Suppose there is a many-body eigenvalue problem $$ H(r_1,\cdots,r_n) = - \frac{1}{2} \nabla^2|_{r_1\cdots,r_n} + V(r_1,\cdots,r_n). $$ This potential is symmetric and the hamiltonian does not depend on spin, but the solution should be for spin 1/2 particles. I am struggling to interpret the symbol $\sigma_i$ mathematically. Perhaps you can help?

It seems many authors take the «non-degenerate ground-state electron density» to be the following symbol $$ n(r) = \sum_{\sigma_1} \sum_{\sigma_2} \cdots \sum_{\sigma_n} \int |\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n)|^2 dr_2 \cdots dr_n. $$ In that context, I will present 3 scenarios of how $\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n)$ might be related to the hamiltonian. Under each scenario, I have entered a few questions. It would be helpful if you could indicate which scenario is correct, and answer the corresponding questions under the correct scenario.

Scenario (1): For every $r_1,\sigma_1,\cdots,r_n,\sigma_n$, $$ H\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n) = E_{\sigma_1,\cdots, \sigma_n} \psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n), $$ and $E_{\sigma_1,\cdots, \sigma_n}$ is different for every combination of $\sigma_1,\cdots,\sigma_n$. (a) Why do all $\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n)$ appear in the ground state density, when the ground state can only be one of the $\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n)$ (by virtue of nondegeneracy)? Do we simply define that the $2^n$ smallest eigenvalues correspond to the ground state? (b) Does $\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n)$ necessarily factor into a spatial part and a spin part for each $\sigma_1,\cdots,\sigma_n$? (c) Is there reason to expect that $\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n)$ antisymmetric for every $\sigma_1,\cdots,\sigma_n$?

Scenario (2) For every $r_1,\sigma_1,\cdots,r_n,\sigma_n$, $$ H\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n) = E_{\sigma_1,\cdots, \sigma_n} \psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n), $$ but $E_{\sigma_1,\cdots, \sigma_n}$ is the same for every $\sigma_1,\cdots, \sigma_n$. (a) Is there any reason to expect that the eigenspace of $E_{\sigma_1,\cdots, \sigma_n}$ is at least $2^n$ or eactly $2^n$? (b) Do physicist say that the ground state is degenerate if and only if the dimension of the eigenspace is larger than $2^n$? (c) Is there reason to expect that $\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n)$ antisymmetric for every $\sigma_1,\cdots,\sigma_n$?

Scenario (3): There a function $\psi$ which is the ground state that is a formal linear combination of terms $$ \psi(r_1,\cdots,r_n) = \sum_{\sigma_1} \sum_{\sigma_2} \cdots \sum_{\sigma_n} \psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n), $$ such that $H\psi = E\psi$, but the individual terms $\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n)$ no not necessarily solve this eigenvalue problem for any $r_1,\sigma_1,\cdots,r_n,\sigma_n$. If this interpretation is correct, I have some questions again: (a) In this case the functions $\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n)$ do not correspond to different eigenvalues for each $\sigma_1,\sigma_2,\cdots,\sigma_n$, so I cannot see a reason why they would be orthogonal. Are they antisymmetric or even normalised for some reason? (b) Is the function $\psi(r_1,\cdots,r_n)$ antisymmetric or normalised in this case? (c) What 'gained' by writing the ground state as as sum of these $2^n$ functions? (d) How can I arrive at the expression for the electron density in this case?

Hoping for some answers!

  • $\begingroup$ Why is the scenario where $E$ doesn't depend on $\sigma$ in exactly the same way that it doesn't depend on $r$ not in this list? $\endgroup$
    – jacob1729
    Commented Jul 12, 2020 at 13:20
  • $\begingroup$ In dealing with electric charges, ρ normally represents the charge density (charge/volume), which can vary from point to point. $\endgroup$
    – R.W. Bird
    Commented Jul 12, 2020 at 17:15
  • $\begingroup$ @jacob1729 That's interesting. Could you give me a couple more hits, and I'll write it in of course. $\endgroup$
    – Mikkel Rev
    Commented Jul 12, 2020 at 17:54

1 Answer 1


Perhaps a very concrete example will be helpful:

Consider the ground state of a hydrogen atom, that is, the electron in a $1s$ orbital. The spatial wavefunction is:

$$\psi(r) = \phi_{1s}(r)$$

But this isn't the electron's wavefunction because the electron also has a spin. Let's suppose the electron is spin up. Then (just like how a particle of definite position has a position wavefunction that is a Dirac delta) the spin wavefunction is:

$$\chi(\sigma) = \delta(\sigma,+1/2)$$

where $\delta(\sigma_1,\sigma_2)$ is a Kronecker delta (we're also specifically working in a basis where spins take values $\pm 1/2$).

There are two degenerate ground states, one of which is:

$$\psi_+(r,\sigma) = \phi_{1s}(r) \delta(\sigma, +1/2)$$

and the other has a minus sign in the spin delta function. These are both individually eigenvalues of $H$ and we can write $H\psi_{\pm}=E^{1s}_{\pm}\psi_{\pm}$. Now in the time-reversal invariant case these are degenerate but we could add a magnetic field to split them via the Zeeman effect. Okay, so with this setup let's quickly go through the OP's scenarios (out of order):

  1. This scenario is that $E^{1s}_+ = E^{1s}_-$ but we've said we can split them with a Zeeman term.

  2. This doesn't really make sense. The Hamiltonian $H$ contains a Zeeman term $H_{Zeeman} \sim \sigma\cdot B$. The reduced wavefunction you suggest no longer has spin space information, it's not in the vector space on which the Hamiltonian acts.

This leaves us with:

  1. This is accurate, but you seem confused about what exactly you're saying when you compute the electron density. In particular, you don't use both the wavefunction $\psi_+$ and $\psi_-$. Suppose the ground state has the spin aligned with our $B$ field so is $\psi_+$. Then $\sum_\sigma |\psi_+|^2 = \sum_\sigma \delta^2(\sigma,1/2)|\phi_{1s}|^2$. A Kronecker delta is either $0$ or $1$ so $\delta^2 = \delta$ and $\sum_\sigma \delta(\sigma,\sigma')=1$ so the net effect of all of this is just to recover $|\psi_{1s}|^2$ which is what we want. Note that we didn't need to consider $\psi_-$ anywhere for the ground state density calculation.
  • $\begingroup$ Thanks for your answer. Have I understood you correct in that scenario 1 is right if there is spin in the Hamiltonian, but 2 is right if there is no spin in the Hamiltonian? Thanks! $\endgroup$
    – Mikkel Rev
    Commented Jul 13, 2020 at 13:23
  • $\begingroup$ 1 is right always, but the point is that in summing over spins to get the density, you're summing over the free indices, which are different to the ones labelling the wavefunction which are what appear in $E_{\sigma(1),\cdots \sigma(k)}$. 2 is not generally correct but if there is time reversal invariant you're guarenteed to have every energy level (at least) doubly degenerate so some of your $E$'s will be equal to some others, but not necessarily to all permutations of spin labels, just to a global flip. $\endgroup$
    – jacob1729
    Commented Jul 13, 2020 at 13:31
  • $\begingroup$ My hope is that the question be read such that the $\sigma$ that appears in $E_{\sigma_1,\cdots,\sigma_n}$ should be the same $\sigma$'s that appears in $\psi$ and therefore also the same sigma that appears in the ground state density. In other words: If $\sigma$ in this expression: $$ n(r) = \sum_{\sigma_1} \sum_{\sigma_2} \cdots \sum_{\sigma_n} \int |\psi(r_1,\sigma_1 , r_2,\sigma_2,\cdots, r_n,\sigma_n)|^2 dr_2 \cdots dr_n. $$ is called 'a free index', I would like the $\sigma$ in $E_{\sigma_1,\cdots,\sigma_n}$ to be interpreted as a free index also. $\endgroup$
    – Mikkel Rev
    Commented Jul 13, 2020 at 20:29

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