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I'm a master student in mathematics currently taking Quantum Mechanics and since the lecture notes provided by the lecturer aren't cutting it I'm reading "Quantum Mechanics and Path Integrals" by R. Feynman and A.R. Hibbs.

In this book there is one line I have been unable to understand. The line in question is:

"Actually this is wrong and, remarkably, electrons obey a different rule. The amplitude for an event in which the identity of a pair of electrons is reversed contributes 180 out of phase."

The line is in chapter "1-3 Interfering Alternatives" under "Some Illustrations".

Now the part I don't get is what an "event in which the identity of a pair of electron is reversed" actually is. The word "identity" is already super weird for me because in that context both of our electrons have the same spin so I don't get why their "identity" would be "reversed".

If anyone can help me it would be highly appreciated.

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  • $\begingroup$ it is making allusion to the fact that multi-particle fermion amplitudes must be antisymmetric under particle exchange, in other words $\phi(x,y) = -\phi(y,x)$ $\endgroup$ – lurscher Sep 30 '19 at 18:59
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I assume he is talking about exchanging particles (try to make you question as self-contained as possible!).

Assume you have two distinguishable particles labeled 1 and 2. With distinguish I mean you could for example paint one of them red and the other blue. Or you could label them in some other way such that you can keep track of them. Suppose we now have a Hamoltonian which behaves the same for both particles: $$H=H_1+H_2$$ where $H_i=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_i^2}+V(x_i)$. If we know the solutions to the individual Hamiltonians $H_i$ we can write the full solution as a product of these single particle solutions: $$\Psi(x_1,x_2)=\psi_m(x_1)\psi_n(x_2).$$ So for example if particle 1 is in the ground state and particle 2 in the third excited state we would get $$\Psi(x_1,x_2)=\psi_0(x_1)\psi_3(x_2).$$ Since the particles are distinguishable we are now done. But we know from experiments that particles are indistinguishable. There is no way to know the difference between two electrons. This means that exchanging two particles ("reversing the identity of a pair of electrons") should give you the same state. At least up to a phase factor since states that differ only by a phase refer to the same physical state. This means our previous state no longer suffices for distinguishable particles. Define the exchange operator as follows $$\hat P\Psi(x_1,x_2)=\Psi(x_2,x_1)$$ By imposing the condition that this should only change the phase and by noting that applying the operator twice gives the same state we get $$\hat P\Psi(x_1,x_2)=e^{i\phi}\Psi(x_1,x_2)\\ \hat P^2\Psi(x_1,x_2)=e^{2i\phi}\Psi(x_1,x_2)=\Psi(x_1,x_2)$$ which means $\phi=0,\pi/2$. So exchanging two indistinguishable particles can change the state at most by introducing a minus sign. It so happens that for electrons, which are fermions, a minus sign is introduced. This gives $$\Psi(x_2,x_1)=-\Psi(x_1,x_2)$$ for electrons. So by exchanging identities he probably meant swapping the particles. This vague definition might be because he wants to avoid the word 'swapping' to emphasize this is a mathematical operation instead of physically moving the particles.

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