# Critique on various ways to think about time reversal transformation on Schrodinger equation?

## Please define how time-reversal symmetry act on Schrodinger equation $$i \frac{\partial}{\partial t} |\Psi{}(t) \rangle = H(t) |\Psi{}(t) \rangle.$$ (for general form: which can be relativistic such as Dirac theory or nonrelativistic.)

Below I present various ways to think about time reversal transformation on Schrodinger equation.

Could you present your critique or criticism on which one make sense? What else is the correct way to do time reversal transformation on Schrodinger equation?

## 1.

The Schroedinger equation governs the quantum system (regardless of relativistic or not) says: $$i \frac{\partial}{\partial t} |\Psi{}(t) \rangle = H(t) |\Psi{}(t) \rangle.$$ It is easy to check the Schroedinger equation is time-reversal $$T$$ invariant if

(1). $$H(t)$$ has no explicit time-dependent, or time-reversal symmetric, then $$H(-t)=H(t)=H$$.

(2). Since we do not want to make $$t=0$$ becomes a special fixed point, we also require the time translational symmetry $$H(t)=H(t+\Delta)$$ for some arbitrary time interval $$\Delta$$.

(3) Under time interval transformation, if we think from the perspective of assign the value of a statevctor $$|\Psi{}(-t) \rangle$$ (or its projected wavefunction) at $$-t$$ to a new statevector $$|\Psi'{}( t) \rangle$$ at $$t$$ following the classical mechanics approach advocated by @Richard Myers https://physics.stackexchange.com/a/633205/42982, we get $$|\Psi{}( t) \rangle \overset{T}{\to} |\Psi'{}( t) \rangle = |\Psi{}(-t) \rangle. \tag{1}$$

We can do a time-reversal transformation on the whole equation acting on the state vector (a ray) in the Hilbert space, $$T i \frac{\partial}{\partial t} |\Psi{}(t) \rangle = T H(t) |\Psi{}(t) \rangle$$ $$=T i T^{-1} T\frac{\partial}{\partial t} |\Psi{}(t) \rangle = T H(t) T^{-1} T|\Psi{}(t) \rangle$$ $$=T i T^{-1} T\lim_{\Delta \to 0}\frac{ |\Psi{}(t+\Delta) \rangle- |\Psi{}(t) \rangle}{\Delta} = H(-t) |\Psi{}(-t) \rangle$$ Then we adopt and plug in eq.(1): $$=Ti T^{-1} \lim_{\Delta \to 0}\frac{ |\Psi{}(-t-\Delta) \rangle- |\Psi{}(-t) \rangle}{\Delta} = H(-t) |\Psi{}(-t) \rangle$$ $$=T i T^{-1} (-)\frac{\partial}{\partial \tilde t} |\Psi{}(\tilde t) \rangle \Big \vert_{\tilde t = -t} = H(-t) |\Psi{}(-t) \rangle.$$

As I show, the Schroedinger equation is time-reversal $$T$$ invariant requiring that (1) $$H(-t)=H(t)$$ and (2) $$T i T^{-1} =-i$$ which is the case that the time-reversal is an antiuntiary and antilinear symmetry.

The above has a problem because we do only map $$|\Psi{}( t) \rangle \overset{T}{\to} |\Psi'{}( t) \rangle = |\Psi{}(-t) \rangle$$ but do not really do a complex conjugation on the (complex valued) wavefunction.

We may also complain that the wavefunction with $$e^{-iEt}$$ time evolution maps to $$e^{+iEt}=e^{-i(-E)t}$$ time evolution. This seems to map $$E>0$$ to $$E<0$$ state!!??

## 2.

Another way to look at the problem maybe this. This follows Sakurai Modern QM book Chap 4.4 approach: We do not act $$T$$ on the Schroedinger equation. But directly find a time-reversal solution $$|\Psi'{}(t) \rangle$$ satisfies the Schroedinger equation. If we define the statevector to be expanded in the projected spatial $$|x \rangle$$ basis: $$|\Psi'{}(t) \rangle = \sum_x |x \rangle \langle x |\Psi'{}(t) \rangle.$$ In this case $$\Theta | x \rangle= |x \rangle$$, so $$|\Psi'{}(t) \rangle = \sum_x |x \rangle \Psi^*{}(x,-t) = \sum_x |x \rangle u^*(x) e^{- i E t}.$$ Then we can test the wavefunction version of Schrodinger equation: $$i \frac{\partial}{\partial t} \Psi{}(x,t) = H(x, -i\frac{\partial}{\partial x}, t) \Psi{}(x,t) .$$ We define $$\Psi'{}(x,t)= \Psi^*{}(x,-t)$$ as $$i \frac{\partial}{\partial t} \Psi'{}(x,t) = i \frac{\partial}{\partial t} \Psi^*{}(x,-t) = i \frac{\partial}{\partial t} u^*(x) e^{-i E t}= E \Psi^*{}(x,-t)= H(x, -i\frac{\partial}{\partial x}, t) \Psi'{}(x,t) .$$

## 3.

Yet another way to look at the problem maybe this. We do not act $$T$$ on the Schroedinger equation.* But directly find a time-reversal solution $$|\Psi'{}(t) \rangle$$ satisfies the Schroedinger equation. If we define the statevector to be expanded in the projected momentum $$|p \rangle$$ basis: $$|\Psi'{}(t) \rangle = \sum_p |p \rangle \langle p |\Psi'{}(t) \rangle$$ In this case $$\Theta | p \rangle= |-p \rangle$$, so $$|\Psi'{}(t) \rangle =\Theta|\Psi {}(t) \rangle = \sum_p |-p \rangle \Psi^*{}(p,-t) = \sum_p |-p \rangle u^*(p) e^{- i E t}= \sum_p | p \rangle u^*(-p) e^{- i E t}.$$ Then we can test the wavefunction version of Schrodinger equation: $$i \frac{\partial}{\partial t} \Psi{}(p,t) = H( i\frac{\partial}{\partial p}, p, t) \Psi{}(x,t) .$$ We define $$\Psi'{}(p,t)= \Psi^*{}(-p,-t)$$.

## Critique or criticism on three ways:

The 1. approach, We do act $$T$$ on the Schroedinger equation, but we fail to see why the wavefunction should be complex conjugated under $$T$$. But the $$T$$ reversed state has the negative energy $$-E$$.

The 2. approach, We do not act $$T$$ on the Schroedinger equation, but we see why the wavefunction should be complex conjugated under $$T$$. Also the $$T$$ reversed state has the same original positive energy $$E$$. The new wavefunction $$\Psi'{}(x,t)= \Psi^*{}(x,-t)$$ satisfies the same form of Schroedinger equation $$i \frac{\partial}{\partial t} \Psi'{}(x,t) = H(x, -i \frac{\partial}{\partial x}, t) \Psi'{}(x,t)$$.

The 3. approach, We do not act $$T$$ on the Schroedinger equation, but we see why the wavefunction should be complex conjugated under $$T$$. Also the $$T$$ reversed state has the same original positive energy $$E$$. The new wavefunction $$\Psi'{}(p,t)= \Psi^*{}(-p,-t)$$ satisfies the same form of Schroedinger equation $$i \frac{\partial}{\partial t} \Psi'{}(p,t) = H( i\frac{\partial}{\partial p}, p, t) \Psi'{}( p, t)$$. I assume it is true?

The three versions of time reversal act differently. Their interpretations seem to be different. What is the correct way to define time reversal on Schroedinger equation?

The equation $$T|\psi(t)\rangle=|\psi(-t)\rangle\tag1$$ is not quite correct.

In order to discuss anti-linear operators we must endow the Hilbert space with a real structure, basically a notion of which vectors are real and which are not. A real structure is a splitting of your Hilbert space $$\mathcal H$$ into two sectors, $$\mathcal H=\mathcal H'\oplus i\mathcal H'\tag2$$ where, by definition, all vectors in $$\mathcal H'$$ are declared to be real. The reality properties of a generic element $$v\in\mathcal H$$ are thus inherited from this decomposition; any $$v$$ can be written as $$v=v'+iv''$$, where $$v',v''\in \mathcal H'$$.

Equation $$(1)$$ is fine, but it implicitly assumes/defines $$|\psi\rangle$$ to be real, i.e., it means that $$|\psi\rangle$$ lies in $$\mathcal H'$$. For general vectors, the correct expression is $$T|\psi(t)\rangle=K|\psi(-t)\rangle\tag3$$ where $$K$$ is the "complex conjugation operation" which, by definition, fixes real vectors and complex-conjugates scalars. In other words, $$K(v'+iv''):=v'-iv''$$.

The position-space equivalent of $$(3)$$ is, non-surprisingly, $$\langle x|K|\psi(-t)\rangle=\psi(x,-t)^*\tag4$$ so that, on wavefunctions, time-reversal sends $$t\to-t$$ and $$i\to-i$$.

A very common example is time-reversal on systems with spin. One often writes $$T=\sigma_yK$$. Note that $$\sigma_y$$ is purely imaginary so $$T^2=\sigma_y\sigma_y^*=-1$$. This immediately leads to e.g. Kramers theorem.

In my simple view the Schrödinger equation is not invariant under time reversal. Instead, it transforms from

$$\frac{\hbar}{i} \frac{\partial}{\partial t} \Psi = H(t) |\Psi$$

Into

$$-\frac{\hbar}{i} \frac{\partial}{\partial t} \Psi = H(-t) |\Psi \,.$$

If H(-t)=H(t) then the solutions of both equations are related by complex conjugation.

• Thanks. 1. you need to also specify the Ψ, which has $t$ dependence. May 25, 2021 at 0:02
• 2. you need to also specify the Ψ, whether this is (1) a state VECTOR in the Hilbert space. (2) Or a projected wavefunction to a basis as a coefficient Ψ? May 25, 2021 at 0:02
• 3. Also can you explain when you do $T$ transforms on Schrödinger equation, do you perform $Ti T^{-1}=-i$? or $T ∂𝑡Ψ T^{-1} =-∂𝑡Ψ$? There are so many possible (-1) sign due to the mod 2 nature of $T$. This needs many clarifications. May 25, 2021 at 0:06
• anyway Thanks! +1 May 25, 2021 at 0:06
• @annmariecœur $\Psi$ is just a complex valued function of space and time. I only replace $\partial_t$ by $-\partial_t$ and $H(t)$ by $H(-t)$. The new $\Psi(t)$ is then equal to the old $\Psi*(-t)$. May 27, 2021 at 16:12