On the question why time isn't an operator, people will usually say that time is a parameter in QM (Time as a Hermitian operator in QM?) and not a variable.

  1. Can someone please distinguish between a parameter, a variable and an operator as it is used in QM?

  2. Oh by the way, if someone can resolve why time cannot be an operator but the derivative of time multiplied by some complex number namely $i\hbar d/dt$ is totally cool.


1 Answer 1


Coordinates and momenta of the system are called variables by convention.

Time is sometimes called variable, after all its nature is to keep changing.

But to distinguish time from coordinates and momenta, people call it parameter in some cases.

Parameter is a word used instead of variable when the quantity is a different kind of argument of a function.

For example, if

$$ \psi_t(q_1,q_2) $$

is a function of both coordinates $q_1,q_2$ and time $t$, but we want to stress that we focus on the dependence on $q_1,q_2$ and think of value of time $t$ as fixed constant during discussion, we call it a parameter.

Operator in quantum theory is used in a specific sense "operator that acts on functions of coordinates", or "operator that acts on functions of momenta." In this sense $i\hbar \partial/\partial t$ is not an interesting QM operator because it always results in function equal to 0 everywhere. It is a differential operator of functions of time, but not a QM operator in the usual sense.

The reason for this is partially that it makes no sense to use this differential operator on functions of $q$ in a typical QT role

$$ \int \psi^* \hat{A}\psi dq $$ to get expected average of quantity modeled by $A$. Sometimes people think that this quantity for $i\hbar \partial/\partial t$ is energy, and then the integral gives expected average energy. But that is correct only if $\psi_t(q)$ obeys time-dependent Schroedinger's equation, while $\hat{H}$ works for any function of $q$.

The difference in treating and naming $t$ stems also from the fact that while coordinates and momenta describe state of the atomic system and are difficult to measure accurately, time $t$ describes state of a clock, which is not microscopic but can be measured simply by taking a look at the clock.

  • $\begingroup$ Note that the Schrodinger equation says precisely that $H = i\hbar d/dt$, which is why it is natural to interpret $i\hbar d/dt$ as energy. (The Hamiltonian is the generator of time translations.) When you say that this is only correct if $\psi$ follows the Schrodinger equation, what exception do you have in mind? $\endgroup$
    – jabirali
    Feb 7, 2015 at 11:32
  • $\begingroup$ Schroedinger's time-dependent equation does not say $H = i\hbar d/dt$, it is an equation for time development of function of coordinates. There are many different Hamiltonian operators, but none is equal to $i\hbar d/dt$ since that would make the equation useless. $\endgroup$ Feb 7, 2015 at 11:43
  • $\begingroup$ I meant the cases when the function $\psi$ does not depend on time, for example when it is a solution of time-independent Schroedinger equation. $\endgroup$ Feb 7, 2015 at 11:51
  • $\begingroup$ I don't see why the operator equation $\hat{H} = i\hbar d/dt$ is any more useless than the equivalent equation $\mathbf{\hat p} = -i\hbar\nabla$ for momentum though; just right-multiply by a wavefunction, and you get the conventional forms of the equations back. In fact, I believe it is common to define the Hamiltonian as the infinitesimal time evolution operator of a given physical system. $\endgroup$
    – jabirali
    Feb 7, 2015 at 12:05
  • $\begingroup$ The $\psi(\mathbf{x})$ in the time-independent Schrodinger equation is not the full wavefunction; the entire wave function $\Psi(\mathbf{x},t) = \psi(\mathbf{x}) \exp(-iEt/\hbar)$ does satisfy the Schroedinger equation. $\endgroup$
    – jabirali
    Feb 7, 2015 at 12:06

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