# Correct quantum numbers interpretation?

1. Jun 24, 2012

### HJ Farnsworth

Greetings everyone,

I'm going through Griffiths QM, and just wanted to make sure I understood the roles (where they come from, etc.) of the quantum numbers n, l, and m. What is written below is basically an outline of my current understanding, of which I am seeking either a confirmation or correction of (please forgive any clumsy equations, I don't have my book with me and don't feel like jumping around the internet too much). At the end of this post is a quick summary of my interpretation of the quantum numbers and the actual questions I have.

Principle quantum number, n:

$\Psi$(r,t)=$\psi$(r)f(t)

If the potential is such that this separation meets the boundary condition of normalizability, then there will be infinitely many solutions (due basically to the fact that increasing the periodicity of the function can result in different solutions that are normalizable, eg., particle in a box), labelled by the principle quantum number, n. Each solution corresponds to a state of definite total energy, and $\Psi$ is the sum of each $\psi$ multiplied by the probability of that $\psi$:

$\Psi$(r,t)=$\sum$cn$\psi$n(r)f(t)

If, on the other hand, the potential is such that separation does not meet the boundary condition, then there are no states of definite total energy, the solutions are labelled by the continuous variable k=$\sqrt{2mE}$/$\hbar$. Instead of cn, a probability density function is used, and $\Psi$ is:

(1/$\sqrt{2π}$)∫$\phi$(k)$\psi$n(r)f(t)dk

Azimuthal and magnetic quantum numbers, l and m:

Start with the TISE. Assume spherically symmetric potential and attempt to solve using separation of variables:

$\psi$(r,$\theta$,$\varphi$)=R(r)Y($\theta$,$\varphi$)

Eventually, this turns into an equation of the form f(r)=g($\theta$,$\varphi$), so that both are equal to a constant, which we label l(l+1). Only the radial part of the equation has an energy term or a potential term, so its solution (and whether or not it turns out that doing separation of variables results in a legal solution) depends on the potential V(r). Assuming it does meet the boundary conditions, the radial solution is now dependent on two quantum numbers, so energy is still E=En, but now:

R(r)=Rnl(r)

For the angular equation, attempt separation of variables once more:

Y($\theta$,$\varphi$)=$\Theta$($\theta$)$\Phi$($\varphi$)

As before, this turns into an equation of the form h($\theta$)=j($\varphi$), so that both are equal to a constant, which we label m2. Solving the two differential equations, the solution to the total angular equation is the spherical harmonics, which from the way we labelled the separation constants, depend on both quantum numbers l and m:

Y($\theta$,$\varphi$)=Y$^{m}_{l}$($\theta$,$\varphi$)

So the solution to the TISE depends on all three quantum numbers:

$\psi$$^{m}_{nl}$(r,$\theta$,$\varphi$)=Rnl(r)Y$^{m}_{l}$($\theta$,$\varphi$)

Bringing angular momentum into the picture, it turns out that solutions to the TISE that are separable as described above represent states of definite L2 and Lz (so basically states where the total angular momentum and one of its components can be known with certainty):

$\widehat{H}$$\psi$$^{m}_{nl}$=E$\psi$$^{m}_{nl}$
$\widehat{L}$2$\psi$$^{m}_{nl}$=$\hbar$2l(l+1)$\psi$$^{m}_{nl}$
$\widehat{L}$z$\psi$$^{m}_{nl}$=$\hbar$m$\psi$$^{m}_{nl}$

Mirroring the principle quantum number construction, the wavefunction for a given energy level can be written as a summation over l and m of each $\psi$$^{m}_{nl}$, multiplied by the probability of being at the L2 labeled by l and the Lz labeled by m:

$\psi$n=$\sum$lmc$^{m}_{l}$Rnl(r)Y$^{m}_{l}$($\theta$,$\varphi$)

SUMMARY

$\Psi$(r,t)=$\psi$(r)f(t)$\Rightarrow$$\widehat{H}$$\psi$n=E$\psi$n, a state of definite total energy, indexed by quantum number n.

$\psi$(r,θ,$\varphi$)=R(r)Y($\theta$,$\varphi$)$\Rightarrow$$\widehat{L}$2$\psi$nl=$\hbar$2l(l+1)$\psi$nl, a state of definite L2, so basically a state of definite total angular momentum, indexed by quantum number l.

Y($\theta$,$\varphi$)=$\Theta$($\theta$)$\Phi$($\varphi$)$\Rightarrow$$\widehat{L}$z$\psi$$^{m}_{nl}$=$\hbar$m$\psi$$^{m}_{nl}$, a state of definite Lz, so basically a state where one component of the angular momentum is known with certainty, indexed by quantum number m.

QUESTIONS

1. What is k=$\sqrt{2mE}$/$\hbar$ called (it seems like there must be a name for it)?

2. If the potential is such that $\psi$(r,$\theta$,$\varphi$)=R(r)Y($\theta$,$\varphi$) ends up not satisfying the boundary condition (ie., a state of definite total energy without definite total angular momentum), what is the integral mirroring (1/$\sqrt{2π}$)∫$\phi$(k)$\psi$n(r)f(t)dk, and what is the variable mirroring k? So basically, could you please rewrite the equation immediately preceding the summary in what would be the appropriate integral form?

3. If the equation $\psi$(r,$\theta$,$\varphi$)=R(r)Y($\theta$,$\varphi$) satisfies the boundary condition, are there situations where separating the angular equation Y($\theta$,$\varphi$)=$\Theta$($\theta$)$\Phi$($\varphi$) will still end up being illegal (ie., are there situations where the total angular momentum can be known with certainty, but none of its components can?) If so, repeat question 2 with the angular separation in place of the TISE separation.

4. This is similar to question 3. In the 2nd and 3rd equations in my summary, I'm saying that as I understand it, if separating the TISE results in legal equations for the wavefunction, then the total angular momentum is known with certainty regardless to whether separating the angular equation results in legal equations for the wavefunction, whereas to know a component of L, you have to be able to separate the angular equation. Is this correct, or do you have to be able to separate both legally to know L2 with certainty?

-HJ Farnsworth

2. Jun 25, 2012

### VortexLattice

Well, that k is typically called the wave number.

3. Jun 25, 2012

### Ken G

The probability amplitude, or just amplitude, not "probability". The amplitude times its complex conjugate is the probability (I'll bet you know that).
I may be wrong here, but I don't think the passage to a continuous energy variable has all that much to do with separability or definite energy states, it's just whether the boundary conditions yield discrete E or not. We can still imagine states of definite total energy, they just require Dirac delta functions to write. Such functions are idealizations, but I would argue that so is any concept of a definite anything, as there are no perfect measurements. It's not a big deal, but to me, to get the f(t), it helps to imagine they multiply "states of definite E", i.e., the E operator still has eigenfunctions even when E is continuous.

On a minor note, in your notation, you would not want to use subscript n, as that implies a countably infinite basis. You could just use k (or E).
It is called "the wave number", which is a lot like the inverse of the "de Broglie wavelength" (though you have to divide k by 2pi before you take its inverse to get that). It is related to momentum measurements (though you have to multiply k by h-bar).
I don't know of any situations where angular momentum is continuous. It seems to me, the expansion you seek refers to the eigenvectors, which are the states of definite angular momentum. Angular momentum has units of action, and quantization of action, in bundles of h-bar, seems like a central postulate of quantum mechanics. So to not have discrete angular momentum eigenvectors, you would need something to prevent the integral around a loop to correspond to an action, like if you had an infinite wall blocking the integral around a loop-- in that case I doubt you'd want to talk about angular momentum eigenvectors at all. The states of definite energy and angular momentum is what you need, it doesn't seem all that important if they are continuous or not, and I can't think of any situation where you'd want to talk about angular momentum where it would not correspond to states of definite discrete values of angular momentum, though someone might correct me. Rather than thinking in terms of the potential, I'd advise thinking in terms of possible outcomes of a measurement. Each of those outcomes must have an eigenstate associated with it, and those are the basis vectors of any expansion you would make over the states of definite angular momentum. When angular momentum doesn't commute with energy, I doubt you'd be able to separate the variables at all, so the whole scheme would break down. You'd need to find quantum numbers of commuting operators.
Since L2 = L2x + L2y + L2z, you can see from the commutation relations that L2 always commutes with Lz, independently from any boundary conditions.
From the above, it seems you are making a false dichotomy. The search for a complete set of quantum numbers involves a set of commuting operators, and if L2 commutes with E, then so does Lz.

Last edited: Jun 25, 2012
4. Jun 25, 2012

### Staff: Mentor

On analogy with ω, the angular [temporal] frequency (which has units of radians per second), a better name for k (which has units of radians per meter) would probably be "angular spatial frequency." But wave number (or wavenumber) is the accepted term.

5. Jun 25, 2012

### HJ Farnsworth

Thanks very much for the responses, they're very helpful.

Ken G, one question:

Ignoring the k stuff, wouldn't the number of energy states (and corresponding solutions to the TISE) be countably infinite, one for each n starting at n=1, and so serve as a countably infinite basis in constructing the solution to the TDSE? If not, could you please briefly explain why?

Again, many thanks.

-HJ Farnworth

6. Jun 25, 2012

### Ken G

It depends on the boundary conditions. You are right for a potential well. But a free particle has uncountably many energy eigenstates (any real frequency, basically). Still, you could put them all in a huge box and just say there's no such thing as a free particle, it wouldn't really change anything because you'd still end up approximating the discrete sums over n with continuous integrals over E. Continuous integrals are just easier to do in a lot of cases, you only want discrete n if there are just a few eigenstates of interest, as in an atom or a square well. There is a kind of hybrid case, between discrete and continuous energies, called a "Rydberg atom," where the n of the valence electron is so large that its energies are nearly continuous.

Last edited: Jun 25, 2012
7. Jun 25, 2012

### HJ Farnsworth

Ken G,

Ah, I see your point. Earlier, when you said that I shouldn't use the subscript since it implies a countably infinite basis, I incorrectly thought you meant that I should deal with just one value for E or k (so a finite basis)- but you were in fact communicating the opposite, namely, that the basis is uncountably infinite since there is an eigenstate for every real number value of E or k.

Further, you're saying integrals end up being much more easily calculable and therefore much more useful than infinite or enormously large sums, analogous to how, in introductory E&M, integrals are used to approximate the superposition principle using a charge density in a material rather than adding the contribution from each individual charged particle to form a discrete sum, since there are enough charged particles that are densely packed to make the integral and the sum approximately equal, but only the integral is an actual feasible calculation to perform.

Correct?

Thanks again.

-HJ Farnsworth

8. Jun 25, 2012

### Dickfore

To avoid the problem of uncountably many possible values for k for a free particle, one usually places the particle in a box with side L (very large), and imposes the boundary condition that the value of the wave function, as well as its derivative to be the same on the opposite boundaries.

The possible stationary states are:
$$\psi_n(x) = \frac{1}{\sqrt{L}} \, e^{i \, k_n \, x}, \ k_n = n \, \frac{2 \pi}{L}, n = \ldots, -1, 0, 1, \ldots$$
Thus, the "distance" between k-values for successive states are:
$$\Delta k =\frac{2 \pi}{L}$$
Or, put differently, there are approximately:
$$\Delta N = L \, \frac{\Delta k}{2 \pi}$$
stationary states in the wave vector interval $(k, k + \Delta k)$. As you can see, this number scales with the length of the box. For an infinite box, you would get infinitely many states. It is customary to take the k-interval to be infinitesimal. Then, the number of states is infinitesimal as well:

This generalizes to D dimensions:
$$dN = V_{D} \, \frac{d^{D}k}{(2\pi)^{D}}, \ V_{D} = L^{D}$$

9. Jun 25, 2012

### Ken G

Yes exactly-- we go back and forth between discrete and continuous kinds of sums at will, depending which is easier to do and what is the desired level of precision.

10. Jun 25, 2012

### HJ Farnsworth

Great.

Thanks for the responses again everyone.

Dickfore, I like that way of looking at it - it's different than the one Griffiths uses, and it's easy to see how it relates to the switching between integrals and summations. Just curious, is there a standard introductory text that lays it out in that perspective?

-HJ Farnsworth

11. Jun 26, 2012

### Dickfore

Yes, of course. You may consider any text on Statistical Physics. This kind of calculation is performed when considering blackbody radiation.

Another topic where a similar calculation is performed is in the Free Electron Model in metals. Look, for example, in

C. Kittel, Introduction to Solid State Physics

Chapter 6, Free Electron Fermi Gas
Look for the sections "Energy Levels in One Dimension", and "Free Electron Gas in Three Dimensions".

If you want a more formal approach, where the transition from summation to integration is performed, you may consult

Ashcroft, Mermin, Solid State Physics

A similar calculation is performed in texts on Scattering in Quantum Mechanics, where one needs to find the number of final free states in a unit solid angle around a particular direction in order to calculate the differential cross-section.

12. Jun 26, 2012

### HJ Farnsworth

Excellent, thank you.

-HJ Farnsworth