Consider a system of 2 identical fermions. $$\psi_{k_1,k_2}(x_1,x_2,m_1, m_2) = \langle x_1\,x_2\,m_1\,m_2\mid \psi \rangle$$ According to what I have read we can construct a state with the right anti-symmetry properties by $$\psi_{k_1,k_2}(x_1,x_2,m_1, m_2) = \frac{1}{\sqrt{2}}\begin{vmatrix}\phi_{k_1}(x_1, m_1) & \phi_{k_2}(x_1, m_1) \\ \phi_{k_1}(x_2, m_2) & \phi_{k_2}(x_2, m_2)\end{vmatrix} $$ $$= \frac{1}{\sqrt{2}}\left(\phi_{k_1}(x_1, m_1) \phi_{k_2}(x_2, m_2) - \phi_{k_2}(x_1, m_1) \phi_{k_1}(x_2, m_2) \right) $$ it seems this vanishes if $k_1 = k_2$ regardless of any of the other variables. I feel like something is amiss and I am either confused or not doing this properly. For $m_i = \pm \frac{1}{2}$, shouldn't we have 2 particles allowed to have the same value of $k$ as long as they do not have the same spin? What about singlet and triplet states? How do I understand and connect these ideas?

  • $\begingroup$ I don't know the answer. But, why did you label the wave function with $k_1$ and $k_2$ in the first equation? Isn't the $k$ dependence given by taking inner product with $k$? $\endgroup$
    – Goobs
    Oct 13, 2015 at 16:34
  • $\begingroup$ Since you're learning about multi-particle quantum states, please take a moment to read this post about the right way to think about particle identity. $\endgroup$
    – DanielSank
    Oct 13, 2015 at 19:31

2 Answers 2


The way you define $\phi_k$, $|\phi_k(x,m)|^2$ gives the probability for having an electron with spin $m$ at position $x$.

Thus, if you want to express that your electron is e.g. spin up with an orbital component $\chi(x)$, you will need to make $\phi_k(x,m)$ depend on $m$, e.g. $\phi_k(x,m)=\chi(x)\delta_{m,\uparrow}$. This property is then contained in the subscript $k$, which therefore holds information both about the orbital and the spin state of the electron.

Thus, if $k_1=k_2$, this means that the electrons have the same orbital and spin state, and thus, there is no antisymmetric state due to the Pauli principle. (Commonly, if spin and orbit are decoubled, one would choose an index $k$ which contains an orbital component $n$ and a spin component $\sigma$, $\phi_{n,\sigma}(x,m) = \chi_n(x)\delta_{\sigma,m}$).


The problem is that your single particle states are not labelled correctly.

A single state of definite momentum and spin needs those two labels: $\phi_{k\sigma}$. For example: $$ \phi_{k\uparrow}(x, m) = \frac 1 {\sqrt V} \delta_{m,\uparrow} e^{ikx}. $$

Then it is no problem to have a two-particle state of the same momentum, but different spins: \begin{align*} \Phi(x_1, m_1, x_2, m_2) &= \frac 1 {\sqrt 2} \begin{vmatrix} \phi_{k\uparrow}(x_1, m_1) & \phi_{k\downarrow}(x_1, m_1) \\ \phi_{k\uparrow}(x_2,m_2) & \phi_{k\downarrow}(x_2,m_2) \end{vmatrix} \\ &= \frac 1 {\sqrt 2} \big(\phi_{k\uparrow}(x_1, m_1)\phi_{k\downarrow}(x_2,m_2) - \phi_{k\uparrow}(x_2, m_2)\phi_{k\downarrow}(x_1,m_1) \big). \end{align*} Which is obviously not zero. (But two electrons with the same spatial wavefunction and antisymmetric (singlet) spin-part, this is necessary as the total wavefuntion must be antisymmetric).

Without the spin label the equation you use is either simply wrong or $k$ does not only label momentum. There are situations where similar stuff appears, but only when the spatial part of the wavefunction and the spin part separate: $\Phi(x_1, m_1, x_2, m_2) = \psi(x_1, x_2)\chi(m_1, m_2)$.

As the total wavefunction must be antisymmetric under exchange of complete variable sets, the following cases arise:

  1. $\psi$ is antisymmetric but $\chi$ is symmetric.

  2. $\chi$ is antisymmetric but $\psi$ is symmetric.

So either you have an antisymmetric space or spin wavefunction. So you antisymmetrize one and symmetrize the other. So you only antisymmetrize the spatial part, if the spin part is symmetric (that is, if the spins are in a triplet state) – but then the Pauli principle forbids the same spatial wave function.

  • $\begingroup$ "The problem is that your single particle states are not labelled correctly." -- This really depends on what the meaning of $k$ is. $\endgroup$ Oct 13, 2015 at 17:12
  • $\begingroup$ @NorbertSchuch Yes it does. But I assumed from the rest of the question, that it does not label the spin (because otherwise the rest of the question would not make sense). Corresponding quote from the question: "shouldn't we have 2 particles allowed to have the same value of k as long as they do not have the same spin". $\endgroup$ Oct 13, 2015 at 17:17
  • $\begingroup$ Fair enough, it is indeed what the OP had in mind. $\endgroup$ Oct 13, 2015 at 17:29

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