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While reading a book about interactions I've come to this paragraph:

In the Schrodinger representation, the state function satisfies the equation $$\mathrm{i}\frac{\partial}{\partial t}\phi(t,x)=H\phi(t,x).$$

Invariance requires the transformed wave function $\mathcal{T}\phi(t,x)$ to satisfy the same equation with $t$ replaced by $t^\prime=-t$. The question is, how is $\mathcal{T}\phi(t,x)$ related to $\phi(t,x)$?

The simplest possible postulate, $\mathcal{T}\phi(t,x)=\phi(-t,x)$, leads to $$\mathrm{i}\frac{\partial}{\partial (-t)}\phi(-t,x)=-H\phi(-t,x)$$

How exactly did we get the minus on the right hand side? The way I think it should go is if we reverse the time and replace all $t$ by -$t$, we get $-t$ in all arguments of the wavefunction and one minus in the derivative with respect to time. Where does the 4th minus on the right hand side of the equation come from? The only explanation I could somehow come up with is that we only changed $t$ to $-t$ inside the argument of wave function and left the derivative untouched and only then mulitplied the whole equation by ($-1$) to get the minus in derivative.

What am I missing? :(

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  • $\begingroup$ The Schrödinger equation is not symmetric under time reversal. $\endgroup$
    – my2cts
    Oct 19, 2019 at 22:11

3 Answers 3

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  1. The Schrödinger equation is not necessarily symmetric under time reversal. For a time reversal operator $\mathcal{T}$, we have $$\mathrm{i}\hbar\frac{\partial}{\partial t}\mathcal{T}\psi(x,t)=H\mathcal{T}\psi(x,t)$$ only when $$\left[H,\mathcal{T}\right]=0.$$

  2. Consider an observable $A$. We have $$\mathcal{T}A\mathcal{T}^{-1}=A'.$$ We require that the following relations hold:

a) Correspondence of position: $$\mathcal{T}x\mathcal{T}^{-1}=x^\prime=x$$ b) Correspondence of momentum (the sign change is because in a classical sense, the particle would be moving in the opposite direction under time reversal): $$\mathcal{T}p\mathcal{T}^{-1}=p^\prime=-p$$ c) Commutation relation: $$[x_i,p_j]=\mathrm{i}\hbar\delta_{i,j}\Rightarrow\mathcal{T}[x_i,p_j]\mathcal{T}^{−1}=[x_i^\prime,p_j^\prime]=[x_i,−p_j ]=−[x_i,p_j]$$ So, $$\mathcal{T}(\mathrm{i}\hbar\delta_{i,j})\mathcal{T}^{-1}=-\mathrm{i}\hbar\delta_{i,j}\Rightarrow\mathcal{T}\mathrm{i}\mathcal{T}^{-1}=-\mathrm{i}.$$

This suggests that we 'construct' the time reversal operator in the following manner, using some linear unitary operator $U$ and an antiunitary operator $\mathcal{K}$ which performs complex conjugation here: $$\mathcal{T}=U\mathcal{K};\quad\mathcal{K}\mathrm{i}\mathcal{K}^{-1}=-\mathrm{i}.$$ In other words, the time-reversal operation in quantum mechanics involves both complex conjugation and negating the time: $$t\rightarrow -t;\quad \mathrm{i}\rightarrow-\mathrm{i}.$$

Let $t^\prime=-t$, such that $\mathcal{T}\psi(x,t)=\psi^\prime(x,t^\prime)$. Note that this is not the same definition as the quoted section in the question; we do not take $\psi(x,t^\prime)=\psi^\prime(x,t^\prime)$ (I suppose that there may be additional commentary regarding that equation beneath the sections included in the quotation). Requiring that $\psi^\prime(x,t^\prime)$ is a solution to the corresponding Schrödinger equation, \begin{align}\mathrm{i}\hbar\frac{\partial}{\partial t^\prime}\psi^\prime(x,t^\prime)=H\psi^\prime(x,t^\prime)&\Leftrightarrow\mathrm{i}\hbar\frac{\partial}{\partial t^\prime}\mathcal{T}\psi(x,t)=H\mathcal{T}\psi(x,t)\\\mathcal{T}\mathrm{i}\hbar\mathcal{T}^{-1}\mathcal{T}\frac{\partial}{\partial t}\mathcal{T}^{-1}\mathcal{T}\psi(x,t)&=\mathcal{T}H\mathcal{T}^{-1}\mathcal{T}\psi(x,t)\\U\mathcal{K}(\mathrm{i}\hbar)U^{-1}\mathcal{K}^{-1}U\mathcal{K}\left(\frac{\partial}{\partial t}\right)U^{-1}\mathcal{K}^{-1}\psi^\prime(x,t^\prime)&=\mathcal{T}H^\prime\mathcal{T}^{-1}\psi^\prime(x,t^\prime)\\&=H^\prime\psi^\prime(x,t^\prime)=H\psi^\prime(x,t^\prime)\end{align}Thus,$$\boxed{-\mathrm{i}\hbar\frac{\partial}{\partial t}\psi^\prime(x,t^\prime)=H\psi^\prime(x,t^\prime);\quad-\mathrm{i}\hbar\frac{\partial}{\partial t}\mathcal{T}\psi(x,t)=H\mathcal{T}\psi(x,t).}\tag{A}$$Equation (A) holds irrespective of $[H,\mathcal{T}]$.

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I think that quoted paragraph is a little ambiguous - as I understand it, when you apply the time reversal operator, you're changing $t \to -t$ and taking the complex conjugate of the whole equation, so the $i$ on the l.h.s. gains a minus sign that's moved to the Hamiltonian.

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The minus sign on the right is an error. The time reversed wave function obeys the conjugate Schrödinger equation.

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