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Source: this video

For a system with two particles (09:30), why is its wave function a product of each particle's wave function? E.g.

$$\psi(x_1,x_2)=\psi_a(x_1)\psi_b(x_2)$$

For indistinguishable particles (16:12), I don't quite understand how the author got to this equation:

$$\psi(x_2,x_1)=\pm\psi(x_1,x_2)$$

He mentions something about complex phases and due to applying the exchange operator twice gets back to where we started, meaning the phase that we have to multiply by is 0 or $\pi$.

Lastly, again for indistinguishable particles, how did he come up with this:

$$\psi(x_1,x_2)=A[\psi_a(x_1)\psi_b(x_2)\pm\psi_a(x_2)\psi_b(x_1)]$$

I understand the sum as the particles are indistinguishable and thus can have either eavefunction of $\psi_a$ or $\psi_b$ but I don't understand the subtraction.

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3 Answers 3

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For question 1, it comes down to probability. I have two distinguishable particles, $a$ and $b$. The probability density to find particle $a$ at $x_1$ is $$P_a (x_1)= \Psi_a(x_1) \Psi_a^*(x_1),$$ and we have a similar expression for particle $b$ at $x_2$. The probability density to find particle $a$ at $x_1$ and particle $b$ at $x_2$ is just the product of probability densities $P_a$, $P_b$. The probability density is then $$\Psi(x_1,x_2)\Psi^*(x_1,x_2)=\Psi_a(x_1) \Psi_a^*(x_1) \Psi_b(x_2) \Psi_b^*(x_2)$$ For any complex number the conjugate is just a multiplication by a phase away: $$(a+b i)^*=e^{i\alpha}(a+b i)$$ $\alpha$ depends on $a$ and $b$. From here I can then write $$\Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2) e^{i\phi}$$ But the last phase is irrelevant, so you get just the product of individual wavefunctions.

For question 2, we go back to probability again. We know that we cannot distinguish particles $a$ and $b$. Then $$\Psi(x_1,x_2)=e^{i\phi}\Psi(x_2,x_1)$$ Repeating the same formula again for $x_2,x_1$ we get $$\Psi(x_2,x_1)=e^{i\phi}\Psi(x_1,x_2)$$. When we plug it into the previous formula, we have $$\Psi(x_1,x_2)=e^{i\phi}\Psi(x_2,x_1)=e^{2i\phi}\Psi(x_1,x_2)$$ This yields $e^{2i\phi}=1$ or $e^{i\phi}=\pm1$. Therefore $\Psi(x_1,x_2)=\pm\Psi(x_2,x_1)$. So the total wavefunction is either symmetric (+) or antisymmetric (-).

For the last question: we start saying that $\Psi(x_1,x_2)$ is a linear combination of $\Psi_a(x_1) \Psi_b(x_2)$ and $\Psi_a(x_2) \Psi_b(x_1)$, so we can write $$\Psi(x_2,x_1)=a\Psi_a(x_1) \Psi_b(x_2)+b\Psi_a(x_2) \Psi_b(x_1)$$ or the equivalent $$\Psi(x_2,x_1)=A[\Psi_a(x_1) \Psi_b(x_2)+e^{i\phi}\Psi_a(x_2) \Psi_b(x_1)]$$ In a similar fashion as for the previous question, we get that $e^{i\phi}$ has to be $+1$ or $-1$. The choice of the sign depends on the symmetry of the total wavefunction (if particles are bosons or fermions)

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  • $\begingroup$ So $b=Ae^{i\phi}$ while a=A? How did you obtain them? $\endgroup$
    – newbie125
    Commented Jul 25, 2016 at 14:48
  • $\begingroup$ I guessed (I did not prove - but it should be easy) that $|a|=|b|$. You will get this by normalizing the wavefunction. I do not care for now what $a$ and $b$ are. In fact, even after getting $b=\pm a$, $a$ is still known only up to a complex phase $\endgroup$
    – Andrei
    Commented Jul 25, 2016 at 18:15
  • $\begingroup$ How did you get $\Psi(x_1,x_2)=e^{i\phi}\Psi(x_2,x_1)$ or $\Psi(x_2,x_1)=e^{i\phi}\Psi(x_1,x_2)$ ? Is it related to $\Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2) $ & $\Psi(x_2,x_1)=\Psi_a(x_2) \Psi_b(x_1) $ and how did you convert between them? $\endgroup$
    – newbie125
    Commented Jul 26, 2016 at 15:03
  • $\begingroup$ It is related to probability. The probability density for the two wavefunctions is the same, so they only differ by a complex constant of magnitude 1 $\endgroup$
    – Andrei
    Commented Jul 27, 2016 at 3:05
  • $\begingroup$ Why do you assume the probabilities are independent so you can multiply them together? $\endgroup$
    – Akababa
    Commented Apr 14, 2018 at 21:49
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If the state of two particles is the tensor product of the two single particle states, then the wave function of the two particles is the product of the two single particle wave functions.

For indistinguishable particles it is an experimental fact that the final state must be either symmetric or antisymmetric with respect to the exchange of the two particles coordinates.

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For first part of your question, you can check my answer here https://physics.stackexchange.com/a/566506/226827

For your second part of question regarding the minus sign, you can get intuition from taking same particles i.e. x1 = x2

when you do this, your wavefunction will become zero, which is exactly the property of fermions, that no two fermions can be in the same state.

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