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I am reading Zee's Quantum field theory in a nutshell. On time reversal he has

Consider the transformation $t\rightarrow t'= -t$. We want to find $\Psi'(t')$ such that $i(\partial/\partial t′)\Psi′( t′) = H\Psi'(t′)$. Write $\Psi′(t′) = T\Psi(t)$, where $T$ is some operator to be determined (up to some arbitrary phase factor $\eta$). Plugging in, we have $i[ \partial/ \partial( − t)] T\Psi( t) = HT\Psi( t)$. Multiply by $T^{-1}$, and we obtain $T^{ − 1}( − i) T( \partial/ \partial t)\Psi( t) = T^{−1}HT( t)\Psi(t)$.

My question is: Has he assumed that $T$ and $\partial/\partial t$ commute and if so why is it valid to do that?

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Consider derivative at $t=0$; denote $\Psi(0)$ as $\Psi_{0}$

$(\partial /\partial t)T\Psi(t)\big|_{t=0}=\displaystyle lim_{h\rightarrow0}((T\Psi_{0})(h)-T\Psi_{0})/h$

Since $T\Psi$ evolves according to $i(\partial /\partial (-t))T\Psi(t)=HT\Psi(t)$

So $(T\Psi_{0}) (h)= exp(ihH)T\Psi_{0}$. Hence we have :

$\displaystyle lim_{h\rightarrow0}((T\Psi_{0})(h)-T\Psi_{0})/h$

$=\displaystyle lim_{h\rightarrow0}(exp(iHh)(T\Psi_{0})-T\Psi_{0})/h$

$=\displaystyle lim_{h\rightarrow0}(Texp(-iHh)\Psi_{0}-T\Psi_{0})/h$ $\:\:\;\;\;$(Since $T$ and $H$ commute, and $T$ is antilinear)

$=\displaystyle lim_{h\rightarrow0}T(exp(-iHh)\Psi_{0}-\Psi_{0})/h$

$=T(\partial /\partial t)\Psi(t)\big|_{t=0}$

Notation : $(T\Psi_{0})(h)$ means we first act $\Psi_{0}$ by $T$ and then time evolve the resulting state by an amount of time $h$.

Another argument : following argument seems more relevant here than above one :-

We have a one parameter family of states $\Psi(t)$ which satisfy

$i(\partial /\partial (t))\Psi(t)=H\Psi(t)$

For definiteness suppose $t\in[0,1]$, and suppose we partition this interval into $N$ equal parts (where $N$ is some large number) as {$0=t_0<t_1<....<t_{N-1}<t_N=1$}. Denote $\Psi (t_j)$ as $\Psi_j$ for $j=0,...,N$, and let $1/N=\delta$ (length of one small interval).Then above differential equation can be written as a set of $N$ linear equations in terms of states $\Psi_j$'s as :

$i(\Psi_1-\Psi_0)/\delta=H\Psi_0$

$i(\Psi_2-\Psi_1)/\delta=H\Psi_1$

....

$i(\Psi_j-\Psi_{j-1})/\delta=H\Psi_{(j-1)}$

....

$i(\Psi_N-\Psi_{(N-1)})/\delta=H\Psi_{(N-1)}$

Now in Zee's book the one parameter family of vectors $T\Psi(t)$ is required to satisfy the differential equation $-i(\partial /\partial t)T\Psi(t)=HT\Psi(t)$. Or in discretised form it is required that the set of vectors $T\Psi_0,T\Psi_1,.....,T\Psi_N$ satisfy following linear equations :

$-i(T\Psi_1-T\Psi_0)/\delta=HT\Psi_0$

$-i(T\Psi_2-T\Psi_1)/\delta=HT\Psi_1$

....

$-i(T\Psi_j-T\Psi_{j-1})/\delta=HT\Psi_{(j-1)}$

....

$-i(T\Psi_N-T\Psi_{(N-1)})/\delta=HT\Psi_{(N-1)}$

Now since $T$ is linear wrt addition of states so it can be taken out:

$-iT(\Psi_1-\Psi_0)/\delta=HT\Psi_0$

$-iT(\Psi_2-\Psi_1)/\delta=HT\Psi_1$

....

$-iT(\Psi_j-\Psi_{j-1})/\delta=HT\Psi_{(j-1)}$

....

$-iT(\Psi_N-\Psi_{(N-1)})/\delta=HT\Psi_{(N-1)}$

In continuum limit these equations are equivalent to :

$-iT(\partial /\partial t)\Psi(t)=HT\Psi(t)$


Edit :

Question: Consider a one parameter family of states $\Psi(t)$ which satisfy Schrodinger equation $i(\partial /\partial t)\Psi(t)=H\Psi(t)$. Is it possible to find an invertible linear operator $T$ that commutes with $H$ and such that for any $\Psi(t)$ as above, $T\Psi(t)$ satisfies $-i(\partial /\partial t)T\Psi(t)=HT\Psi(t)$ ?

Our previous argument (2nd one) extends to one proof that it is not possible; Here is another one :

If $T$ is such an operator then $T\Psi(t)=exp(itH)T\Psi(0)$. (because $T\Psi(t)$ solves time reversed Schr. equation) -----(1)

Also $\Psi(t)=exp(-itH)\Psi(0)$ (Because $\Psi(t)$ solves usual Schr. equation). -------(2)

Substituting (2) into (1) we get $Texp(-itH)\Psi(0)=exp(itH)T\Psi(0)$

Now using the fact that T is invertible we get :

$exp(-itH)\Psi(0)=T^{-1}exp(itH)T\Psi(0)$

Again using the fact that $T$ is linear and commutes with $H$ we get

$exp(-itH)\Psi(0)=exp(itH)\Psi(0)$ (note that if $T$ were antilinear then in place of $exp(itH)$ on RHS we would have $exp(-itH)$, and hence there would be no problem)

Now multiplying on both sides with $exp(-itH)$ we get

$exp(-2itH)\Psi(0)=\Psi(0)$

Differentiating with respect to $t$ and putting $t=0$ we get

$H\Psi(0)=0$

But $\Psi(0)$ was any arbitrary state in our space of states. So we have $H=0$ identically. Hence the required linear operator is not possible unless $H$ vanishes identically.

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@physicsphile i have made some corrections. –  user10001 Sep 3 '12 at 1:31
    
I think that is right, I suppose the bottom line is that T takes the complex conjugate of what it is operating on and $\partial/\partial t$ is real. But I don't quite follow the logic of Zee's presentation of it. I think it would be worth removing your first argument as you pointed out it is not correct as T will change the sign of the exponential –  physicsphile Sep 3 '12 at 4:34
    
i don't understand why first argument is incorrect. i have checked it many times and it seems to be correct; only problem is that it will go through for linear $T$ as well. Anyway, if i found that its incorrect i will remove it. –  user10001 Sep 3 '12 at 5:41
    
$\Psi(t)$ evolves according to usual Schrodinger equation and so $\Psi(t)=exp(-itH)\Psi_0$. –  user10001 Sep 3 '12 at 6:26
    
Here $T\Psi$ is required to satisfy time reversed form of Schrodinger equation, which is of the form $-i(\partial/\partial t)(T\Psi)(t) = HT\Psi(t)$. So solution for this equation would be of the form $T\Psi(t)=exp(itH)T\Psi_0$. –  user10001 Sep 3 '12 at 6:42
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