The Hamiltonian is given as

$H=-\frac{\hbar^2}{2 m_e r^2}\displaystyle\sum_{n=1}^N \dfrac{\partial^2}{\partial \theta_n^2}$

In the first part we show that the $\psi_k=\frac{1}{\sqrt{2\pi}}\exp(i k \theta)$ is a solution to $H\psi_k = E_k \psi_k$.

$E_k=\frac{k^2 \hbar^2}{2 m_e r^2}$

All fine so far.

$r$ is given to be $4 \times 10^{-10}\text{m}$ and $N=18$ and it asks for the ground state energy.

I am unsure how the $k=0$ state can be occupied since zero energy is forbidden?

But for each quantum number $k=0,\pm 1,\pm 2\cdots$ we have an increasing energy so the ground state will consist of


Where $(k=0)^2$ denotes 2 electrons being in the state with $k=0$, one with spin up and one with spin down.

The corresponding energy: $\displaystyle\sum_{k=-4}^4 2E_k = \frac{60 \hbar^2}{m_e r^2} = 4.57 \times 10^{-18} \text{J}$

But in the solution the energy is given as $\frac{30 \hbar^2}{m_e r^2} = 2.29 \times 10^{-18} \text{J}$, half of my value. Is this an error or have I missed something?


2 Answers 2


You're double counting the degenacy in $E_{k}$ by multiplying by $2$ and summing from $-4$ to $4$. Either you sum from $0$ to $4$ $2E_{k}$ or you sum from $-4$ to $4$ $E_{k}$ but not both

  • $\begingroup$ Are you saying that $\pm k$ having the same energy is equivalent to spin-up and spin-down? There are 2 electrons in the $k=-1$ state so surely for $k=-1$ the energy is $2\times E_{-1}$ just as there are two electrons in the $k=+1$ state, spin up and spin down so the energy is $2 E_1$ for $k=1$? $\endgroup$
    – shilov
    Commented Mar 25, 2013 at 16:19
  • $\begingroup$ Yes, in this level of calculation your energy depends on $k^2$ and $k^2$ only, as you can see in the $E_{k}$ explicit expression $\endgroup$
    – J L
    Commented Mar 25, 2013 at 16:21
  • $\begingroup$ I find this strange because I didn't think the constraint was on the energy but on the quantum numbers... here the numbers are $k$ and $S_z$? Perhaps I don't understand degeneracy correctly! $\endgroup$
    – shilov
    Commented Mar 25, 2013 at 16:25

I cannot comment on your question because I have too few credits - so I am posting this as an answer.

I am unsure how the $k=0$ state can be occupied since zero energy is forbidden?

  1. This appears to be the free particle hamiltonian - on the circumference of a circle of radius $r$ - the state with zero energy is not forbidden - you might be confusing this with the Harmonic Oscillator problem. The problem admits a continuous spectrum of $k$ as solutions.

  2. Given that Hamiltonian is in 2D polar coordinates - the particles should not have any spin states here .. unless this is a part of a larger 3D problem. Consult your textbook again.

  3. @Nivalth: the degeneracy OP considers is a spin degeneracy and not the $k\rightarrow E$ jacobian value - but you got the math correct.


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