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I don't know if this question has already been resolved but considering that $i\hbar\partial_t$ is the energy operator, and $\partial^2_t$ is the waves operator (or helmholtz), I can't accept that $t$ itself isn't an operator

What is the argument here that says $t$ is not an operator?

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  • $\begingroup$ This is basically a duplicate of this: physics.stackexchange.com/questions/163279/… See my answer there. $\endgroup$ – hft Feb 16 '15 at 8:55
  • $\begingroup$ Yes this question has already been resolved. $\endgroup$ – hft Feb 16 '15 at 8:57
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To say that something is a (linear) operator you have to specify the space where it acts. You may say that, for example, wavefunctions of quantum mechanics are maps: $t\to \psi(t)$ that are continuous in $t$ with values in $L^2(\mathbb{R}^d)$. If we restrict to compact time intervals $[0,T]$, we may denote the space of these maps by $C^0([0,T],L^2(\mathbb{R}^d))$.

On $C^0([0,T],L^2(\mathbb{R}^d))$, with norm $\lVert \psi(\cdot)\rVert_{C^0}=\sup_{t\in [0,T]}\lVert\psi(t)\rVert_{L^2}$, both $t$ and $-i\partial_t$ are densely defined linear operators [actually the multiplication by $t$ is bounded, with norm $T$; the derivation has domain $C^1([0,T],L^2(\mathbb{R}^d))$].

On $C^0(\mathbb{R},L^2)$, also the multiplication by $t$ is unbounded, and since every wavefunction of QM satisfies, by means of Schrödinger equation $\lVert\psi(t)\rVert_{L^2}=\lVert\psi(t_0)\rVert_{L^2}=k$ for any $t,t_0\in \mathbb{R}$ (where $k\geq 0$, usually $k=1$); we see that every nonzero wavefunction is outside the domain of definition of $t$, as a multiplication operator on $C^0(\mathbb{R},L^2)$ [because $\sup_{\mathbb{R}}\lvert t\rvert \lVert\psi(t)\rVert_{L^2}=k\sup_{\mathbb{R}}\lvert t\rvert=+\infty$].

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You are touching on the subject of relativistic quantum mechanics where time and space $(t,x)$ are handled on the same footing as operators. The accepted description is to not use quantum wavefunctions as describing one particle but rather the state of a quantum field. Doing this turns into the subject of quantum field theory and is the basis of modern quantum experimentation/theory.

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    $\begingroup$ In QFT both $x$ and $t$ are parameters not operators. $\endgroup$ – Constandinos Damalas Feb 16 '15 at 7:31

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