Time reversal on the field operator

Question

What I roughly know about time reversal operator T is,

$$ T[a] = a^\dagger, T[a^\dagger] = a\\ T[i] = -i, T[t] = -t. $$

For the number operator $n=a^\dagger a$, I expect $T[n]=n$. Does that mean time reversal is NOT conjugating the field operator, which gives $T[a^\dagger a]=aa^\dagger$, but conjugating the whole thing? i.e. $T[a^\dagger a] = (a^\dagger a)^\dagger = a^\dagger a$?

But if that's the case, I don't understand $T[a|\psi(t)\rangle] = \langle\psi(-t)|a^\dagger$, since I expect $T[a|\psi(t)\rangle]=a^\dagger|\psi(-t)\rangle$.

Could somebody explain how to apply the time reversal to e.g. Heisenberg equation of motion?

Answer 1

The $T[t]=-t$ does not make sense formally, even if I understand what you are trying to say. Time is not an operator, a mere real parameter.

The mistake is in your original calculation. Remember that $T[x]=x$ and $T[p]=-p$ as it should (by definition), consistent with classical expectations. This gives: $$ T[a] = T[\frac{x+ip}{\sqrt 2}]\\ = \frac{T[x]-iT[p]}{\sqrt 2}\\ = \frac{x+ip}{\sqrt 2}\\ = a $$ (first use anti-linearity and the effect on $x,p$, both of which cancel out). This gives the expected $T[n] = T[a^\dagger]T[a] = n$.

In the Schrödinger picture, the time evolution of the kets is governed by the equation: $$ \frac{d}{dt}|\psi(t)\rangle = -iH|\psi(t)\rangle $$ Since $H = \frac{1}{2}(p^2+x^2) = n+\frac{1}{2}$ is reversible, writing $|\psi(t)\rangle$ (resp $|\phi(t)\rangle$) the evolution of the ket $|\psi\rangle$ (resp $|\phi\rangle$) at $t=0$ with $|\phi\rangle = T|\psi\rangle$, you have: $$ T|\psi(t)\rangle = |\phi(-t)\rangle $$ (same initial condition and same equation of motion), in particular if $|\psi\rangle=|\phi\rangle$: $$ T|\psi(t)\rangle = |\psi(-t)\rangle $$ Applying this to your example, I think you forgot to mention that $|\psi(0)\rangle$ is time reversible (important assumption that is not always verified, for example a non-zero momentum eigenfunctions), in which case you indeed find: $$ Ta^\dagger|\psi(t)\rangle = a^\dagger|\psi(-t)\rangle $$

In the Heisenberg picture, the time dependence of an observable $O(t)$ with initial value $O$ is governed by the equation: $$ \dot O(t) = -i[O(t),H] $$ Similarily, considering the time evolved $\tilde O(t)$ with initial value $\tilde O = T[O]$, you'd get: $$ T[O(t)] = \tilde O(-t) $$ in particular, if $O = \tilde O$: $$ T[O(t)] = O(-t) $$ For example, applying this to $a$, with $a(t) = e^{-it}a$ (from solving Heisenberg's equations), you indeed get (not forgetting anti-linearity): $$ T[a(t)] = a(-t) (= e^{it}a) $$

Hope this helps, tell me if something's not clear.

Edit (answers to comment)

1. Yes with usual composition. Note that since you’re conjugating by an anti-linear operator, the resulting operator is linear.

2. In your example, I meant $T|\psi(0)\rangle= |\psi(0)\rangle$.

3. No it is not. Even classically you can see the problem. With $x(t)$ evolving under a time reversible force, you have $x(-t)=\tilde x(t)$ with $\tilde x$ evolving under the same equations of motion, same initial position but opposing initial velocity. In this example, you can see it explicitly. Since $T|x\rangle= |x\rangle$, you have: $$ T|p\rangle=T\int dx e^{ipx}|x\rangle \\ =\int dx e^{-ipx}T|x\rangle \\ =\int dx e^{-ipx}|x\rangle \\ = |-p\rangle $$ (using anti-linearity and we obtain the expected result)

4. Yes sorry typo, corrected it.

5. As I stated above, your rule $t\to-t$ is false. $t$ is a real parameter, and $T$ is still real linear. This is usually guides you for what to expect when applying $T$ to a time dependent object, but you cannot use it for an actual computation