1
$\begingroup$

We construct identical particle state by symmetrizing or antisymmetrizing the tensor product of single partice states. When considering spin, a two fermions state should be $$|\psi\rangle=\frac{1}{\sqrt{2}}(|\psi_1\rangle_{\sigma_1}\otimes|\psi_2\rangle_{\sigma_2}-|\psi_2\rangle_{\sigma_2}\otimes|\psi_1\rangle_{\sigma_1}),$$ where $|\psi_1\rangle_{\sigma_1}=|\psi_1\rangle\otimes|\sigma_1\rangle$. However, many text books said that $|\psi\rangle=|\phi\rangle\otimes|\chi\rangle$, where $|\phi\rangle$ is the spatial part and $|\chi\rangle$ is the spin part. My question is which one is correct, are the two answers coherent?

$\endgroup$
0
$\begingroup$

I think the question already has the answer. In this case $|\psi_i \rangle$ is the spatial part, and $|\sigma_i \rangle$ is the spin part of $i^{\text{th}}$ particle. These two are different properties and lie in seperate spaces. One is position living in a Hilbert space of infinite dimensions and the other is spin Hilbert space of two dimensions, representing an internal property. Combined wavefunction is the direct product of each space for just one particle.

Since there are two fermions associated with different spaces, we shall take the direct product of each of the particles. Anti-symmetrizing the wavefunction using Slater determinant yields the equation mentioned by OP.

See the below for detailed explanation.


  • The state vector $|\Psi^{(n)} \rangle$ of a system of $N$ particles is an element of the direct product space, $$ |\Psi^{(n)} \rangle = |\psi{(1)} \rangle \otimes |\psi{(2)} \rangle \otimes |\psi{(3)} \rangle \otimes­ \cdots \otimes­ |\psi{(n)} \rangle $$ formed from the $N$ single-particle spaces $|\psi(i) \rangle$ associated with each particle.

  • There are various properties (more precisely, degrees of freedom) associated to each particle, e.g. position, spin etc. and each of them lives in a different Hilbert space of certain dimension. For two properties $P_1$ and $P_2$ that are independent of each other for a particle, dimension $N_1$ and $N_2$, respectively (either or both of which may be infinite), the combined system should be represented as, $$P_{12} = P_1 \otimes P_2$$ of dimension $N_1 \times N_2$.

  • The main point of this digression is that the state space of one spin-$1/2$ particle can be represented as direct product, $$|\Psi \rangle = |\psi_{\text{spatial}}\rangle ­ \otimes |\sigma_{\text{spin}}\rangle$$ of a quantum space $|\psi_{\text{spatial}}\rangle$ describing the particle’s spatial state (which is spanned, e.g., by an infinite set of position eigenstates $|\mathbf{r}\rangle$), and a two-dimensional quantum space $|\sigma_{\text{spin}}\rangle$ describing the particle’s internal structure.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Neither answer is right in general.

  • The only actual requirement is that the state be [anti]symmetric with respect to exchange. That's it.

  • The first state you give, $|\psi\rangle=\frac{1}{\sqrt{2}}(|\psi_1\rangle_{\sigma_1} \otimes|\psi_2\rangle_{\sigma_2}-|\psi_2\rangle_{\sigma_2} \otimes |\psi_1 \rangle_{ \sigma_1})$, is a special case, but it is not general. This can be thought of as a single 'configuration', or Slater determinant, and those are often good approximations in the absence of strong correlations. But there are valid states which are superpositions of multiple different Slater determinants and cannot be reduced into this form.

  • The second state you give, $|\psi\rangle=|\phi\rangle\otimes|\chi\rangle$, generally with spatial and spin parts of definite exchange symmetry, is again a special case, in which the spatial and spin parts are factorizable. If the system's hamiltonian has no spin-orbit coupling, then you are guaranteed an eigenbasis of states of this form. But in general (i) states need not have this form, (ii) your system's hamiltonian may well have spin-orbit coupling, and thus (iii) the eigenstates won't have this form.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a lot! I think the point is that both of them are basis of the fock space, then definitely they are not eigenstates for a general Hamiltonian with interaction, but any state can be expressed as linear combination of them. $\endgroup$ – Daniel YUE Sep 13 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.