In an Introduction to Quantum Mechanics by Griffiths (pg. 180), he claims that

"P and H are compatible observables, and hence we can find a complete set of functions that are simultaneous eigenstates of both. That is to say, we can find solutions to the Schrodinger equation that are either symmetric (eigenvalue +1) or antisymmetric (eigenvalue -1) under exchange"

I understand why commuting (or as he calls them, compatible) observables share a common eigenbasis. What I don't see is why P, the exchange operator, and H, the Hamiltonian, need to commute for the second sentence to be true. If P is an observable, then supposedly it is a Hermitian operator whose eigenstates span the L2 Hilbert space. Isn't that a sufficient condition for us to say that we can construct any solution to the Schrodinger equation via a linear combination of those eigenstates, and that the eigenstates themselves are solutions to the Schrodinger equation? What is he trying to show by stating the P and H commute?

  • 2
    $\begingroup$ Yes. But they need not be energy eigenstates. And if you take a linear combination of them, they are no longer (in general) permutation eigenstates. However, if $P$ and $H$ commute, we can form eigenstates that simultaneously diagonalise $P$ and $H$ $\endgroup$ – Superfast Jellyfish Feb 28 '20 at 6:16
  • $\begingroup$ Loose language. It might be clearer if he said solutions of the TISE (time-independent SE), in which case all solutions may be classified this way. $\endgroup$ – Cosmas Zachos Feb 28 '20 at 16:00

$[H,P]=0$ means the Hamiltonian does not mix symmetric and antisymmetric states, i.e. for any $|\psi_\mathrm{S}\rangle$ and $|\psi_\mathrm{A}\rangle$ we have $$ \langle\psi_\mathrm{S}|H|\psi_\mathrm{A}\rangle=0. $$ There are two consequences:

1) If we try to find the eigenstate of $H$ as a mixture of symmetric and antisymmetric states, $$ H(\alpha|\psi_\mathrm{S}\rangle+\beta|\psi_\mathrm{A}\rangle)=E(\alpha|\psi_\mathrm{S}\rangle+\beta|\psi_\mathrm{A}\rangle), $$ then, by multiplying it by $\langle\psi_\mathrm{S}|$ or $\langle\psi_\mathrm{A}|$ from the left, we get: $$ \alpha(E-\langle\psi_\mathrm{S}|H|\psi_\mathrm{S}\rangle)=0,\qquad\beta(E-\langle\psi_\mathrm{A}|H|\psi_\mathrm{A}\rangle)=0. $$ So $\alpha$ and $\beta$ cannot be simultaneously nonzero if the energies of symmetric and antisymmetric states are different: $\langle\psi_\mathrm{S}|H|\psi_\mathrm{S}\rangle\neq\langle\psi_\mathrm{A}|H|\psi_\mathrm{A}\rangle$. For any realistic system of interacting particles it is the case.

Thus the eigenstates of interacting many-particle system are either symmetric or antisymmetric.

2) If $|\psi(t)\rangle$ is symmetric/antisymmetric at $t=0$, then, after the Schrodinger time evolution $$ i\hbar\frac{\partial|\psi\rangle}{\partial t}=H|\psi\rangle, $$ $|\psi(t)\rangle$ will remain symmetric/antisymmetric at $t>0$.

So we need to choose, depending on physical reasons, which sector of the Hilbert space (symmetric or antisymmetric) we are working in with the system of particles of given kind. There will be no physically realistic problem (both stationary $-$ when you find the Hamiltonian eigenstates, or nonstationary $-$ when you calculate time evolution) where you will need to mix symmetric and antisymmetric states.

Update: Symmetric and antisymmetric states have different energies, at least, by two reasons. First, the fermions need to fill the single-particle states with higher and higher energies up to the Fermi energy because of the Pauli exclusion principle, while bosons can condense in a single lowest-energy state. Second, the exchange energy of many-particle system, appearing in the Hartree-Fock approximation, is positive for bosons and negative for fermions. So it is reasonable to assume that the energies of ground as well as excite states of a Hamiltonian are different in symmetric and antisymmetric sectors (except some rare random coincidences), although I don't know the proof of this statement. Therefore the combined spectrum $\{E_n,P_n=\pm1\}$ of $H$ and $P$ is nondegenerate, and each eigenstate of $H$ is automatically an eigenstate of $P$ and vise versa.

  • $\begingroup$ Can you explain why the energies of symmetric & antisymmetric states are different? Just because [P,H] = 0, doesn't mean that every eigenstate of H is an eigenstate of P (this would only be true if they both had only non-degenerate eigenvalues). Rather, it means we can construct a basis in which each state is an eigenstate of both. So it could be possible that an eigenstate of H is constructed with two different eigenstates of P - it simply would not appear in the common basis. In that case, that eigenstate would have one value of E, and so the symmetric&antisymmetric states in it would too. $\endgroup$ – Sophia Tevosyan Feb 29 '20 at 21:34
  • $\begingroup$ @SophiaTevosyan See the update to my answer. It is generally believed that, at least, almost all symmetric and antisymmetric states have different energies (perhaps there exist some strict proof), and the main body of my answer was intended essentially to explain what is the practical significance of this fact. $\endgroup$ – Alexey Sokolik Mar 1 '20 at 10:46

Let $|\psi\rangle$ be an eigenstate of $\hat H$. Since $[\hat P, \hat H]=0$, the state $|\psi\rangle$ is also an eigenstate of $\hat P$, i.e., $$\hat P |\psi\rangle = \lambda|\psi\rangle,$$ where $\lambda$ is either $+1$ or $-1$. Therefore all such $|\psi\rangle$ that are eigenstates of $\hat H$, are either symmetric ($\lambda=1$) or anti-symmetric ($\lambda=-1$).

Therefore, the vanishing commutation of the Hamiltonian $H$ with the exchange operator $P$, i.e., $\left[P, H\right] = 0$ leads to the existence of either symmetric or anti-symmetric energy eigenstates $-$ as the Griffiths claims.

However, Griffiths did not claim the other ways around $-$ existence of either symmetric and anti-symmetric energy eigenstates guarantees to the vanishing commutation of $P$ and $H$.


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