The permutation operator $\hat{P}_{xyz}$, for a system of three particles, assigns the quantum numbers of particle 1 to particle $x$, of particle 2 to particle $y$ and so on. That is, given a ket, e.g. $|q_i;q_j;q_k\rangle$ and one possible permutation operator $\hat{P}_{312}$ $$ \hat{P}_{312} |q_i;q_j;q_k\rangle = | q_j;q_k;q_i\rangle $$ So far, this I get. But can I apply the permutation operator to a wave function $\psi (q_i, q_j, q_k)$, which I would write in bra-ket notation as $\langle q_i, q_j, q_k | \psi \rangle$. Applying the operator would be written like this $$ \langle q_i, q_j, q_k | \hat{P}_{312} | \psi \rangle $$ if I am not mistaken. Do I work from "right to left" then? Unfortunately my textbook only works with states/kets and doesn't touch on the subject of this question.

  • 2
    $\begingroup$ Hint: $\langle x | y \rangle = \langle y | x \rangle^*$ $\endgroup$
    – Aaron
    Oct 6 '17 at 15:44
  • $\begingroup$ Lets see... $$ \begin{align} \langle q_i q_j q_k | P_{312} | \psi \rangle &= \langle \psi | P_{312} | q_i q_j q_k \rangle^* \\ &= \langle \psi | q_j q_k q_i \rangle^* \\ &= \langle q_j q_k q_i | \psi \rangle \\ &= \psi (q_j q_k q_i) \end{align} $$ Is that correct? $\endgroup$
    – John W.
    Oct 6 '17 at 16:01
  • $\begingroup$ Almost. You need to remember the to take the hermitian conjugate of $P_{312}$. That is to say, $\langle q_i q_j q_k|P_{312}|\psi\rangle = \langle \psi|P_{312}^\dagger | q_i q_j q_k\rangle$. In this case, $P^\dagger_{312} = P_{312}$ so you get the right answer at the end. The short way to remember this is that if you want to apply operators right to left, you apply the hermitian conjugate of the operator instead. $\endgroup$
    – Aaron
    Oct 6 '17 at 16:11
  • $\begingroup$ That was my concern, whether or not the conjugate of $P$ was $P$ itself. Thank you very much for the input and correction! $\endgroup$
    – John W.
    Oct 6 '17 at 16:13
  • $\begingroup$ Sorry, I made a mistake. $P^\dagger_{312} = P_{321}$. So the answer should be $\psi(q_j q_k q_i)$. $\endgroup$
    – Aaron
    Oct 6 '17 at 16:19

If we know nothing about $P_{312}$ except its definition, we should do it this way: \begin{align} \langle q_i q_j q_k| P_{312} |\psi\rangle &= \sum_{a,b,c}\langle q_i q_j q_k| P_{312}|q_a q_b q_c\rangle\langle q_a q_b q_c|\psi\rangle \,\,\,\, \text{(identity insertion)}\\ &= \sum_{a,b,c} \langle q_i q_j q_k|q_c q_a q_b\rangle\langle q_a q_b q_c |\psi\rangle \\ &= \sum_{a,b,c}\delta_{ic} \delta_{ja}\delta_{kb} \langle q_a q_b q_c|\psi\rangle \\ &= \langle q_j q_k q_i|\psi\rangle \end{align}

This also proves the fact that $P^\dagger_{312} = P_{321}$. Let me be explicit about how we get this result. We know from above that \begin{align} \langle q_i q_j q_k|P_{312}| q_a q_b q_c\rangle = \delta_{ic}\delta_{ja}\delta_{kb} \end{align} Now take the complex conjugate of this. We get that \begin{align} \langle q_a q_b q_c| P^\dagger_{312} |q_i q_j q_k \rangle = \delta_{ic} \delta_{ja} \delta_{kb} \end{align} We also know that \begin{align} \langle q_a q_b q_c| P_{321} |q_i q_j q_k \rangle = \delta_{ic} \delta_{ja} \delta_{kb} \end{align} Since the kets $|q_a q_b q_c\rangle$ form a basis, this tells us everything that the operators $P_{312}^\dagger$ and $P_{321}$ do. Therefore, we can remove the bras and kets and say that $P^\dagger_{312} = P_{321}$.

The fast way I remember this is to look at the permutation matrix that corresponds to the permutation operator. We know that permutation matrices are real and orthogonal, which is also unitary so that $P^\dagger = P^{-1}$ for any permutation matrix. Clearly $P^{-1}_{312} = P_{321}$. This is not really a proof, just a way to remember.

Of course, if you already knew this fact then one can skip the algebra and get to the answer much quicker via \begin{align} \langle q_i q_j q_k| P_{312} |\psi\rangle &= \langle \psi| P_{312}^\dagger | q_i q_j q_k\rangle^* \\ &= \langle \psi| P_{321} | q_j q_k q_i \rangle \\ &= \langle q_j q_k q_i|\psi\rangle \end{align}

  • $\begingroup$ A follow-up question: what happens for $P_{123}$? If I apply that to a state it's just the identity operator, i.e., $P_{123}| q_i q_j q_k\rangle = | q_i q_j q_k \rangle$. For the wave function case, using your method with the identity insertion, this yields $$ \begin{align} [\dots] &= \sum_{abc} \langle q_i q_j q_k | q_a q_b q_c \rangle \langle q_a q_b q_c | \psi \rangle \\ &= \sum_{abc} \delta_{ia} \delta_{jb} \delta_{kc} \langle q_a q_b q_c | \psi \rangle \\ &= \langle q_i q_j q_k | \psi \rangle \end{align} $$ Which would imply $P_{123}=P_{123}^\dagger$. Is that correct? $\endgroup$
    – John W.
    Oct 6 '17 at 17:55
  • $\begingroup$ This is not correct, primarily because you should not call the eigenstate of $P_{123}$ as $|q_i q_j q_k\rangle$. The fact that $\langle q_i q_j q_k| q_a q_b q_c \rangle = \delta_{ia} \delta_{jb} \delta_{kc}$ is a statement that these kets are orthogonal (in fact, the identity insertion is a statement that these kets form an orthogonal basis). In general, one would not expect the eigenstate $|A\rangle$ of $P_{123}$ to be orthogonal to the kets $|q_i q_j q_k\rangle$. $\endgroup$
    – Aaron
    Oct 6 '17 at 18:48
  • $\begingroup$ Now I'm even more confused. How can you justify the result then? If I'm completely honest, I'd rather just know how to find the conjugate of some $P$ with some rules and doing the "quick" way. $\endgroup$
    – John W.
    Oct 6 '17 at 18:54
  • 1
    $\begingroup$ @JohnW. I have edited the answer for more clarity. $\endgroup$
    – Aaron
    Oct 6 '17 at 22:50

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