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In the time dependent Schrodinger equation $\displaystyle, H\Psi = i\hbar\frac{\partial}{\partial t}\Psi$ , the Hamiltonian operator is given by

$\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2+V$

Why can't we consider $\displaystyle i\hbar\frac{\partial}{\partial t}$ as an operator for the Hamiltonian as well?

My answer (which I am not sure about) is the following:

$\displaystyle H\Psi = i\hbar\frac{\partial}{\partial t}\Psi$ is not an equation for defining $H$. This situation is similar to $\displaystyle F=ma$. Newton's 2nd law is not an equation for defining $F$, $F$ must be provided independently.

Is my reasoning (and the analogy) correct, or the answer is deeper than that?

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Yes, you are correct. The physics is in the Hamiltonian and the Schrodinger equation describes how the Hamiltonian causes the wavefunction to change as a function of time. – FrankH Nov 26 '11 at 14:34
Related: – Qmechanic Nov 26 '11 at 14:57

1) If one a priori wrongly declares that the Hamiltonian operator $\hat{H}$ is the time derivative $i\hbar \frac{\partial}{\partial t}$, then the Schrödinger equation

$$\hat{H}\Psi~=~i\hbar \frac{\partial\Psi}{\partial t}$$

would become a tautology. Such trivial Schrödinger equation could not be used to determine the future (nor past) time evolution of the wavefunction $\Psi({\bf r},t)$.

2) On the contrary, the Hamiltonian operator $\hat{H}$ is typically a function of the operators $\hat{\bf r}$ and $\hat{\bf p}$, and the Schrödinger equation

$$\hat{H}\Psi~=~i\hbar \frac{\partial\Psi}{\partial t}$$

is a non-trivial requirement for the wavefunction $\Psi({\bf r},t)$.

3) One may then ask why is it then okay to assign the momentum operator as a gradient

$$\hat{p}_{i}~=~ \frac{\hbar}{i}\frac{\partial}{\partial r^i} ?$$

(This is known as the Schrödinger representation.) The answer is because of the canonical commutation relations

$$[\hat{r}^i, \hat{p}_j]~=~i\hbar\delta^i_j.$$

4) On the other hand, the corresponding commutation relation for time $t$ is

$$[\hat{H}, t]~=~0, $$

because time $t$ is a parameter not an operator in quantum mechanics, see also this question. Note that in contrast

$$[i\hbar \frac{\partial}{\partial t}, ~t]~=~i\hbar,$$

which also shows that one should not identify $\hat{H}$ and $i\hbar \frac{\partial}{\partial t}$.

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You cannot "cancel" the wave function in Schrödinger's equation because the wave function is the main variable of it. It is an equation for the wave function.

The time-derivative can't be considered an operator because an operator is, by definition, a well-defined unique map $$L:\,{\mathcal H}\to {\mathcal H}$$ from the Hilbert space to the same Hilbert space. It is a map: for each choice of a vector $|\psi\rangle$, it must tell you what is $L|\psi\rangle$. Linear operators are uniquely determined by a particular matrix with respect to a particular basis. The time derivative is nothing of the sort. It is only well-defined when you tell me what $|\psi(t)\rangle$ is: the input (information one needs to know) isn't just a vector; it must be a vector-valued function of time.

There's no analogy between Newton's $F=ma$ and Schrödinger's equation, except that both of them are equations. A better quantum counterpart of Newton's equations would be the Heisenberg equations for the operators rather than Schrödinger's equation. Well, a very mild analogy – which would exist in probably any equation – is that one needs to have a particular $x,p$-dependent formula for force $F$ to calculate a particular $x(t)$; in the same way, one needs a particular choice of the Hamiltonian to calculate $|\psi(t)\rangle$. But it's true in any equation: all shortcuts have to be fully explained for the equation to make a really well-defined sense and to specifically apply to a particular system.

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The purpose of the Hamiltonian is to determine the time evolution $\frac{\partial}{\partial t}$, and therefor using $\frac{\partial}{\partial t}$ itself as Hamiltonian is "of no use" The more so because all systems regardless of the underlying physics would have the same Hamiltonian.

What you want is an expression which, depending on the particular physics, predicts the time evolution using quantities which are already known before the time evolution actually happens.

Regards, Hans

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Dear Hans, $\partial /\partial t$ is surely not a "Hamiltonian by definition". It is by definition the limit of $[object(t+dt)-object(t)] / dt$ in the limit $dt\to 0$. Schrödinger's equation is only valid for state vectors satisfying the right dynamical equations - not for all objects and not even for time-dependent elements of the Hilbert space in physics - and it is a nontrivial law of physics (a constraint), not a "definition" of anything. You could perhaps say the opposite thing, the Hamiltonian is by definition the operator generating time translations - but "is" isn't symmetric here. – Luboš Motl Nov 26 '11 at 16:16
Dear Lubos, The simple misunderstanding of the OP is in my opinion best served by a simple answer. That is: the purpose of the Hamiltonian is to determine the time evolution $\partial/\partial t$, or as you say: The Hamiltonian is by definition the operator generating time translations. Once that is understood he'll see that it is of no use to define $\partial/\partial t$ as the Hamiltonian. QMechanic's answer points this out as well but does so still in the context of the limited application area of Schrödinger's equation. – Hans de Vries Nov 26 '11 at 16:54
Adapted the wording of the main post as to express what I said in the comment. – Hans de Vries Nov 26 '11 at 17:09
Right, your point understood, Hans. When one says "Hamiltonian is by definition XY", there are different angles what the "definition" may mean. Of course, one may constructively define the Hamiltonian for particular systems, like $p^2/2m+V(x)$, in which case it's this expression by definition. More generally, we want to define it as whatever is needed for Schrödinger's or Heisenberg's equations to hold. The latter approach is more general. Still, when we say "the equation holds", it is not the same thing as saying "the operators are the same" because $\psi$ can't be canceled. – Luboš Motl Nov 26 '11 at 18:11
Also, it's true that it would be "useless" to define the operator to be the same because then you would have no nontrivial dynamical equations that could predict the future. That's what you are probably saying. However, even if it would be useless, one may still ask whether it would be legitimate. I think the answer is No, see e.g. Qmechanic's answer. $[H,t]=0$ is different from the nonzero $\partial_t,t=1$ so the operators can't be the same. ($\partial_t$ isn't really an operator acting on the states like $\psi(x,y,z)$ but "extended" states $\psi(x,y,z,t)$ etc.: another "distortion".) – Luboš Motl Nov 26 '11 at 18:14

Mathematically $i\hbar\frac{\partial}{\partial t}$ is a differential operator. Let's call it $\hat{E}$: $$\hat{E}:= i\hbar\frac{\partial}{\partial t}$$

However, saying that $\hat{E}\psi=i\hbar\frac{\partial}{\partial t}\psi$ is just saying that $\hat{E}\psi\equiv i\hbar\frac{\partial}{\partial t}\psi$ and it is not yet an equation (it's a tautology as Qmechanic pointed out). From differential equations you know that, for example, for $\hat{L}:= \frac{d}{dx}$, $\hat{L}\psi(x)\equiv \frac{d\psi(x)}{dx}$ is not an equation. Instead, $\hat{L}\psi(x)=0=0\cdot \psi$ is an equation and of course it doesn't mean that $\frac{d}{dx}= 0$. Or better, take the Lapalacian in two dimensions $\nabla^2\equiv\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$. Then the Laplace's equation is $$\nabla^2 \psi(x,y)\equiv\frac{\partial^2 \psi(x,y)}{\partial x^2}+\frac{\partial^2 \psi(x,y)}{\partial y^2}=0$$

You can rewrite it as $$\frac{\partial^2 \psi(x,y)}{\partial x^2}=-\frac{\partial^2 \psi(x,y)}{\partial y^2}$$

Obviously it doesn't mean that $\frac{\partial^2}{\partial x^2}= -\frac{\partial^2}{\partial y^2}$, it means that acting by $\hat{L}_1:=\frac{\partial^2}{\partial x^2}$ on $\psi$ gives you the same function as acting on $\psi$ by $\hat{L}_2:=-\frac{\partial^2}{\partial y^2}$: $$\hat{L}_1\psi(x,y)=\phi(x,y)$$


i.e. $\hat{L}_1\equiv\hat{L}_2$, but not in general, only on a specific function space of functions $\psi$ such that $\hat{L}_1\psi=\phi=\hat{L}_2\psi$.

In the case of time dependent Schroedinger equation we have two operators $\hat{E}$ and $\hat{H}$ (as $\hat{L}_1$ and $\hat{L}_2$ in our previous example) acting on $\psi$ leading to the same result $\phi$: $$\hat{E}\psi\equiv i\hbar\frac{\partial}{\partial t}\psi=\hat{H}\psi\equiv \left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right)\psi=\phi$$ This comes from the fact that $\hat{E}\psi=E\psi=\phi$, $E=\frac{p^2}{2m}+V(x)$ and replacing $p$ by $\hat{p}$ and $x$ by $\hat{x}$ gives us the Hamiltonian operator $H(\hat{x},\hat{p})$ such that $\hat{H}\psi=E\psi=\phi$. So we can consider that $\hat{E}\equiv \hat{H}$ only on a specific function space of functions $\psi(x,t)$, though they are different $\hat{E}\neq \hat{H}$ (like the $\hat{L}_1$, $\hat{L}_2$ are).

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Although not directly related to the question at hand, I would like to make the comment that the momentum operator does not necessarily arise from imposing a commutator. It arises as follows:

Start by defining a translation operator that acts on a field (consider the simple case first):


Expand as:

$\hat{T}_a\psi(x)=\psi(x)+a\psi'(x)+ \frac{a^2\psi''(x)}{2!}+...$

=$[I+a\psi'(x)+ \frac{a^2\psi''(x)}{2!}+...]\psi(x)$

Call the derivative operator $\hat{D}$. Using the notation for an exponential we can write that as:


Now we have that the differential operator is the infinitesimal generator of translation.

To keep the translation operator Hermitian we redefine define a new operator $\hat{p}=i\hat{D}$.

This is then identified with the physical quantity "momentum" if the variable x describes "position". There is a lot more to this and maybe I will edit this post when I have the time.

The point I wish to make is that $i$'s are not thrown in by hand in an ad-hoc manner, but there is a purpose for making such substitutions.

For a long time I was unsatisfied with the way QM textbooks approach the topic of operators. Nobody told me this in the intro QM class. I was really lucky to have a superb mathematical physics teacher to explain this to me. Great class and a terrific teacher!

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Can we get the notes of this class ? :) – Revo Dec 20 '11 at 18:37
@Revo You should check out Sakurai's book on quantum mechanics. – becko Mar 18 '13 at 1:37

In any formalism in which a Hilbert space of states is associated with a space-like hyperplane, which is certainly the case in your Schrödinger equation example, time is a parameter that picks out a Hilbert space $\mathcal{H}_t$. In that case, the Answers from Lubos and Qmechanic describe the situation pretty well.

In quantum field theory, certainly in the most widely used formalisms, the Hilbert space of states is still associated with a space-like hyperplane (and the Lorentz covariance of these formalisms is somewhat troubled), so that again time is a parameter, and again the Answers from Lubos and Qmechanic are good. It is possible, however, to construct formalisms in which a Hilbert space is associated with all of space-time, in which case time-like and space-like translations are much more directly comparable. There is a difference between time-like and space-like translations because of the different sign of the metric for the two cases, however time-like translation can be presented as an operator that acts on one Hilbert space, just as space-like translations do. It's arguable, however, that there is no Schödinger equation in such formalisms, which rather steps outside your Question (and hence that this is just going to be a confusing digression ---but, if you're wondering, this mathematics is out there ...).

This is all nice enough as mathematics, and I hope that seeing the contrast will help, but the alternative construction, as I outline it above, has found essentially no useful application as a Physical description.

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The short answer is, because it is identically zero.

If you say «operator», you have to be able to specify, ¿on what space does it operate?

Simplifying Peter Morgan's answer a little, here the Hamiltonian is supposed to be an operator on the Hilbert space of state vectors (or wave functions) of the system. In your case, this Hilbert space is a space of functions of three variables, $x$, $y$, and $z$. They could be denoted $\psi(x,y,x)$, for example, $$e^{-x^2-y^2-z^2}$$ or, for another example, $v$. They are constant in time so if you take their time derivative you get zero....I am not joking. You got confused because the vector can vary in time, but then it is a different vector, i.e., if you consider $$v_t=\psi_t(x,y,z)$$ this describes a path in HIlbert space. But operators get applied only to individual vectors in the Hilbert space, not to paths... this is what some of the other posters were referring to when they pointed out that time is a parameter here. An example of a path in Hilbert space could be given by a concrete formula, but then it would invite you to get confused again about the difference between a parameter and a I won't write any examples.

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I think, you answer is correct. The definition is matter. The Hamiltonian is a total energy operator by definition. Assuming $i\hbar\frac{\partial}{\partial t}$ is a Hamiltonian leads to problems since this expression does not contain the information about system's energy which is consist of kinetic and potential parts depending on the system's configuration.

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