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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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4 Answers 4

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

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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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You're asking how certain quantities act under a transform that reverses time $t → -t$. Let's deal with this question in general. Associate with each quantity $q$ its parity $[q]$, which we assume to be non-zero. The associated transform is assumed to be $q → [q] q$, and these are the only ones we'll consider here.

We'll assume that it's product-preserving, i.e. $[qq'] = [q][q']$. Similarly, we assume compatibility with differential quotients $[dq/dq'] = [q]/[q']$, similarly for partial derivatives $[∂q/∂q'] = [q]/[q']$, along with compatibility under differentials: $[dq] = [q]$. We also require compatibility for linear combinations, which is that we must have $[q] = [q']$, whenever linear combinations, such as $q + q'$, arise; and we'll only treat transforms that have this kind of compatibility.

Consider an action given by the integral $S = \int L(q,v) dt$, where the system whose dynamics are described by the action has configuration variables $q = \left(q^a: 0≤a<N\right)$ and first derivatives $v = \left(v^a = dq^a/dt: 0≤a<N\right)$. Assume that all of the $q$'s have the same parity: $[q^a] = A$, for $0≤a<N$. Set $[L] = B$ and $[t] = C$, so that $[dt] = C$ and $[S] = [L dt] = BC$. If the action is invariant then $BC = 1$ - which is already your answer in disguise.

The velocities satisfy $v^a = dq^a/dt$, so $\left[v^a\right] = A/C$. The conjugate momenta are given by $p_a = ∂L/∂v^a$, so $\left[p_a\right] = BC/A$. Thus, we have for each product $\left[v^a p_a\right] = B$, and for the Hamiltonian $H = \sum_{0≤a<N} p_a v^a - L$: $[H] = B$.

Also: for the coefficients of inertia $m_{ab} = ∂^2L/∂v^a∂v^b$, we have $\left[m_{ab}\right] = BC^2/A^2$; and for the conjugate forces $f_a = ∂L/∂q^a$, we have $\left[f_a\right] = B/A$.

Now, consider the following transforms:

  • $\widehat{A}$: where $(A,B,C) = (-1,+1,+1)$,
  • $\widehat{B}$: where $(A,B,C) = (+1,-1,+1)$,
  • $\widehat{C}$: where $(A,B,C) = (+1,+1,-1)$,

as well as their products, e.g. $\widehat{AB} = \widehat{A}\widehat{B}$, where $(A,B,C) = (-1,-1,+1)$.

List the quantities whose signs are reversed by the respective transforms:

  • $\widehat{A}$: $(q,v,p,f)$,
  • $\widehat{B}$: $(S,L,H,p,f,m)$,
  • $\widehat{C}$: $(t,S,v,p)$,
  • $\widehat{AB}$: $(S,L,H,q,v,m)$,
  • $\widehat{AC}$: $(t,S,q,f)$,
  • $\widehat{BC}$: $(t,L,H,v,f,m)$,
  • $\widehat{ABC}$: $(t,L,H,q,p,m)$.

The transform $\widehat{A}$ is "configuration reversal", which generalizes parity reversal; while $\widehat{B}$ is "dynamics reversal", which reverses all of the dynamical variables, as well as the action. The transforms that reverse time and the action are $\widehat{C}$ and $\widehat{AC}$, while the ones that reverse time and leave the action invariant are $\widehat{BC}$ and $\widehat{ABC}$.

Which one of these is "time reversal"? If you require that $[v] = -1$ under time reversal, in addition to leaving the action invariant, then that singles out $\widehat{BC}$. Look at what happened to $m$: $[m] = -1$, and $f$ too: $[f] = -1$. Let that sink in. If you don't want the coefficients of inertia and forces to reverse, but still want the velocities to reverse, then you're stuck with $\widehat{C}$, in which case the action reverses: $[S] = -1$.

Since the combination $2πiS/h$ appears in quantum theory, then if you want $[2πiS/h] = 1$, (with $[h] = 1$) then $[S] = -1$ requires $[i] = -1$. So, action-reversal goes with $i → -i$.

I can see a lot of other questions coming out of this: what is the effect of time reversal on the action, the mass, the Lagrangian, the force, the energy? A little search shows that some of these other issues have also popped up in other threads on the forum. With the choices presented, something has to give, so it might be a good idea to do a search on these other questions, too.

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