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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.

Community
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Olórin
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1 Answers1

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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.

Ján Lalinský
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  • 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? – jabirali Feb 07 '15 at 11:32
  • 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. – Ján Lalinský Feb 07 '15 at 11:43
  • 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. – Ján Lalinský Feb 07 '15 at 11:51
  • 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. – jabirali Feb 07 '15 at 12:05
  • 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. – jabirali Feb 07 '15 at 12:06
  • @jabirali, the sequence of characters $\hat{H} = i\hbar d/dt$ is useless as a definition of $\hat{H}$, because it is not a definition but a short-hand for the general form of time-dependent Schroedinger equation, which is already simple enough to remember. The Hamiltonian still needs a definition in terms of coordinates and momenta. The relation $\hat{p}k = -i\hbar\partial/\partial{x_k} $, on the other hand, is useful short-hand for the definition of the operators of canonical momentum. – Ján Lalinský Feb 07 '15 at 12:24
  • I still don't agree that $\hat H = i\hbar d/dt$ is a useless definition. Essentially, we just say that $\hat H(\mathbf x, \mathbf p, t)$ is whatever operator that is equivalent to $i\hbar d/dt$ for a given physical system; and then, writing an explicit form of $H$ is equivalent to defining the time evolution of that given physical system. In that way, the interpretation of $\hat H$ as $i\hbar d/dt$ is more fundamental than the interpretation of $\hat H$ as an energy operator (since the latter is only valid if $\hat H(\mathbf x,\mathbf p,t)$ is not an explicit function of $t$.) – jabirali Feb 07 '15 at 13:16
  • Within your view Schroedinger's t-d equation becomes a tautology. It is then irrefutable by experiment. That may seem like building something solid, but it is useless as a definition of $\hat{H}$ for concrete physical system. Why? Because to use it, say, to find the emission frequencies of hydrogen atom, you still need to define further what $\hat{H}$ is in terms of coordinates and momenta of the system. In fact your definition defines no Hamiltonian, it defines what evolution of a physical system is. And thus it restricts your idea of evolution to Hamiltonian operators. – Ján Lalinský Feb 07 '15 at 18:20
  • For example why that is important, if Einstein considered $\mathbf F=m\mathbf a$ to be a definition of force, it would be a tautology and it would never be possible to consider particle accelerating in the direction of the $x$-axis due to force directed along the $y$-axis. But this happens all the time and we understood it not by changing the explicit form of $\mathbf F$, but by updating the equation to $\mathbf F = m\frac{d\gamma \mathbf v}{dt}$. – Ján Lalinský Feb 07 '15 at 18:36
  • Actually, $\mathbf F = d\mathbf p/dt$ is precisely just a definition of force, in the same way $H = i\hbar d/dt$ is a definition of the Hamiltonian. The underlying physical law here is the conservation of momentum, and Newtons laws are just a convenient way to reexpress that underlying conservation law. And then, yes, the definition of a force was updated to a more convenient form when relativity came around. – jabirali Feb 07 '15 at 18:47
  • But I think I'll stop the argument there; I think we're starting to get a bit off-topic :). My original remark was just that I don't think it is correct to say that $i \hbar d/dt$ isn't an operator in the same way that $-i\hbar \nabla$ is an operator, and this was relevant for the original posters question. – jabirali Feb 07 '15 at 18:49
  • Your definition of force is usable for a very restricted class of situations and fails for others. To close this by helpful words, let me say your way of thinking about concepts is very useful in mathematics, where you'll apply it with more success. It is however not sufficient in physics, where the complexity of the subject and empirical updates of knowledge do not allow us insistence on one definition of force or time evolution but requires broader view of these concepts independent of the equations that are currently known to approximately quantify them. – Ján Lalinský Feb 07 '15 at 19:20