I Quantum numbers for energy levels

[TheCelt]

Hello

Can some one explain how you work out the combinations of quantum numbers for infinite wells in higher dimensions?

For example if i have an energy level $$E_4$$ In a 2D well, then for quantum numbers does this mean the combinations allowed must be:

$$4^2 + 1^2$$
$$1^2 + 4^2$$

So then you have:

$$ E_4 = 17 \pi^2 \hbar^2 / 2mL^2 $$

Or this not how it works ?

 

[PeroK]

Hello

Can some one explain how you work out the combinations of quantum numbers for infinite wells in higher dimensions?

For example if i have an energy level $$E_4$$ In a 2D well, then for quantum numbers does this mean the combinations allowed must be:

$$4^2 + 1^2$$
$$1^2 + 4^2$$

So then you have:

$$ E_4 = 17 \pi^2 \hbar^2 / 2mL^2 $$

Or this not how it works ?

Why would that be $E_4$? What are $E_1, E_2, E_3$?

 

[TheCelt]

I'm not given that info, I'm trying to find the difference in energy to the ground state. Why would it not be E4?

 

[PeroK]

I'm not given that info, I'm trying to find the difference in energy to the ground state. Why would it not be E4?

If you say it's $E_4$ it's up to you to justify why. I think it's $E_6$.

 

[TheCelt]

Well i presmued for an energy level $$E_n$$ there must be one quantum number that must be n the others are 1 ? Otherwise I am confused on how the quantum numbers increase for each energy level ?

 

[PeroK]

Well i presmued for an energy level $$E_n$$ there must be one quantum number that must be n the others are 1 ? Otherwise I am confused on how the quantum numbers increase for each energy level ?

What source are you using to learn this? In general, you have all combinations of energy states in each dimension: $E_{n_x} + E_{n_y}$. For a square well, the energy levels will be: $E_1 + E_1, E_1 + E_2, E_2 + E_2, E_1 + E_3 \dots$.

Note also that the eneregy levels $E_1 + E_2$ etc. are degenerate, as you can also have $E_2 + E_1$.

 

[vanhees71]

I do not understand the notation to begin with. To solve the energy eigenvalue problem completely you need to find a complete set of orthonormal eigenvectors of the Hamiltonian and the corresponding eigenvalues.

If I understand it right, you discuss here the infinite square potential for a particle in two dimensions, i.e., a particle restricted to $(x,y) \in [0,a] \times [0,b]$. Obviously the most simple complete set of energy eigenvectors are the momentum eigenstates, which are given by
$$u_{i,j}(\vec{x})=A \sin(k_{xi} x) \sin(k_{yj} y), \quad k_{xi}=\frac{\pi}{a} i, \quad k_{yj} = \frac{\pi}{b} j, \quad i,j \in \mathbb{N}=\{1,2,\ldots \}.$$
The $A$ can be chosen to normlize the eigenfunctions and make the a complete orthonormal set of energy eigenvectors.

The corresponding energy eigenvalues are labelled with the two indices $i$ and $j$:
$$E_{i,j}=\frac{1}{2 m} \hbar^2 (k_{xi}^2+k_{yj}^2).$$

 

[jtbell]

Yes, for e.g. a 2-dimensional infinite square well you don't have E₁, E₂, etc. Instead you have E_1,1, E_1,2, E_2,1, E_2,2, E_3,1, E_3,2, etc.

 

[Vanadium 50]

Yes, for e.g. a 2-dimensional infinite square well you don't have E1, E2, etc. Instead you have E1,1, E1,2, E2,1, E2,2, E3,1, E3,2, etc.

Well...

The way you do it is the way I do it, but not the way everyone would do it. You can have (in the 2D case)

E₀ = 2, which is the (1,1) case.
E₁ = 5, which is a doublet (2-fold degenerate), (1,2) and (2,1)
E₂ = 8, again a singlet: (2,2)
E₃ = 10, which is another doublet, (1,3) and (3,1)
E₄ = 13, which is another doublet, (2,3) and (3,1)

While this looks pointless for the infinite square well, it makes more sense for the hydrogen atom and the simple harmonic oscillator.

 

[vanhees71]

Well, labelling in this way in the degenerate case is not unique, which is why it's called degenerate. If you want the energy eigenstates and the energy eigenvalues are degnerate, you can always find other observables than the energy, which are conserved and compatible to each other. If you have enough such observables (in the ideal case you have a complete compatible set of observables) the energy eigenstates are labelled uniquely and it's also useful to label the energy eigenvalues by these unique "labels".

E.g., for the hydrogen atom you have $(n,\ell,m,\sigma_3)$ (main quantum number, angular momementum, magnetic quantum number, and spin projection). Here the energy eigenvalues are $2n^2$-fold degenerate (in the most simple approximation in QM 1 neglecting fine structure etc).

 

[Vanadium 50]

But by labeling the hydrogen atom that way, you've made a choice on how to think about the problem: spherical coordinates. Everybody knows the natural coordinates are parabolic.

 

[jtbell]

E₁ = 5, which is a doublet (2-fold degenerate), (1,2) and (2,1)

I see now that the OP used only one L, which presumably indicates that L_x = L_y. I was thinking of the more general case in which L_x ≠ L_y and the levels are not degenerate.

 

[vanhees71]

But by labeling the hydrogen atom that way, you've made a choice on how to think about the problem: spherical coordinates. Everybody knows the natural coordinates are parabolic.

Sure, that's natural if you have a degeneracy. Which additional observables you choose to uniquely define the energy eigenstates is arbitrary, but of course the physics doesn't change. You get from one description to the other by a unitary transformation between the resulting orthonormal bases of energy eigenvectors.