# Show that the minimum energy value for a given $l$ increases as $l$ increases

"Consider a particle in a central field and assume that the system has a discrete spectrum. Each orbital quantum number $l$ has a minimum energy value. Show that this minimum value increases as $l$ increases."

First, couldn't one just say that we know that for a given $l$, the lowest energy is (for hydrogen) $\require{enclose} \enclose{horizontalstrike}{-13.6 / (l+1)^2}$ eVolts?

But in their (Schaum's) solution they insist on doing it with integrals..

$\enclose{horizontalstrike} {\hat{H} = \dfrac{-\hbar^2}{2mr^2} \frac{d}{dr} (r^2 \frac{d}{dr}) + \dfrac{\hbar^2}{2m} \dfrac{l(l+1)}{r^2} + V(r)}$

$\enclose{horizontalstrike} {\hat{H_1} = \dfrac{-\hbar^2}{2mr^2} \frac{d}{dr} (r^2 \frac{d}{dr}) + V(r)}$

$\enclose{horizontalstrike} {\hat{H} = \hat{H_1} + \dfrac{\hbar^2}{2m} \dfrac{l(l+1)}{r^2}}$

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The minimum energy value for $l$ is

$\enclose{horizontalstrike} {E_l = \iiint \Psi_{l}^* (\hat{H_1} + \dfrac{\hbar^2}{2m} \dfrac{(l)(l+1)}{r^2}) \Psi_{l} \:\: dr \: d\theta \: d\phi}$

The minimum energy value for $l+1$ is

$\enclose{horizontalstrike} {E_{l+1} = \iiint \Psi_{l+1}^* (\hat{H_1} + \dfrac{\hbar^2}{2m} \dfrac{(l+1)(l+2)}{r^2}) \Psi_{l+1} \:\: dr \: d\theta \: d\phi}$

(These $\Psi_{l}$ and $\Psi_{l+1}$ apparently encapsulate any possible value of $n$.)

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By splitting up $\enclose{horizontalstrike} {(l+1)(l+2) = (2l+2) + (l)(l+1)}$,

$\enclose{horizontalstrike} {E_{l+1} = \iiint \Psi_{l+1}^* (\dfrac{\hbar^2}{2m} \dfrac{(2l+2)}{r^2}) \Psi_{l+1} \:\: dr \: d\theta \: d\phi} \\ + \enclose{horizontalstrike}{\iiint \Psi_{l+1}^* (\hat{H_1} + \dfrac{\hbar^2}{2m} \dfrac{(l)(l+1)}{r^2}) \Psi_{l+1} \:\: dr \: d\theta \: d\phi}$

The first term is always positive because $\Psi_{l+1}^* \Psi_{l+1}$ is always positive and $(2l+2)$ is always positive.

Then if we can show that the second term is $\ge$

$\enclose{horizontalstrike} {\iiint \Psi_{l}^* (\hat{H_1} + \dfrac{\hbar^2}{2m} \dfrac{(l)(l+1)}{r^2}) \Psi_{l} \:\: dr \: d\theta \: d\phi}$

, we are done.

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So to show that $\enclose{horizontalstrike} {\iiint \Psi_{l+1}^* (\hat{H_1} + \dfrac{\hbar^2}{2m} \dfrac{(l)(l+1)}{r^2}) \Psi_{l+1} \:\: dr \: d\theta \: d\phi \ge} \\ \enclose{horizontalstrike}{\iiint \Psi_{l}^* (\hat{H_1} + \dfrac{\hbar^2}{2m} \dfrac{(l)(l+1)}{r^2}) \Psi_{l} \:\: dr \: d\theta \: d\phi}$

I split it up into showing that:

(1) $\enclose{horizontalstrike} {\iiint \Psi_{l+1}^* (\hat{H_1}) \Psi_{l+1} \:\: dr \: d\theta \: d\phi} \ge \\ \enclose{horizontalstrike}{\iiint \Psi_{l}^* (\hat{H_1}) \Psi_{l} \:\: dr \: d\theta \: d\phi}$

and

(2) $\enclose{horizontalstrike} {\iiint \Psi_{l+1}^* (\dfrac{\hbar^2}{2m} \dfrac{(l)(l+1)}{r^2}) \Psi_{l+1} \:\: dr \: d\theta \: d\phi} \ge \\ \enclose{horizontalstrike}{\iiint \Psi_{l}^* (\dfrac{\hbar^2}{2m} \dfrac{(l)(l+1)}{r^2}) \Psi_{l} \:\: dr \: d\theta \: d\phi}$

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I rewrite (1) as

$\enclose{horizontalstrike} {\iiint R_{l+1}^* Y_{l+1}^* (\hat{H_1}) R_{l+1} Y_{l+1} \:\: dr \: d\theta \: d\phi} \ge \\ \enclose{horizontalstrike}{\iiint R_{l}^* Y_{l}^* (\hat{H_1}) R_l Y_l \:\: dr \: d\theta \: d\phi}$

Since $\hat{H_1}$ depends only on $r$,

$\enclose{horizontalstrike} {\int R_{l+1}^* (\hat{H_1}) R_{l+1} \iint Y_{l+1}^* Y_{l+1} \:\: dr \: d\theta \: d\phi \ge} \enclose{horizontalstrike} {\int R_{l}^* (\hat{H_1}) R_{l} dr \iint Y_{l}^* Y_{l} \:\: dr \: d\theta \: d\phi}$

Due to normalization,

$\enclose{horizontalstrike} {\int R_{l+1}^* (\hat{H_1}) R_{l+1} \cdot 1 \: dr \ge \int R_{l}^* (\hat{H_1}) R_{l} \cdot 1 \: dr}$

I think the above is true but I'm kind of stuck.

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Going on, I rewrite (2):

$\enclose{horizontalstrike}{\iiint R_{l+1}^* Y_{l+1}^* (\dfrac{\hbar^2}{2m} \dfrac{(l)(l+1)}{r^2}) R_{l+1} Y_{l+1} \:\:dr \:d\theta \:d\phi \ge \iiint R_{l}^* Y_{l}^* (\dfrac{\hbar^2}{2m} \dfrac{(l)(l+1)}{r^2}) R_{l} Y_{l} \:\:dr \:d\theta \:d\phi}$

and through a similar argument about normalization of $Y$

$\enclose{horizontalstrike} {\int R_{l+1}^* (\dfrac{\hbar^2}{2m} \dfrac{(l)(l+1)}{r^2}) R_{l+1} \:dr \:\ge\: \int R_{l}^* (\dfrac{\hbar^2}{2m} \dfrac{(l)(l+1)}{r^2}) R_{l} \:dr}$

which again seems to be fair, but I'm stuck.

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(aug.29,2015)

Their solution:

Start by defining

$E_{min.}^{l_0} \equiv min_{l=l_0} ( \iiint \Psi_{l_0}^* [\hat{H_1} + \dfrac {\hbar^2}{2m} \dfrac {l_0(l_0+1)}{r^2}] \Psi_{l_0} \:dr\:d\theta\:d\phi )$

where $l_0$ is just some fixed value for $l$, for example 3.

So we have defined $E_{min.}^{l_0}$ to be "out of all the possible wave functions with $l=l_0$, pick the one with the lowest energy, calculate the displayed integral, and let that scalar value be known as $E_{min.}^{l_0}$."

Then compare to this expression:

$\iiint \Psi_{l_0+1}^* [\hat{H_1} + \dfrac {\hbar^2}{2m} \dfrac {l_0(l_0+1)}{r^2}] \Psi_{l_0+1} \:dr\:d\theta\:d\phi$

(i.e., the same expression but just with now we are letting the wave function be any wave function with orbital quantum number $l_0+1$ instead of $l_0$.)

They say that "The minimum eigenvalue of the Hamiltonian $\hat{H} = \hat{H_1} + \dfrac {\hbar^2}{2m} \dfrac {l_0(l_0+1)}{r^2}$ corresponds to the eigenfunction $\Psi_{l_0}$. Thus, the second expression $> E_{min.}^{l_0}$." Which would complete the proof, but I interject-

Certainly the operator $\hat{H} = \hat{H_1} + \dfrac {\hbar^2}{2m} \dfrac {l_0(l_0+1)}{r^2}$ is not the Hamiltonian for a wave function with orbital number $(l_0+1)$. So why couldn't that operator, when operating on $\Psi_{l_0+1}$ in the triple integral expression, produce some "New, Small" wave function $\Psi_{New,Small}$ such that

$\iiint \Psi_{l_0+1}^* \Psi_{New,Small} \:dr\:d\theta\:d\phi$

actually drops below $E_{min.}^{l_0}$?

• Hi. First, couldn't one just say that we know that for a given l, the lowest energy is (for hydrogen).... Why would you say that, i.e., they are looking for a general solution, you seem to want to tie it specifically to Hydrogen. Sorry if I have misunderstood your reasoning. – user81619 Aug 27 '15 at 17:56
• I mean that it seems like we could reason that since the energy of a particular level is proportional to $-1/n^2$, then certainly the "minimum energy level for a given $l$" must increase as $l$ increases. I worded my argument confusingly. – a00 Aug 27 '15 at 18:10
• But $n$ and $l$ are not the same quantum numbers. – march Aug 27 '15 at 18:27
• I see what you mean about why we can't use the hydrogen atom solutions to show this. "Central potential" may not refer to a hydrogen-atom-like (i.e.point charge) system. – a00 Aug 28 '15 at 12:56
• This is not an answer, but Jan Jensen on chemistry.stackexchange points out that n is the number of nodes in a wave function. This is why n >= l+1. Assuming that this holds even if the V(r) central potential is NOT as simple as a point charge at the origin, then this tells us why "as l increases, the n giving the minimum energy must follow suit." Put another way, we know that energy depends on the "fast oscillation-ness" of the wave function. Faster oscillations (more curvature of the wave function when you project out from r=0 to r=infinity, – a00 Aug 30 '15 at 15:21

Because the potential only depends on $r$, it can be shown that the wave function is separable into $\Psi=RY$. (This might be shown by David Miller in his online quantum mechanics course at Stanford in his lectures in section 7 on the hydrogen atom.)
Then we can say that the $Y(\theta,\phi)$ portion is the spherical harmonics, same as for the hydrogen atom. (This just arises out of the form of the 3-d Laplacian: when you separate the equation and apply the Hamiltonian this is just what the solutions that you get are.)
Now the spherical harmonics have eigenvalues $\sqrt{(l)(l+1)}$ in the Hamiltonian. What this is saying is that the spherical harmonics, the more kinky they get (higher frequency oscillation as you go around the sphere) the bigger the eigenvalue, which translates (when you're talking about the Hamiltonian) into monotonically strictly greater energy.
So this gives an idea of why the radial portion "doesn't matter." Whatever that minimal energy radial function looks like, it's treated separately from the angular portion (thanks to separability of the wave function) and does its own contribution to the energy (via the Hamiltonian) and so we only need to look at the angular portion of the Hamiltonian to judge whether or not an increase in $l$ will lead to strictly greater energy.