}^{\dagger}U_{l}$ $\displaystyle =$ $\displaystyle \left(\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)U_{l}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(-U_{l}^{\prime}+\frac{l+1}{\rho}U_{l}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}-\frac{l+1}{\rho^{2}}U_{l}+\frac{l+1}{\rho}U_{l}^{\prime}-\frac{l+1}{\rho}U_{l}^{\prime}+\frac{\left(l+1\right)^{2}}{\rho^{2}}U_{l}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{l\left(l+1\right)}{\rho^{2}}U_{l} \ \ \ \ \ (17)$

Comparing with 6 we see that

$\displaystyle d_{l}d_{l}^{\dagger}U_{l}=U_{l} \ \ \ \ \ (18)$

We can also show that

 $\displaystyle d_{l}^{\dagger}d_{l}U_{l}$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(\frac{d}{d\rho}+\frac{l+1}{\rho}\right)U_{l}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(U_{l}^{\prime}+\frac{l+1}{\rho}U_{l}\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{l+1}{\rho^{2}}U_{l}-\frac{l+1}{\rho}U_{l}^{\prime}+\frac{l+1}{\rho}U_{l}^{\prime}+\frac{\left(l+1\right)^{2}}{\rho^{2}}U_{l}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{\left(l+1\right)^{2}+l+1}{\rho^{2}}U_{l}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -U_{l}^{\prime\prime}+\frac{\left(l+1\right)\left(l+2\right)}{\rho^{2}}U_{l}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle d_{l+1}d_{l+1}^{\dagger}U_{l} \ \ \ \ \ (24)$

Starting from 18 we multiply on the left by ${d_{l}^{\dagger}}$ to get

$\displaystyle d_{l}^{\dagger}d_{l}\left(d_{l}^{\dagger}U_{l}\right)=d_{l}^{\dagger}U_{l} \ \ \ \ \ (25)$

Comparing this with 24 we see that

$\displaystyle d_{l}^{\dagger}U_{l}=c_{l}U_{l+1} \ \ \ \ \ (26)$

where ${c_{l}}$ is a constant.

Thus ${d_{l}^{\dagger}}$ is a raising operator, in that it raises the angular momentum number ${l}$ by 1 when it acts on ${U_{l}}$. By convention, ${c_{l}=1}$ (any adjustments to the constant can be made when normalizing).

We can start the process by looking at 6 with ${l=0}$ which is

$\displaystyle \frac{d^{2}}{d\rho^{2}}U_{l}=-U_{l} \ \ \ \ \ (27)$

This has the two solutions

 $\displaystyle U_{0}^{A}\left(\rho\right)$ $\displaystyle =$ $\displaystyle \sin\rho\ \ \ \ \ (28)$ $\displaystyle U_{0}^{B}\left(\rho\right)$ $\displaystyle =$ $\displaystyle -\cos\rho \ \ \ \ \ (29)$

The minus sign in front of ${\cos\rho}$ is just conventional. Since we require ${U_{0}\left(0\right)=0}$, ${U_{0}^{B}}$ is unacceptable if the region we’re considering include ${\rho=0}$, so we have

$\displaystyle U_{0}\left(\rho\right)=\sin\rho \ \ \ \ \ (30)$

For the general case that excludes ${\rho=0}$, we must include the cosine term as well.

From here, we can generate solutions for higher values of ${l}$ by applying 26. Actually, the radial function that appears in the wave function is given by 2, so it is ${R_{l}}$ that we really want. That is, we want

$\displaystyle R_{l}=\frac{U_{l}}{r}=k\frac{U_{l}}{\rho} \ \ \ \ \ (31)$

As with the constant ${c_{l}}$ in 26, we can absorb ${k}$ into normalization to be done later, so we can generate functions

$\displaystyle R_{l}=\frac{U_{l}}{\rho} \ \ \ \ \ (32)$

Applying 26 we have

 $\displaystyle \rho R_{l+1}$ $\displaystyle =$ $\displaystyle d_{l}^{\dagger}\left(\rho R_{l}\right)\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l+1}{\rho}\right)\left(\rho R_{l}\right)\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{l}-\rho R_{l}^{\prime}+\left(l+1\right)R_{l}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\rho R_{l}^{\prime}+lR_{l}\ \ \ \ \ (36)$ $\displaystyle R_{l+1}$ $\displaystyle =$ $\displaystyle \left(-\frac{d}{d\rho}+\frac{l}{\rho}\right)R_{l}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\rho^{l}\frac{d}{d\rho}\left(\frac{R_{l}}{\rho^{l}}\right) \ \ \ \ \ (38)$

We can convert this into a general formula by writing

$\displaystyle \frac{R_{l+1}}{\rho^{l+1}}=\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{l}}{\rho^{l}} \ \ \ \ \ (39)$

Starting at ${l=0}$, we have

$\displaystyle \frac{R_{1}}{\rho^{1}}=\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{0}}{\rho^{0}} \ \ \ \ \ (40)$

For the next step, we have

 $\displaystyle \frac{R_{2}}{\rho^{2}}$ $\displaystyle =$ $\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{1}}{\rho^{1}}\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)\frac{R_{0}}{\rho^{0}}\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)^{2}\frac{R_{0}}{\rho^{0}} \ \ \ \ \ (43)$

Thus in general

$\displaystyle \frac{R_{l+1}}{\rho^{l+1}}=\left(-\frac{1}{\rho}\frac{d}{d\rho}\right)^{l+1}\frac{R_{0}}{\rho^{0}} \ \ \ \ \ (44)$

Note that

$\displaystyle \left(-\frac{1}{\rho}\frac{d}{d\rho}\right)^{l+1}\ne\left(-\frac{1}{\rho}\right)^{l+1}\frac{d^{l+1}}{d\rho^{l+1}} \ \ \ \ \ (45)$

since the factor of ${\frac{1}{\rho}}$ has to be included when taking the derivative.

We’ll explore the nature of these solutions in the next post.

# Nondegenerate states in 3-d spherically symmetric systems

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.5.

[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 solving the Schrödinger equation for spherically symmetric potentials, we found that we could reduce the problem to the equation

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (1)$

where ${U_{El}\left(r\right)}$ is related to the radial function by

$\displaystyle R_{El}\left(r\right)=\frac{U_{El}\left(r\right)}{r} \ \ \ \ \ (2)$

We can write 1 as an eigenvalue equation for the operator ${D_{l}}$ in the form

$\displaystyle D_{l}\left(r\right)U_{El}=EU_{El} \ \ \ \ \ (3)$

with

$\displaystyle D_{l}\left(r\right)\equiv-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}} \ \ \ \ \ (4)$

We can show that, provided ${U_{El}\left(r\right)\rightarrow0}$ as ${r\rightarrow0}$, there are no degenerate eigenstates (that is, any state ${U_{El}}$ that is an eigenstate with energy ${E}$ is unique up to a scaling factor). The proof is similar to that in 1-d quantum mechanics, and goes by contradiction.

We suppose that there are two different functions ${U_{1}}$ and ${U_{2}}$ that satisfy 1 for the same energy ${E}$ (and same angular momentum number ${l}$). We then have

 $\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{1}$ $\displaystyle =$ $\displaystyle EU_{1}\ \ \ \ \ (5)$ $\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{2}$ $\displaystyle =$ $\displaystyle EU_{2} \ \ \ \ \ (6)$

Multiply the first by ${U_{2}}$ and the second by ${U_{1}}$ and subtract to get

$\displaystyle U_{2}U_{1}^{\prime\prime}-U_{1}U_{2}^{\prime\prime}=0 \ \ \ \ \ (7)$

This expression is

$\displaystyle U_{2}U_{1}^{\prime\prime}-U_{1}U_{2}^{\prime\prime}=\frac{d}{dr}\left(U_{2}U_{1}^{\prime}-U_{1}U_{2}^{\prime}\right)=0 \ \ \ \ \ (8)$

which we can integrate to get

$\displaystyle U_{2}U_{1}^{\prime}-U_{1}U_{2}^{\prime}=C \ \ \ \ \ (9)$

for some constant ${C}$. This relation is valid for all ${r}$, so we can choose ${r=0}$ where ${U_{2}\left(0\right)=U_{1}\left(0\right)=0}$, which shows that ${C=0}$. Therefore

$\displaystyle \frac{U_{1}^{\prime}}{U_{1}}=\frac{U_{2}^{\prime}}{U_{2}} \ \ \ \ \ (10)$

Integrating gives us

$\displaystyle \ln U_{1}=\ln U_{2}+K \ \ \ \ \ (11)$

for some other constant ${K}$, so

$\displaystyle U_{1}=e^{K}U_{2} \ \ \ \ \ (12)$

That is, any two eigenfunctions with the same eigenvalue ${E}$ are multiples of each other, so represent the same state, which is nondegenerate.

Note that the derivation didn’t rely on the value of ${U}$ anywhere except at ${r=0}$, so there is no requirement that, for example, ${U\rightarrow0}$ as ${r\rightarrow\infty}$. Also, the derivation is valid whatever the sign of ${E}$.

# Spherically symmetric potentials: hermiticity of the radial function

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.3.

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The Schrödinger equation in 3-d for a potential that depends only on ${r}$ is

$\displaystyle -\frac{\hbar^{2}}{2\mu}\left[\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial\psi}{\partial r}\right)+\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{r^{2}\sin^{2}\theta}\left(\frac{\partial^{2}\psi}{\partial\phi^{2}}\right)\right]+V\psi=E\psi \ \ \ \ \ (1)$

Eigenfunctions in this equation satisfy

$\displaystyle \psi=R_{Elm}\left(r\right)Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (2)$

where the subscript ${Elm}$ refers to the energy ${E}$ and the angular momentum quantum numbers ${l}$ and ${m}$. ${Y_{l}^{m}}$ is a spherical harmonic and ${R_{Elm}}$ is the radial function which depends on the potential ${V}$. With the substitution

$\displaystyle R_{El}\left(r\right)=\frac{U_{El}\left(r\right)}{r} \ \ \ \ \ (3)$

the differential equation reduces to

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (4)$

The quantity in the square brackets is an operator which will call ${D_{l}\left(r\right)}$:

$\displaystyle D_{l}\left(r\right)\equiv-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}} \ \ \ \ \ (5)$

Equation 4 is similar to the 1-d Schrödinger equation except that the variable ${r}$ goes from 0 to ${\infty}$ rather than from ${-\infty}$ to ${\infty}$, and the potential is modified by the ‘centrifugal term’ ${\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}}$. Because ${r}$ begins at 0 rather than ${-\infty}$, the usual boundary conditions on ${U}$ (that it tend to zero at ${\pm\infty}$) must also be modified. We can get the new boundary conditions by imposing the hermiticity condition, which says that

 $\displaystyle \int_{0}^{\infty}U_{1}^*\left(D_{l}U_{2}\right)dr$ $\displaystyle =$ $\displaystyle \left[\int_{0}^{\infty}U_{2}^*\left(D_{l}U_{1}\right)dr\right]^*\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}\left(D_{l}U_{1}\right)^*U_{2}dr \ \ \ \ \ (7)$

The two terms ${V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}}$ in 5 are real and multiplicative, so the hermiticity condition is automatically satisfied for them. For the derivative term, we can use the usual integration by parts.

 $\displaystyle \int_{0}^{\infty}U_{1}^*\left(\frac{d^{2}}{dr^{2}}U_{2}\right)dr$ $\displaystyle =$ $\displaystyle \left.U_{1}^*\frac{dU_{2}}{dr}\right|_{0}^{\infty}-\int_{0}^{\infty}\frac{dU_{1}^*}{dr}\frac{dU_{2}}{dr}dr\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left.U_{1}^*\frac{dU_{2}}{dr}\right|_{0}^{\infty}-\left.U_{2}\frac{dU_{1}^*}{dr}\right|_{0}^{\infty}+\int_{0}^{\infty}U_{2}\left(\frac{d^{2}}{dr^{2}}U_{1}^*\right)dr \ \ \ \ \ (9)$

If we require

$\displaystyle \left.U_{1}^*\frac{dU_{2}}{dr}\right|_{0}^{\infty}-\left.U_{2}\frac{dU_{1}^*}{dr}\right|_{0}^{\infty}=0 \ \ \ \ \ (10)$

then we have

 $\displaystyle \int_{0}^{\infty}U_{1}^*\left(\frac{d^{2}}{dr^{2}}U_{2}\right)dr$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}U_{2}\left(\frac{d^{2}}{dr^{2}}U_{1}^*\right)dr\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\int_{0}^{\infty}U_{2}^*\left(\frac{d^{2}}{dr^{2}}U_{1}\right)dr\right]^* \ \ \ \ \ (12)$

and the hermiticity condition 6 is satisfied.