With the occupation number representation of multi-particle states I am happy with the meaning $$ \left| 1, 1 \right> $$ to represent a system with $1$ particle in the first state, $1$ particles in the second state so that the corresponding wave function of the associated single particle states would be (for spinless fermions) would be $$ \psi_1 ( {\bf{r}}_1)) \cdot \psi_2 ({\bf{r}}_2) $$ that would need to be made anti-symmetric

$$\frac{1}{\sqrt{2}} \bigg( \psi_1 ( {\bf{r}}_1)) \cdot \psi_2 ({\bf{r}}_2) - \psi_1 ( {\bf{r}}_2)) \cdot \psi_2 ({\bf{r}}_1)\bigg). $$

However, what can I make of the following

$$\frac{1}{\sqrt{2}} \bigg( \left|1, 0 \right> + \left| 0, 1 \right> \bigg)~?$$

Does it represent a state with one particle definitely in state 1 and the other particle definitely in state 2 so that the corresponding wave function would be

$$ \frac{1}{\sqrt{2}} \bigg( \psi_1 ( {\bf{r}}_1) + \psi_2 ( {\bf{r}}_2)\bigg)~?$$

  • $\begingroup$ Can you clean up the first part of your question a bit? I think I understand your question but there seem to be a few confusing typos. After the first $\vert 1,1\rangle$, you refer to $n_1$ and $n_2$ so you probably meant $\vert n_1,n_2\rangle$ or that $n_1=n_2=1$? Next, you write neither symmetric nor anti-symmetric state in your second expression (but you say that you are writing a Fermionic state -- which ought to be anti-symmetrized). Further, you write an anti-symmetrized state in the third expression but you preface that by saying that you are writing a symmetrized state. $\endgroup$
    – user87745
    Mar 11, 2021 at 16:42
  • $\begingroup$ @DvijD.C. I thought I had put down symmetrized with the meaning "appropriately symmetrization ", but there seems to be an issue about its spelling. $\endgroup$
    – jim
    Mar 11, 2021 at 20:15

1 Answer 1


$\vert 0,1\rangle$ and $\vert 1,0\rangle$ are both just one-particle states and they live in the one-particle sector of the Fock space. If you want to write them in the position basis they would be $$\langle \hat{\mathbf{r}}\vert 0,1\rangle = \psi_2(\mathbf{r})$$$$\langle \hat{\mathbf{r}}\vert 1,0\rangle = \psi_1(\mathbf{r})$$ Thus, their superposition that you are interested in would simply be the following in the position basis $$\frac{1}{\sqrt{2}}\langle\hat{\mathbf{r}}\vert0,1\rangle+\frac{1}{\sqrt{2}}\langle\hat{\mathbf{r}}\vert1,0\rangle=\frac{1}{\sqrt{2}}\big(\psi_2(\mathbf{r})+\psi_1(\mathbf{r})\big)$$

In particular, the position operator is simply $\hat{\mathbf{r}}$ in the single-particle sector and not $\hat{\mathbf{r}}\otimes\hat{\mathbf{r}}$. The Fock space is constructed by taking the direct sum of (symmetrized/anti-symmetrized) products of copies of a single-particle Hilbert space. Thus, the operators on the Fock space are also constructed as the direct sum of the operators in the corresponding ingredients of the direct sum. So, if you wish, the position operator on the Fock space is $\hat{\mathbf{r}} \oplus (\hat{\mathbf{r}}\otimes\hat{\mathbf{r}})\oplus(\hat{\mathbf{r}}\otimes\hat{\mathbf{r}}\otimes\hat{\mathbf{r}})\oplus...$. Thus, when you apply it over a state that purely lives in, say, the $n-$particle sector, the relevant position operator would be simply $\hat{\mathbf{r}}\otimes...n\text{ times}...\otimes\hat{\mathbf{r}}$. However, if you were to express a state like $\vert 0,1\rangle+\vert 1,1\rangle+\vert 1,0\rangle$, you would see that the relevant part of the position operator on the full Fock space would be $\hat{\mathbf{r}}\oplus(\hat{\mathbf{r}}\otimes\hat{\mathbf{r}})$. And thus, its position basis wavefunction would be written as $$\psi_1(\mathbf{r})+\psi_2(\mathbf{r})+\frac{1}{\sqrt{2}}(\psi_1(\mathbf{r}_1)\psi_2(\mathbf{r}_2)\pm\psi_2(\mathbf{r}_1)\psi_1(\mathbf{r}_2))$$

Postscript: Perhaps, it should be clarified where the $\psi_1$ and $\psi_2$ come from in the wavefunction of $\vert 1,0\rangle$, $\vert 0,1\rangle$ -- as one might be confused and thinking that the $1,2$ in the subscript are somehow supposed to refer to as to whether it is the "first particle" or "the second". That is not the purpose of these subscripts. They simply refer to the eigenvalues of the operator corresponding to which we have constructed the occupation number representation. In particular, the occupation number representation always refers to some operator whose eigenvalues are being occupied by the number of particles equal to that occupation number. For example, if the occupation number representation is constructed in reference to the momentum operator then $n_1,n_2$ are usually written as $n_{p_1}, n_{p_2}$ -- meaning that they represent the number of particles with momentum $p_1$ and $p_2$ respectively. In this scenario, $\vert \psi_{1,2}\rangle = \vert p_{1,2}\rangle$. So, a better notation for us to adopt would be to explicitly clarify as to which operator $\hat{O}$ is being referred to by the occupation numbers and then write things like $\psi_{o_1}$, $\psi_{o_2}$, etc.


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