How exactly do quantum numbers increase in relation to energy levels in more than one dimension?

When you have increasing energy levels in 2 or 3 dimensions how does the values of quantum numbers $$n$$ increase for each dimension?

For example if you have ground state $$E_1$$ then you have for $$(n_x,n_y,n_z)$$ is $$(1,1,1)$$ in 3 dimensions.

So does this mean for energy $$E_2$$ you have $$(2,2,2)$$ ?

I've only learnt it in 1 dimension so i don't understand how it works in higher dimensions?

• For which class of systems? What is the Hamiltonian? What are $n_{x,y,z}?$
– user87745
Commented Nov 15, 2020 at 5:20
• Oh this was in relation to an infinite square (cube) well. I was just unclear on how the quantum numbers increase as energy levels increase.
– WDUK
Commented Nov 15, 2020 at 5:24

For a particle in a box (also known as a particle in an infinite potential well) in $$d$$ dimensions, the Hamiltonian inside the box is given by $$\hat{H}=\frac{1}{2m}\sum_{i=1}^d{p}^2_i$$where $$p_i$$ is the momentum operator in the $$i^{\text{th}}$$ direction. As you already know (I infer so from your question) that using the boundary conditions, we arrive at the conclusion that $$p_i$$ must be of the form $$p_i=\frac{n_i\pi}{L_i}$$where $$L_i$$ is the length of the box in the $$i^{\text{th}}$$ direction and $$n_i\in\mathbb{N}$$. Thus, the energy of an eigenstate is given by $$E=\frac{\pi^2}{2m}\sum_{i=1}^d\frac{n_i^2}{L_i^2}$$.

Now, in the case of one--dimensional box, $$E$$ only depends on $$n_x$$ and is monotonically increasing with $$n_x$$. This means that the spectrum of a one--dimensional particle in a box is non--degenerate, and the value of the energy depends only on one quantum number, namely, $$n_{x}$$. One can simply write that $$E=E_{n_x}$$

However, in the case of a multidimensional box, $$E$$ depends on the multiple quantum numbers, namely, $$\{n_i\}$$ where $$i>1$$. And, we have to write $$E=E_{\{n_i\}}$$ to signify that $$E$$ depends on multiple quantum numbers.

Moreoever, multiple different sets of values of $$n_i$$ can produce the same value of energy. For example, let's consider the case where $$L_i=L$$ and $$i=3$$. All the following configurations of $$n_i$$ produce the same value of energy:$$n_x=2,n_y=1,n_z=1$$ $$n_x=1,n_y=2,n_z=1$$ $$n_x=1,n_y=1,n_z=2$$ Thus, we see that the spectrum of a particle in a box can be degenerate in a higher dimension.

So, to answer your question, the first excited energy state in a three-dimensional particle in a box (with equal sides) can correspond to all of the above configurations of quantum numbers.

• When i read this article: tinyurl.com/yxpvz8br Table 3.9.1 about half way down seems to suggest you cannot have $n=0$ in any of the dimensions for $E_1$ that you are suggesting ? It starts at $(1,1,1)$
– WDUK
Commented Nov 15, 2020 at 6:04
• @WDUK Yes, I apologize, my bad. As I already had written, the $n$s should be natural numbers. If any of them is $0$ then the eigenfunction is just zero which is not normalizable and thus, not physical. I have corrected my examples.
– user87745
Commented Nov 15, 2020 at 6:08

So does this mean for energy $$E_2$$ you have $$(2,2,2)$$?

This is wrong! The 3-Dimensional box (and 2-Dimensional) have degeneracy is energy state that there 2 or more level for which the energy is same. Like for $$E_2$$, We have $$(2,1,1),(1,2,1),(1,1,2)$$.

Note that for 3-D box, The energy given by $$E_{n_1,n_2,n_3}=(n^2_1+n^2_2+n^2_3)\frac{\pi^2\hbar^2}{2mL^2}$$

• So is the general rule that for an increase in $E_n$ you get only one increase in quantum number but it can be in any combination of those 3 dimensions as a degeneracy ? So the sum of quantum numbers in $d$ dimensions for some energy $E_n = d+(n-1)$ ?
– WDUK
Commented Nov 15, 2020 at 5:46