# Time reversal on the field operator

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?

• lt seems that you are working entirely in the Schrodinger picture, then under $T$, $a$ and $a^\dagger$ do not change. Jun 6, 2022 at 16:40
• @MengCheng What happens for the Heisenberg picture operators a and $a^\dagger$ under T? Jun 7, 2022 at 0:34

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.

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

• 1. Is T[AB] = T[A]T[B] proved by $T[AB]=TABT^{-1}=TAT^{-1}TBT^{-1}=T[A]T[B]$? 2. What do you mean $|\psi(0)\rangle$ is time reversal? You meant $|\psi(0)\rangle=|\phi(0)\rangle$? 3.Isn't non-zero momentum eigenfunction $e^{ipx}$ time-reversal symmetric? Jun 7, 2022 at 0:42
• 4. Shouldn't the second last equation be $T[O(t)]=O(-t)$? Jun 7, 2022 at 0:53
• 5. From $T[e^{-it}a]=e^{it}a$ (last equation), the rule of thumb $i\rightarrow -i$, $t\rightarrow -t$ does not work (this gives $T[e^{-it}a]=e^{-it}a$)... I guess you are right, but what makes this rule of thumb not working? Jun 7, 2022 at 0:54