4 A Quantum Particle in Three Dimensions

Although it goes beyond the scope of this course, I should point out that the commutation relations (4.84) are very famous in mathematics and appear in many places that have nothing to do with quantum mechanics. These equations define what mathematicians call a Lie algebra, in this case the algebra $su(2)$. It is very closely related to the group of $SU(2)$ which, in turn, is almost (but not quite!) the same thing as the group of rotations $SO(3)$. In some sense, the commutation relations (4.84) distill the essence of rotation. The “almost but not quite” of the previous sentence opens up a number of weird and wonderful loopholes that the quantum world exploits. (Like, for example, the fact that there are particles called fermions that don’t come back to themselves when rotated by ${360}^{\circ }$.). We discussed some of these issues briefly when exploring special relativity in the lectures on Dynamics and Relativity and we’ll meet them many more times in later courses.

Let’s prove the commutation relations (4.84). It will suffice to prove just one case, say

$[{\widehat{L}}_1,{\widehat{L}}_2]=i\mathrm{\hslash }{\widehat{L}}_3$

(4.85)

To check this, we recall that operator equations of this kind should be viewed in terms of their action on functions and then just plug in the appropriate definitions,

	$[{\widehat{L}}_1,{\widehat{L}}_2]\psi$	$=$	$-{\mathrm{\hslash }}^2\left(x_2{\displaystyle \frac{\partial }{\partial x_3}}-x_3{\displaystyle \frac{\partial }{\partial x_2}}\right)\left(x_3{\displaystyle \frac{\partial }{\partial x_1}}-x_1{\displaystyle \frac{\partial }{\partial x_3}}\right)\psi$
			$\mathrm{ }\mathit{\hspace{1em}\hspace{1em} }+{\mathrm{\hslash }}^2\left(x_3{\displaystyle \frac{\partial }{\partial x_1}}-x_1{\displaystyle \frac{\partial }{\partial x_3}}\right)\left(x_2{\displaystyle \frac{\partial }{\partial x_3}}-x_3{\displaystyle \frac{\partial }{\partial x_2}}\right)\psi$

A little algebra will convince you that nearly all terms cancel. The only ones that survive are

$[{\widehat{L}}_1,{\widehat{L}}_2]\psi ={\mathrm{\hslash }}^2\left(x_1{\displaystyle \frac{\partial }{\partial x_2}}-x_2{\displaystyle \frac{\partial }{\partial x_1}}\right)\psi =i\mathrm{\hslash }{\widehat{L}}_3\psi$

Because this holds for all $\psi (𝐱)$, it is equivalent to the claimed result (4.85). The other commutation relations (4.84) follow in the same vein.

To prove (4.86) we will first need a simple lemma. For any operators $\widehat{O}$ and $\widehat{M}$,

$[{\widehat{O}}^2,\widehat{M}]=\widehat{O}[\widehat{O},\widehat{M}]+[\widehat{O},\widehat{M}]\widehat{O}$

(4.87)

To prove this, we start from the right-hand side and expand out

	$\widehat{O}[\widehat{O},\widehat{M}]+[\widehat{O},\widehat{M}]\widehat{O}$	$=$	${\widehat{O}}^2\widehat{M}-\widehat{O}\widehat{M}\widehat{O}+\widehat{O}\widehat{M}\widehat{O}-\widehat{M}{\widehat{O}}^2$
		$=$	${\widehat{O}}^2\widehat{M}-\widehat{M}{\widehat{O}}^2=[{\widehat{O}}^2,\widehat{M}]$

We now just use this result repeatedly to prove (4.86). We want to show that, for example,

$[{\widehat{L}}_1^2,{\widehat{L}}_1]+[{\widehat{L}}_2^2,{\widehat{L}}_1]+[{\widehat{L}}_2^2,{\widehat{L}}_1]=0$

The first of these is trivial: all operators commute with themselves so $[{\widehat{L}}_1^2,{\widehat{L}}_1]=0$. Using (4.87), the next two can be written as

	$[{\widehat{L}}_2^2,L_1]$	$=$	${\widehat{L}}_2[{\widehat{L}}_2,L_1]+[{\widehat{L}}_2,{\widehat{L}}_1]{\widehat{L}}_2$		(4.88)
		$=$	$-i\mathrm{\hslash }({\widehat{L}}_2{\widehat{L}}_3+i\mathrm{\hslash }{\widehat{L}}_3{\widehat{L}}_2)$

where, to get to the second line, we’ve used the commutation relations (4.84). We also have

	$[{\widehat{L}}_3^2,L_1]$	$=$	${\widehat{L}}_3[{\widehat{L}}_3,L_1]+[{\widehat{L}}_3,{\widehat{L}}_1]{\widehat{L}}_3$		(4.89)
		$=$	$i\mathrm{\hslash }({\widehat{L}}_3{\widehat{L}}_2+i\mathrm{\hslash }{\widehat{L}}_2{\widehat{L}}_3)$

Adding (4.88) and (4.89) gives us the result (4.86) that we wanted.

The corresponding statement in quantum mechanics is that, for a central potential, we have

$[\widehat{H},{\widehat{L}}_i]=[\widehat{H},{\widehat{𝐋}}^2]=0$

(4.90)

Again, this is important. It means that, for eigenstates of the Hamiltonian, total angular momentum and angular momentum in one chosen direction will be shared. A quantum particle can have simultaneous values for all three of these quantities. As we move forward, this is the result that we will allow us to solve the 3d Schrödinger equation.

Before we can do that, we first have to prove (4.90). It follows from some straightforward algebra that is very similar in spirit to what we’ve done so far. You can first show that

$[{\widehat{L}}_i,{\widehat{x}}_j]=i\mathrm{\hslash }{ϵ}_{ijk}{\widehat{x}}_k\mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }[{\widehat{L}}_i,{\widehat{p}}_j]=i\mathrm{\hslash }{ϵ}_{ijk}{\widehat{p}}_k$

From this, it’s a small step to show that

	$[{\widehat{L}}_i,{\widehat{𝐱}}^2]=[{\widehat{L}}_i,{\widehat{x}}_1^2+{\widehat{x}}_2^2+{\widehat{x}}_3^2]$	$=$	$0$
	$\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }[{\widehat{L}}_i,{\widehat{𝐩}}^2]=[{\widehat{L}}_i,{\widehat{p}}_1^2+{\widehat{p}}_2^2+{\widehat{p}}_3^2]$	$=$	$0$

But this is all we need. The Hamiltonian for a central potential is a function only of ${\widehat{𝐩}}^2$ and ${\widehat{𝐱}}^2$,

$\widehat{H}={\displaystyle \frac{{\widehat{𝐩}}^2}{2m}}+V(r)$

where $r^2={𝐱}^2$. So we’re guaranteed to get $[\widehat{H},{\widehat{L}}_i]=0$. Since the Hamiltonian commutes with each angular momentum operator, it also commutes with the total angular momentum operator ${\widehat{𝐋}}^2$.

The upshot of this analysis is that a particle moving in a central potential can sit in simultaneous eigenstates of three operators, which we usually take to be:

$\widehat{H},{\widehat{𝐋}}^2,\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }{\widehat{L}}_3$

The associated eigenvalues are energy, total angular momentum, and the angular momentum in the $z=x_3$ direction. In what follows, all energy eigenstates are labelled by these three numbers.

Note also, that this is the maximum set of mutually commuting operators. For a generic central potential $V(r)$, there are no further operators lurking out there that also commute with these three.

It’s straightforward to write the angular momentum operators in spherical polar coordinates. We just need to use the chain rule to get, for example,

	${\displaystyle \frac{\partial }{\partial x_1}}$	$=$	${\displaystyle \frac{\partial r}{\partial x_1}}{\displaystyle \frac{\partial }{\partial r}}+{\displaystyle \frac{\partial \theta }{\partial x_1}}{\displaystyle \frac{\partial }{\partial \theta }}+{\displaystyle \frac{\partial \varphi }{\partial x_1}}{\displaystyle \frac{\partial }{\partial \varphi }}$
		$=$	$\sin \theta \cos \varphi {\displaystyle \frac{\partial }{\partial r}}+\cos \theta \cos \varphi {\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial }{\partial \theta }}-{\displaystyle \frac{\sin \varphi }{\sin \theta }}{\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial }{\partial \varphi }}$

with similar results for $\partial /\partial x_2$ and $\partial /\partial x_3$.

Combining these, we can write the angular momentum operators as

	${\widehat{L}}_1$	$=$	$i\mathrm{\hslash }\left(\cot \theta \cos \varphi {\displaystyle \frac{\partial }{\partial \varphi }}+\sin \varphi {\displaystyle \frac{\partial }{\partial \theta }}\right)$
	${\widehat{L}}_2$	$=$	$i\mathrm{\hslash }\left(\cot \theta \sin \varphi {\displaystyle \frac{\partial }{\partial \varphi }}-\cos \varphi {\displaystyle \frac{\partial }{\partial \theta }}\right)$
	${\widehat{L}}_3$	$=$	$-i\mathrm{\hslash }{\displaystyle \frac{\partial }{\partial \varphi }}$		(4.92)

Note that the angular momentum operators care nothing for how the wavefunctions depend on the radial coordinate $r$. The angular momentum of a state is, perhaps unsurprisingly, encoded in how the wavefunction varies in the angular directions $\theta$ and $\varphi$. Furthermore, ${\widehat{L}}_3$ takes a particularly straightforward form, simply because $x_3$ is singled out as a special direction in spherical polar coordinates.

We also need an expression for the total angular momentum operator. This can be written most simply as

${\widehat{𝐋}}^2={\widehat{L}}_1^2+{\widehat{L}}_2^2+{\widehat{L}}_3^2=-{\displaystyle \frac{{\mathrm{\hslash }}^2}{{\sin }^2\theta }}\left[\sin \theta {\displaystyle \frac{\partial }{\partial \theta }}\left(\sin \theta {\displaystyle \frac{\partial }{\partial \theta }}\right)+{\displaystyle \frac{\partial {}^2}{\partial {\varphi }^2}}\right]$

(4.93)

Again, we should be careful in interpreting what this means. In particular, the first $\partial /\partial \theta$ acts both on the $\sin \theta$ in round brackets, and also on whatever function the operator ${\widehat{𝐋}}^2$ hits.

Our goal is to find simultaneous eigenfunctions for ${\widehat{L}}_3$ and ${\widehat{𝐋}}^2$. We’ll ignore all $r$ dependence for now, and reconsider it in the following sections when we solve the Schrödinger equation. The ${\widehat{L}}_3$ eigenfunction is straightforward: we can take

$\psi (\theta ,\varphi )=Y(\theta )e^{im\varphi }$

(4.94)

for any function $Y(\theta )$. The wavefunction is single valued in $\varphi$, meaning $\psi (\theta ,\varphi )=\psi (\theta ,\varphi +2\pi )$, only if $m$ is an integer. Using (4.92), we see that the corresponding eigenvalues are

${\widehat{L}}_3{e}^{im\varphi }=m\mathrm{\hslash }{e}^{im\varphi }\mathit{\hspace{1em}\hspace{1em}}m\in 𝐙$

We learn that the angular momentum in the $z=x_3$ direction is quantised in units of $\mathrm{\hslash }$. So too, of course, is the angular momentum in any other direction. But, as we’ve stressed above, if we sit in an eigenstate of ${\widehat{L}}_3$, then we can’t also sit in an eigenstate of $𝐧\cdot \widehat{𝐋}$ for any direction $𝐧$ other than the $z$-direction, with $𝐧=(0,0,1)$.

Next we turn to the ${\widehat{𝐋}}^2$ eigenvalue equation. We keep the ansatz (4.94) and look for solutions

${\widehat{𝐋}}^{2}Y(\theta )e^{im\varphi }=\lambda Y(\theta )e^{im\varphi }$

Using our expression (4.93), this becomes

			$-{\displaystyle \frac{{\mathrm{\hslash }}^2}{{\sin }^2\theta }}\left[\sin \theta {\displaystyle \frac{\partial }{\partial \theta }}\left(\sin \theta {\displaystyle \frac{\partial }{\partial \theta }}\right)+{\displaystyle \frac{\partial {}^2}{\partial {\varphi }^2}}\right]Y(\theta )e^{im\varphi }=\lambda Y(\theta )e^{im\varphi }$
	$⟹$		$-{\displaystyle \frac{{\mathrm{\hslash }}^2}{{\sin }^2\theta }}\left[\sin \theta {\displaystyle \frac{\partial }{\partial \theta }}\left(\sin \theta {\displaystyle \frac{\partial }{\partial \theta }}\right)-m^2\right]Y(\theta )=\lambda Y(\theta )$		(4.95)

Solutions to this equation are a well-studied class of mathematical functions called as associated Legendre Polynomials. Their construction comes in two steps. First, we introduce an associated set of functions known as (ordinary) Legendre Polynomials, $P_l(x)$. These obey the differential equation

${\displaystyle \frac{d}{dx}}\left[(1-x^2){\displaystyle \frac{d{P}_l}{dx}}\right]+l(l+1)P_l(x)=0\mathit{\hspace{1em}\hspace{1em}\hspace{1em} }l=0,1,2,\mathrm{\dots }$

The $P_l(x)$, with $l=0,1,2,\mathrm{\dots }$ are polynomials of degree $l$. The eigenfunction solutions (4.95) are then associated Legendre polynomials $P_{l,m}(\cos \theta )$, defined by

$Y(\theta )=P_{l,m}(\cos \theta )={(\sin \theta )}^{|m|}{\displaystyle \frac{d^{|m|}}{d{(\cos \theta )}^{|m|}}}{P}_l(\cos \theta )$

The corresponding eigenvalue is

$\lambda =l(l+1){\mathrm{\hslash }}^2$

Because $P_l(\cos \theta )$ is a polynomial of degree $l$, you only get to differentiate it $l$ times before it vanishes. This means that the ${\widehat{L}}_3$ angular momentum is constrained to take values that lie in the range

$-l\le m\le l$

The requirement that $|m|\le l$ is simply telling us that the angular momentum in a any given direction (in this case $z$) can’t be more than the total angular momentum. When $m=+l$ (or $m=-l$) you should think of the object as having angular momentum as aligned (or anti-aligned) with the $z$-axis. When , you should think of the angular momentum as aligned in some intermediate direction. Of course, all of these thoughts should be quantum thoughts: even when $m=l$, if you measure that angular momentum in a perpendicular direction like ${\widehat{L}}_1$ then you’re not guaranteed to get zero. Instead there will be a probability distribution of answers centred around zero.

To build some intuition, here are some of the first spherical harmonics. The $l=0$ case is just a constant

$Y_{0,0}=1$

This is the case when a particle has no angular momentum and, correspondingly, its wavefunction has no angular dependence. In the context of atomic physics, it is referred to as an s-wave. The next collection of spherical harmonics all have total angular momentum $l=1$,

$Y_{1,-1}=\sin \theta e^{-i\varphi }\mathit{\hspace{1em}\hspace{1em} },Y_{1,0}=\cos \theta \mathit{\hspace{1em}\hspace{1em} },Y_{1,1}=\sin \theta e^{i\varphi }$

In the context of atomic physics, these are called p-wave states. The next, $l=2$ collection are d-waves,

$Y_{2,\pm 2}={\sin }^2\theta e^{\pm 2i\varphi }\mathit{\hspace{1em}\hspace{1em} },Y_{2,\pm 1}=\sin \theta \cos \theta e^{\pm i\varphi }\mathit{\hspace{1em}\hspace{1em} },Y_{2,0}=3{\cos }^2\theta -1$

There are various ways to visualise spherical harmonics. For example, in Figure 17,we plot the absolute value $|Y_{l,m}(\theta ,\varphi )|$ as the distance from the origin in the $(\theta ,\varphi )$ direction. Taking the absolute value removes the $e^{im\varphi }$ dependence, so these plots are symmetric around the $z$-axis. Alternatively, in Figure 18, we plot $\mathrm{Re}(Y_{l,m}(\theta ,\varphi ))$ on the sphere, with the value denoted by colour.

How does this look in polar coordinates? It’s best to work with the form of the Schrödinger equation (4.96) which is now

$-{\displaystyle \frac{{\mathrm{\hslash }}^2}{2m{r}^2}}{\displaystyle \frac{d}{dr}}\left(r^2{\displaystyle \frac{dR}{dr}}\right)+\left({\displaystyle \frac{1}{2}}m{\omega }^2{r}^2+{\displaystyle \frac{l(l+1){\mathrm{\hslash }}^2}{2m{r}^2}}\right)R=ER$

We’re now awash with various constants and life will be easier if we can get rid of a few. We make the same scaling (2.29) that we used for the 1d oscillator,

$y=\sqrt{{\displaystyle \frac{m\omega }{\mathrm{\hslash }}}}r\mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }\tilde{E}={\displaystyle \frac{2E}{\mathrm{\hslash }\omega }}$

The Schrödinger equation then becomes

${\displaystyle \frac{1}{y^2}}{\displaystyle \frac{d}{dy}}\left(y^2{\displaystyle \frac{dR}{dy}}\right)-\left(y^2+{\displaystyle \frac{l(l+1)}{y^2}}\right)R=-\tilde{E}\psi$

(4.99)

We’re going to solve it using the same kind of polynomial expansion that we used for the 1d harmonic oscillator. First, look at the large distance $y\to \mathrm{\infty }$ behaviour. Here the dominant terms are

${\displaystyle \frac{d^{2}R}{d{y}^2}}-y^{2}R\approx 0\mathit{\hspace{1em}\hspace{1em} }⟹\mathit{\hspace{1em}\hspace{1em} }R\approx e^{-y^2/2}\mathit{\hspace{1em}\hspace{1em} }\mathrm{as}y\to \mathrm{\infty }$

Next, look at the behaviour near the origin $y\ll 1$. Here the dominant terms in the equation are

${\displaystyle \frac{d^{2}R}{d{y}^2}}+{\displaystyle \frac{2}{y}}{\displaystyle \frac{dR}{dy}}-{\displaystyle \frac{l(l+1)}{y^2}}R\approx 0\mathit{\hspace{1em}\hspace{1em} }\mathrm{for}y\ll 1$

If we make the power-law ansatz $R\sim y^{\alpha }$, we find

$\alpha (\alpha -1)+2\alpha -l(l+1)=0\mathit{\hspace{1em}\hspace{1em} }⟹\mathit{\hspace{1em}\hspace{1em} }R\sim y^l\mathit{\hspace{1em}}\mathrm{as}y\to 0$

Here we’ve discarded the second solution $\alpha =-(l+1)$ on the grounds that the resulting wavefunction diverges at the origin. Putting this together, our ansatz for the general solution is

$R(y)=y^{l}h(y)e^{-y^2/2}$

where we take

$h(y)={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{a}_k{y}^k$

At this stage we’ve got a little bit of algebra to do, plugging our ansatz into the Schrödinger equation (4.99). Once the dust settles, you’ll find the result

${\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{a}_k\left(k(2l+k+1)y^{l+k-2}+(\tilde{E}-(2l+2k+3))y^{l+k}\right)=0$

(4.100)

We start our recursion relation by setting

$a_0=1$

This is just the statement that the small $y$ behaviour is indeed $R\sim y^l$ rather than some higher power. Now note that the first term in our equation will have term proportional to $a_1{y}^{l-1}$. But there’s nothing to cancel this in the second term. This means that we must set

$a_1=0$

As we’ll see, our recursion relation will then set $a_k=0$ for all odd $k$. To find this recursion relation, we shift the summation variable $k\to k+2$ in the first term in (4.100) so that all terms have the same power of $y^{l+k}$,

$\left[{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{a}_{k+2}(k+2)(2l+k+3)+{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{a}_k\left(\tilde{E}-(2l+2k+3)\right)\right]y^{l+k}=0$

For this to hold for all $y$, we must have the following recursion relation

$a_{k+2}={\displaystyle \frac{\tilde{E}-(2l+2k+3)}{(k+2)(2l+k+3)}}{a}_k$

At this stage, the argument is the same as we used for the 1d harmonic oscillator. If the recursion relation fails to terminate then $a_{k+2}\to 2a_k/k$ for large $k$, but that’s bad: it messes up the $e^{-y^2/2}$ behaviour at large $y$ and turns it into a non-normalisable $e^{+y^2/2}$ behaviour. This means that we must find solutions where the recursion relation terminates. This immediately allows us to read off the energy. If we are to have $a_{2q+2}=0$ (recall that all $a_k$ for odd $k$ already vanish) then the energy of the state must be

$E=\mathrm{\hslash }\omega \left({\displaystyle \frac{3}{2}}+l+2q\right)\mathit{\hspace{1em}\hspace{1em}\hspace{1em} }l,q=0,1,2,\mathrm{\dots }$

(4.101)

Let’s compare this to our Cartesian result (4.98). Clearly they agree, but the extra work we did in polar coordinates brings more information: it tells us how the energy depends on the total angular momentum $l$. In particular, the lowest energy state with angular momentum $l$ has $E=\mathrm{\hslash }\omega (\frac{3}{2}+l)$.

We also have a better understanding of the degeneracy of states. When $N$ is even, the states with energy $E=\mathrm{\hslash }\omega (\frac{1}{2}+N)$ comprise of all even angular momentum states $l=0,2,4,\mathrm{\dots },N$. Since there are $2l+1$ states (4.97) associated to each angular momentum $l$, the total number of states at level $N$ must be

$\text{Total Degeneracy}={\displaystyle \sum _{k=0}^{N/2}}4k+1=N\left({\displaystyle \frac{N}{2}}+1\right)+{\displaystyle \frac{N}{2}}+1={\displaystyle \frac{1}{2}}(N+1)(N+2)$

which agrees with the count from the Cartesian calculation. Similarly, when $N$ is odd the states at energy $E=\mathrm{\hslash }\omega (\frac{1}{2}+N)$ comprise of all odd angular momentum states $l=1,3,5,\mathrm{\dots },N$: again, the total count is $\frac{1}{2}(N+1)(N+2)$.

Consider the spherical finite potential well,

Clearly this is the 3d version of the 1d finite well that we considered in Section 2.3. The associated 3d radial Schrödinger equation (4.96) is

$-{\displaystyle \frac{{\mathrm{\hslash }}^2}{2m{r}^2}}{\displaystyle \frac{d}{dr}}\left(r^2{\displaystyle \frac{dR}{dr}}\right)+\left[V(r)+{\displaystyle \frac{l(l+1){\mathrm{\hslash }}^2}{2m{r}^2}}\right]R=ER$

Our strategy is to massage the Schrödinger equation into the same form we saw in 1d and then simply invoke some of our earlier results. The only thing we need to fix is the unusual looking derivative terms. But this is easily achieved: we redefine the radial wavefunction as

$\chi (r)=rR(r)$

(4.102)

Then, written in terms of the new function $\chi (r)$, the Schrödinger equation takes the 1d form

$-{\displaystyle \frac{{\mathrm{\hslash }}^2}{2m}}{\displaystyle \frac{d^2\chi }{d{r}^2}}+\left[V(r)+{\displaystyle \frac{l(l+1){\mathrm{\hslash }}^2}{2m{r}^2}}\right]\chi =E\chi$

(4.103)

However, we haven’t quite washed away all traces of our 3d origin. Some remnant survives in the boundary conditions that we must impose on $\chi$. First, the wavefunction must be normalisable. Since the $Y_{l,m}$’s don’t cause us any trouble, we’re left with the requirement

where the $r^2$ factor in the middle equation comes from the integration measure for polar coordinates. We see that the normalisability condition restricts the asymptotic $\to \mathrm{\infty }$ behaviour of $\chi (r)$ as it would for a 1d wavefunction. The only novelty is what happens at the origin $r=0$. Since $R(r)$ is finite at the origin, clearly we need

$\chi (0)=0$

(4.104)

The derivative ${\chi }^{\prime }(r=0)$ must then be finite.

Here we’ll look just for s-wave bound states, with $l=0$. In this case, the Schrödinger equation (4.103) is identical to those we solved for the 1d problem, but with the novelty that the coordinate $r\in [0,\mathrm{\infty })$ rather than the $x\in (-\mathrm{\infty },+\mathrm{\infty })$, together with the additional requirement (4.104).

There is a simple fix to both these issues. We solve the Schrödinger equation (4.103) pretending that $r\in (-\mathrm{\infty },+\mathrm{\infty })$, defining $V(-r)=V(+r)$. But we then look for odd solutions with

$\chi (-r)=-\chi (+r)$

This condition ensures that we have $\chi (0)=0$ as required. Moreover, we’re not missing any solutions. As we saw in Section 2, with an even potential all solutions are either even or odd, but the even solutions have $\chi (0)\ne 0$ and so are illegitimate for our present purposes.

This is all we need. We solved for the parity odd solutions in Section 2.3.1. In contrast to the parity even solutions, their existence is not guaranteed. They only arise when the potential is suitably deep or suitably wide, meaning (2.45)

${\displaystyle \frac{8m{V}_0}{{\mathrm{\hslash }}^2{\pi }^2}}>{\displaystyle \frac{1}{a^2}}$

This is our minor punchline. In 1d, there are potentials (like the finite well) where a bound state is always guaranteed. This is no longer the case in 3d where very shallow potentials do not house bound states.

Perhaps surprisingly, the case of $d=2$ turns out to be that same as $d=1$ with bound states guaranteed for certain classes of potentials, no matter how shallow. It is only in $d\ge 3$ that you need a deep, wide potential to trap a quantum particle.

All of this motivates us to look for solutions of the form

$R(r)=r^{l}f(r)e^{-r/a}$

(4.107)

where $f(r)$ is a polynomial

$f(r)={\displaystyle \sum _{k=0}}{c}_k{r}^k$

(4.108)

where we must have $c_0\ne 0$ so that we get the right behaviour at small $r$.

Now we’ve got to roll up our sleeves and substitute this into the Schrödinger equation to get a recursion relation for the coefficients $c_k$. It’s best to go slowly. First you can check that if you plug the ansatz (4.107) into (4.106), then the function $f(r)$ must satisfy

${\displaystyle \frac{d^{2}f}{d{r}^2}}+2\left({\displaystyle \frac{l+1}{r}}-{\displaystyle \frac{1}{a}}\right){\displaystyle \frac{df}{dr}}-{\displaystyle \frac{1}{ar}}\left(2(l+1)-\beta a\right)f=0$

Next, substitute the power-law ansatz (4.108) into this differential equation to find

${\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{c}_k\left[(k(k-1)+2k(l+1))r^{k-2}-{\displaystyle \frac{1}{a}}(2k+2(l+1)-\beta a)r^{k-1}\right]=0$

Finally, shift the dummy summation variable $k\to k-1$ in the second term to get a uniform power of $r^{k-2}$. This then gives us our desired recursion relation

$c_k={\displaystyle \frac{1}{ak}}{\displaystyle \frac{2(k+l)-\beta a}{k+2l+1}}{c}_{k-1}$

(4.109)

The argument that we now make follows that of the harmonic oscillator. Suppose that this recursion relation fails to terminate. Then, as $k\to \mathrm{\infty }$, we have

$c_k\to {\displaystyle \frac{2}{ak}}{c}_{k-1}$

But, as we explained in Section 2.2, this is the kind of expansion that we get from exponentially divergent functions of the form $f(r)=e^{2r/a}$. On a mathematical level, there’s nothing wrong with these solutions: they just change the asymptotic form of (4.107) from $e^{-r/a}$ to $e^{+r/a}$. But we’ve no use of such solutions for physics because they’re badly non-normalisable.

This means that, once again, we need to restrict attention to series that terminate. Which, in turn, means that there should be a positive integer $q$ for which $c_q=0$, while $c_k\ne 0$ for all . Clearly this holds only if $a$ takes special values,

$a={\displaystyle \frac{2}{\beta }}(q+l)\mathit{\hspace{1em}\hspace{1em}}\mathrm{with}q=1,2,\mathrm{\dots }$

Alternatively, since both $q$ and $l$ are integers, we usually define the integer

$n=q+l$

which obviously obeys $n>l$. We then have

$a={\displaystyle \frac{2}{\beta }}n$

All that’s left is to chase through the definition of the various variables in this expression. These can be found in (4.105). The length scale $a$ is related to the energy scale $E$ which is given by

$E=-{\displaystyle \frac{e^4{m}_e}{32{\pi }^2{ϵ}_0^2{\mathrm{\hslash }}^2}}{\displaystyle \frac{1}{n^2}}\mathit{\hspace{1em}\hspace{1em}}n=1,2,\mathrm{\dots }$

(4.110)

This is our final result for the energy spectrum of the hydrogen atom. The integer $n$ coincides with what we previously called the principal quantum number.

There’s a whole lot to unpick in this expression. First, there’s the ugly morass of constants sitting on the right-hand side. Clearly they set the energy scale of the bound states. There’s actually a better way to organise them to see what’s going on. To this end, we introduce the fine structure constant,

$\alpha ={\displaystyle \frac{e^2}{4\pi {ϵ}_0\mathrm{\hslash }c}}$

At first glance, you may think that this doesn’t help matters because, rather than simplifying things, we’ve introduced a new physical constant $c$, the speed of light. However, the advantage of the fine structure constant is that it’s a dimensionless quantity. As you can see, it includes the expression $e^2/4\pi {ϵ}_0$ familiar from the Coulomb force and it should be viewed as the convention-independent way to characterise the strength of electromagnetism. Moreover, it takes a value that’s easy to remember:

$\alpha \approx {\displaystyle \frac{1}{137}}\approx 0.0073$

Written in terms of the fine structure constant, the energy eigenvalues are

$E=-{\displaystyle \frac{{\alpha }^2{m}_e{c}^2}{2}}{\displaystyle \frac{1}{n^2}}$

This makes it clear what’s really setting the energy scale of the hydrogen atom: it is the rest mass energy of the electron! This is given by

$m_e{c}^2\approx 0.51\times {10}^6\mathrm{eV}$

A million electronvolts is an energy scale of relativistic particle physics. It is much larger than the typical energy scales of atomic physics but we can see why: there are two powers of the fine structure constant that interpolate between the two. The energy scale of the hydrogen atom is set by

$\mathrm{Ry}={\displaystyle \frac{{\alpha }^2{m}_e{c}^2}{2}}\approx 13.7\mathrm{eV}$

This is called the Rydberg. As you can see, the (negative) ground state energy of the hydrogen atom is exactly one Rydberg; the higher excited states have energy $E=-\mathrm{Ry}/n^2$. Note that the hydrogen atom has an infinite number of negative energy bound states.

Next there’s a surprise. The energy spectrum of hydrogen does not depend on the angular momentum $l$. At least, that’s almost true: it depends only in the sense that the states for a given energy $E=-\mathrm{Ry}/n^2$ only include angular momentum . We saw a similar phenomenon for the 3d harmonic oscillator in Section 4.2 and this might lead you to think that this is common. It’s not. In fact, the only two systems where the energy spectrum doesn’t explicitly depend on angular momentum are the harmonic oscillator and the hydrogen atom!

The angular momentum does affect the degeneracy of the spectrum. Recall that there are $2l+1$ states (4.97) associated to each total angular momentum $l$. These are the states labelled by $m=-l,\mathrm{\dots },+l$. For a fixed energy level $n$, we can have any total angular momentum $l=0,\mathrm{\dots }n-1$. This means that the degeneracy at level of the state at level $n$ is

$\text{Total Degeneracy}={\displaystyle \sum _{l=0}^{n-1}}(2l+1)=n^2$

The $n=1$ ground state is unique: it sits in $l=0$ s-wave. There are four $n=2$ excited states corresponding to the s-wave and three $l=1$ states of the p-wave. There are nine $n=3$ excited states, corresponding to the s-wave, the three $l=1$ p-wave states and five $l=2$ d-wave states. And so on.