In quantum mechanics, there is no operator for time instead of this we we have $\psi(t)$ which is related to $\psi(0)$.

In a book, I read that if there is an operator for time, then it will not be Hermitian". I am not able to understand how exactly why? What is the reasoning behind this statement.

  • $\begingroup$ Hm, time is not an observable like energy or impulse. $\endgroup$ – Karl Apr 21 at 16:51
  • $\begingroup$ Time is not just a physical value like any other. What would you want this operator to return, had it existed? Time in milliseconds since the Big Bang? $\endgroup$ – Ivan Neretin Apr 21 at 16:59
  • $\begingroup$ No, I mean time elapsed. $\endgroup$ – Manu Apr 21 at 17:01
  • 4
    $\begingroup$ If you go to Physics and search for "time operator" you will find plenty of answers... $\endgroup$ – orthocresol Apr 21 at 17:10
  • $\begingroup$ Define "Hermitian". $\endgroup$ – Karl Apr 21 at 19:43

On the not Hermitian point.

Let $\hat{F}$ be a non-stationary observable (but does not depend explicitly on time $t$, hence $\partial \hat{F}/\partial t = 0)$. in the Heisenberg picture it satisfies the Heisenberg equation of motion for $\hat{F}$, namely $$i \frac{d \hat{F}}{dt} = [\hat{F},\hat{H}] \tag{1}$$.

Now assume that whenever $d \hat{F}/dt \neq 0$ then it has an inverse $$(d \hat{F}/dt)^{-1} = i([\hat{F},\hat{H}])^{-1}\tag{2}$$ Now, if \begin{align} [(d \hat{F}/dt)^{-1}, \hat{H}] &= 0 \\ \implies [([\hat{F},\hat{H}])^{-1},\hat{H}] &=0. \end{align} Now introduce a Time operator, $\hat{T}$ as \begin{align} \hat{T} &= \frac{1}{2}\left(\hat{F}\left(\frac{d \hat{F}}{dt}\right)^{-1} + \left(\frac{d \hat{F}}{dt}\right)^{-1} \hat{F}\right) \\ &=\frac{1}{2}\left(\hat{F} ([\hat{F},\hat{H}])^{-1}+([\hat{F},\hat{H}])^{-1}\hat{F}\right)\tag{3} \end{align} Now using equations $(1)$ and $(2)$ into $(3)$ one finds that $$[\hat{H}, \hat{T}] = i \tag{4}$$

That is, $\hat{T}$ and $\hat{H}$ form a canonically conjugate pair. Owing to the fact that the Hamiltonian spectrum is bounded from below, the time operator $\hat{T}$ is not self adjoint, in other words, not Hermitian.

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