I'm struggeling to solve this question. Can anyone help me please?

Let us consider a generic quantum mechanical system governed by the Hamiltonian $\ H(t) $.

In what follows we denote the evolution operator by $\ u(t, t_0) $. Hence, $\ |Ψ(t)> = u(t, t_0)|Ψ_0> $ satisfies the time-dependent Schrodinger equation, where $\ |Ψ_0> $ represents the wave function at $\ t = t_0$.

(a) Prove the unitarity condition $\ u^† (t, t_0)u(t, t_0) = u(t, t_0)u^†(t, t0) = I $ .

(b) Now let us assume that the system under study exhibits a symmetry represented by an antilinear (and antiunitary) operator $U$ which has nothing to do with the time reversal. Show that in this case $\ [u(t, t0), U] = 0 $, and thus the system is unstable since its spectrum is unbounded from below.

(c) Show that instability disappears if antilinear $U$ includes time reversal.

The conclusion from (b) and (c) is that symmetries represented by antilinear operators are possible, but they necessarily involve time reversal.

  • $\begingroup$ Hint: H is hermitian. Take ${t_0} =0$. Also: $u(t)=e^{\frac{iHt}{h}}$. $\endgroup$ Feb 10, 2017 at 12:17

1 Answer 1


Let's make $u(t)=e^{it}$ ($H$ is hermitian)

Then $u^\dagger u=e^{-it}e^{it}=e^{it}e^{-it}=u u^{\dagger}=1$

Let $e^{it}$ and $U$ act on $\phi$:

$$(e^{it} U-U e^{it})\phi=(-e^{it}U+ e^{-it}U)\phi,\quad \text{or}\quad (-e^{it}+e^{-it})U\phi.$$

Now $U\phi=-U\phi$, so $(-e^{it}+ e^{-it})=-(-e^{it}+e^{-it})$ and therefore $0$, so (b) holds

When we include time reversal in $U$, let $U e^{it}$ act on $\phi(0)$, which becomes $-\phi(t)$.

Try it from there yourself.

  • $\begingroup$ $H$ doesn't matter because it's Hermitian. And $\hbar$ is a constant. $\endgroup$ Feb 10, 2017 at 15:22
  • $\begingroup$ The hermitian conjugate of $H$ is H. Of course, you can let it in, but for the end result, it doesn't matter: $(-e^{iHt}+e^{-iHt})=-(-e^{iHt}+e^{-iHt})$, which implies also zero. $\endgroup$ Feb 10, 2017 at 15:30
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    $\begingroup$ @AccidentalFourierTransform-Why did you delete your two comments, in which you asked me why I didn't take $H$ into consideration in the $e^{it}$ expression? $\endgroup$ Feb 11, 2017 at 16:42

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