Suppose I have two bosons with symmetric wave function (I guess there should be tensor products?): $$\psi(x_1,x_2)=\psi_a(x_1)\psi_b(x_2)+\psi_b(x_1)\psi_a(x_2).$$

Suppose now that I perform a measurement of a particle, and I find its location to be $y$. What does the wave function "collapse" to?

Since they are identical bosons, I know it should collapse to something symmetric. It seems to me that I should somehow change one of the $\psi_a$ or $\psi_b$ to be the Dirac $\delta$ function at $y$, but I also don't know which particle I measured! So how would I determine what the resulting wave funciton is, or is my problem ill-posed?

  • $\begingroup$ You haven't measured anything about the other particle? $\endgroup$ Commented Dec 14, 2021 at 23:10
  • $\begingroup$ I think my idea is that we have some device that detected a particle at $y$. But we don't know which particle it is. $\endgroup$
    – Vasting
    Commented Dec 15, 2021 at 0:17
  • $\begingroup$ What do the subscripts $a$ and $b$ represent? Spin? If $a\neq b$, then the particles are not identical. $\endgroup$ Commented Dec 15, 2021 at 3:29

1 Answer 1


The eigenstate which $\psi(x_1,x_2)$ collapses to is ill-posed because you provided/have information of only one particle whereas the system requires two to be described.

Collapse means the wave function $|\psi\rangle=\sum_i a_i|\psi_i\rangle$, sum of every eigenstate, falling into one single eigenstate $\sum a_i|\psi_i\rangle\to1\cdot|\psi_\sphericalangle\rangle$. So that, for instance, a measurement, say $\hat{x}$, reads $x=\langle\psi_\sphericalangle|\;\hat{x}\;|\psi_\sphericalangle\rangle$ where $x$ represents the eigenvalue of $\hat{x}$.

Just for the idea, for two-particle bosonic system, the general form of a wave function can be expressed as $$|\psi(x_1,x_2)\rangle=\sum_j\sum_i \frac{a_{ij}}{\sqrt{2}}\left(|\psi_i(x_1)\rangle\otimes|\psi_j(x_2)\rangle+|\psi_j(x_1)\rangle\otimes|\psi_i(x_2)\rangle\right)\;.$$

Regardless of its lengthy expression, the collapse just means that the wave function falls into one single eigenstate whose position eigenvalue is of the form of $(x_1,x_2)$.

  • $\begingroup$ I see. So are you saying that it is physically not possible to measure exactly one particle? It seems to me that just knowing the location of one particle should change the wave-function, even if it is incomplete information of the entire system. Do I not project onto some subspace that accommodates such a measurement? $\endgroup$
    – Vasting
    Commented Dec 15, 2021 at 1:07

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