# Non-existence of time operator in quantum mechanics

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.

• Hm, time is not an observable like energy or impulse. – Karl Apr 21 at 16:51
• 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? – Ivan Neretin Apr 21 at 16:59
• No, I mean time elapsed. – Manu Apr 21 at 17:01
• If you go to Physics and search for "time operator" you will find plenty of answers... – orthocresol Apr 21 at 17:10
• Define "Hermitian". – Karl Apr 21 at 19:43

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.