Why can't $ i\hbar\frac{\partial}{\partial t}$ be considered the Hamiltonian operator?

Question

In the time-dependent Schrodinger equation, $ 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 second law is not an equation for defining $F$; $F$ must be provided independently.

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

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
I think you last comment is spot-on, this just the same thing as Newton's 2nd law; nothing more, nothing less. — gatsu, Jan 02 '17 at 20:47
Well, if we know $\psi$ then it might be possible to solve the operator $H.$ I don't know if it is possible or not. — Hulkster, Jul 08 '21 at 13:58
You re right, phylosophycally the thing is the same as F= ma: both F and H are prescribed models to close the equations of motion. See https://physics.stackexchange.com/a/697000/226902 — Quillo, Mar 01 '23 at 00:13

score 143 · Answer 1 · edited Jan 28 '20 at 09:40

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 $$\hat{H}\Psi~=~i\hbar \frac{\partial\Psi}{\partial t}\tag{1}$$ would become a tautology. Such trivial Schrödinger equation could not be used to determine the future (nor past) time evolution of the wavefunction $\Psi({\bf r},t)$.
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 $$\hat{H}\Psi~=~i\hbar \frac{\partial\Psi}{\partial t}\tag{2}$$ is a non-trivial requirement for the wavefunction $\Psi({\bf r},t)$.
One may then ask why is it then okay to assign the momentum operator as a gradient $$\hat{p}_k~=~ \frac{\hbar}{i}\frac{\partial}{\partial r^k}~?\tag{3}$$ (This is known as the Schrödinger representation.) The answer is because of the canonical commutation relations $$[\hat{r}^j, \hat{p}_k]~=~i\hbar~\delta^j_k~\hat{\bf 1}.\tag{4}$$
On the other hand, the corresponding commutation relation for time $t$ is $$[\hat{H}, t]~=~0, \tag{5}$$ because time $t$ is a parameter not an operator in quantum mechanics, see also this & this Phys.SE posts. Note that in contrast $$\left[i\hbar \frac{\partial}{\partial t}, ~t\right]~=~i\hbar,\tag{6}$$ which also shows that one should not identify $\hat{H}$ and $i\hbar \dfrac{\partial}{\partial t}$.

score 39 · Answer 2 · answered Nov 26 '11 at 14:38

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.

score 21 · Answer 3 · edited May 02 '13 at 13:02

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 $\hat{L}_2:=-\frac{\partial^2}{\partial y^2}$: $$\hat{L}_1\psi(x,y)=\phi(x,y)$$

$$\hat{L}_2\psi(x,y)=\phi(x,y)$$

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).

Hans de Vries · Answer 4 · 2011-11-26T17:04:02.907

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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

edited Nov 26 '11 at 17:04

answered Nov 26 '11 at 15:31

Hans de Vries

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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
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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

score 8 · Answer 5 · answered Oct 08 '16 at 03:35

The Schrödinger equation isn't really a PDE. It's an ODE. The Schrödinger equation is $i\hbar \frac{d}{dt} | \psi\rangle = \hat H | \psi\rangle$. Here the state vector $| \psi\rangle$ is an $\mathcal H$-valued function of a single independent variable $t$ and $\mathcal H$is some Hilbert space and $\hat H$ is an operator in that Hilbert space. There is only one independent variable, so it's an ODE.

Now, the operator $\hat H$ is usually a function of operators $\hat x, \hat p$ that satisfy $[\hat x, \hat p] = i\hbar$ (and also other operators like the spin, and possibly also $t$). Because of the Stone-von Neumann theorem, such operators can always be put in a form where for eigenstates of the operator $\hat x$, say $|x\rangle$, $$\begin{align} \langle x | \hat x | \psi \rangle & = x\langle x|\psi\rangle \\ \langle x | \hat p |\psi\rangle & = -i\hbar \left.\frac{\partial}{\partial x'} \langle x' | \psi\rangle \right|_{x' = x}\end{align} \tag{1} $$ and if we introduce $\psi(x,t) = \langle x | \psi\rangle$ (remember $|\psi\rangle$ depends on $t$) we get the usual way it's written, $\hat x = x, \hat p = -i\hbar \frac{\partial}{\partial x}$. Sometimes people even write $\hat p \psi(x) = -i\hbar \frac{\partial \psi(x)}{\partial x}$ but this is complete abuse of notation because $\psi(x)$ is a number and $\hat p$ acts in the Hilbert space, not on numbers. What they mean is $(\hat p \psi)(x)$ but this is not what they write.

Anyway, if you use (1), you can often write the Schrödinger equation like $$i\hbar \partial_t \psi(x,t) = -\frac{\hbar^2}{2m} \partial_x^2 \psi(x,t) + V(x) \psi(x,t) $$ and this looks like a PDE because you decided to write the vector ODE $ih\frac{d}{dt} |\psi\rangle = \hat H |\psi\rangle$ in components in a particular basis. $x$ isn't really an independent variable. It's an index labeling components.

score 7 · Answer 6 · edited Feb 28 '23 at 18:31

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):

$\hat{T}_a\psi(x)=\psi(x+a)$

Expand as:

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

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

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

$\hat{T}_a\psi(x)=e^{a\hat{D}}\psi(x)$

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!

@Revo You should check out Sakurai's book on quantum mechanics. — a06e, Mar 18 '13 at 01:37

score 5 · Answer 7 · edited Jul 10 '17 at 10:28

Many answers gave you the in-depth explanations which are good ones. However, there are simple reasons too. Contrary to what has been said, I think your analogy puts you on the right track.

Just as you mention Newton’s second law is not intended to find $F$. Still, $F$ should not be taken as the same as $ma$. $F$ is an abstraction of something else that should be plugged in there, which represents the amount of interactions or disturbance external elements exert on the particle. According to one's setup, $F$ will be replaced by the correct function $f(x, \dot x,t)$. For instance, if one's setup is elastic, Newton’s second law’s equation would be plugged in with Hooke’s law. The same Happening if the interaction is gravitational, electric, etc. So, $F$ could have many distinct flavours, sometimes so fundamentally different as gravity and electromagnetism (at least for the time being) depending on your experiment.

Now in $\hat{H}\Psi~=~i\hbar \frac{\partial\Psi}{\partial t}$, The right side of that equation should not be taken as the same as the left side of that equation. $\hat H$ is an abstraction of something else that should be plugged in there too. For instance, if one deals with non-relativistic simple particle, $\hat H$ should be replaced by Schroedinger’s Hamiltonian. For a non-relativistic electromagnetic field, one replaces it by Pauli’s operator, the same goes for Dirac’s Hamiltonian.

score 4 · Answer 8 · answered Mar 13 '19 at 23:14

The deeper problem with this supposition is that it assumes a conceptual identity between the notions of Hamiltonian and energy, and this is an identity that is not correct. That is, discernment needs to be applied to separate the two of these things.

Conceptually, energy is a physical quantity that is, in a sense, "nature's money" - the "currency" that you have to expend to produce physical changes in the world. On a somewhat deeper level, energy is to time what momentum is to space. This can be seen across many areas, such as Noether's theorem, which relates the law of conservation of energy to the fact that the history of a system can be translated back and forth in time and still work the same way, i.e. that there is no preferred point in time in the laws of physics, and likewise, the same for momentum with it being translated around in space and still working the same way. It also occurs in relativity, in which the "four-momentum" incorporates energy as its temporal component.

The Hamiltonian, on the other hand, is a mathematically modified version of the Lagrangian, through what is called the Legendre transform. The Lagrangian is a way to describe how that forces impact the time evolution of a physical system in terms of an optimization process, and the Hamiltonian converts this directly into an often more useful/intuitive differential equation process. In many cases, the Hamiltonian is equal to, the system total mechanical energy $E_\mathrm{mech}$, i.e. $K + U$, but this is not always so even in classical Hamiltonian mechanics, a fact which indicates and underscores the basic conceptual separation between the two.

In quantum mechanics, the "energy is to time what momentum is to space" concept manifests in that it is the generator of temporal translation, or the generator of evolution, in the same way that momentum is the generator of spatial translation. In particular, just as we have a "momentum operator"

$$\hat{p} := -i\hbar \frac{\partial}{\partial x}$$

which translates a position-space (here using one dimension for simplicity) wave function (mathematical representation of restricted information regarding the particle position on the part of an agent) $\psi$ via the somewhat-loose "infinitesimal equation"

$$\psi(x - dx) = \psi(x) + \left(\frac{i}{\hbar} \hat{p} \psi\right)(x)$$

for translating it by a tiny forward nudge $dx$, likewise we would want to have an energy operator

$$\hat{E} := i\hbar \frac{\partial}{\partial t}$$

which does the same but for translation with regard to time (the sign change is because we usually consider a temporal advance from $t$ to $t + dt$, as opposed to psychologically [perhaps also psycho-culturally] preferring spatial motions to be directed rightward, in our descriptions of things.). The problem here is that wave functions generally do not contain a time parameter, and at least non-relativistic quantum mechanics treats space and time separately, so the above cannot be a true operator on the system state space. Rather, it is more of a "pseudo-operator" that we'd "like" to have but can't "really" for this reason. One should note that this is the expression that appears on the right of the Schrodinger equation, which we could thus "better" write as

$$\hat{H}[\psi(t)] = [\hat{E}\psi](t)$$

where $\psi$ is now a temporal sequence of wave functions (viz. a "curried function", which becomes an "ordinary" function when you consider the wave functions as the basis-independent Hilbert vectors). The Hamiltonian operator $\hat{H}$ is a bona fide operator, which acts only on the "present" configuration information for the system. What this equation is "really" saying is that in order for such a time series to represent a valid physical evolution, the Hamiltonian must also be able to translate it through time. The distinction between Hamiltonian and energy manifests in that the Hamiltonian will not translate every time sequence, while the energy pseudo-operator will, just as the momentum operator will translate every spatial wave function. Moreover, many Hamiltonians may be possible that give rise to the same energy spectrum.

Because these two things are different, it makes no sense to equate them as operators, like suggested. You can, and should, have $\hat{H}[\psi(t)] = [\hat{E}\psi](t)$, but you should not have $\hat{H} = \hat{E}$!

score 3 · Answer 9 · answered Nov 26 '11 at 17:29

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.

joseph f. johnson · Answer 10 · 2011-12-20T09:03:01.617

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 variable...so I won't write any examples.

score 3 · Answer 11 · answered Dec 28 '16 at 13:06

Answer to the top question is actually very short. Time is an external parameter in conventional QM; parametrizing a unitary evolution. It, as well as $i\partial_t$, nothing has in common with operators, observables etc. In other words $t$ in $\psi(t)$ does not enumerate some basis vectors of an observable like $x$ does in $\psi(x)$.

score 2 · Answer 12 · answered Aug 23 '19 at 17:20

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Asking why $$i \hbar \frac{\partial}{\partial t}$$ isn't the hamiltonian operator in QM is the same as asking why the time derivative isn't the hamiltonian in Hamilton's equations: $$ \frac{d p_i}{dt} = -\, \frac{\partial H}{\partial q_i}, \qquad \dots $$

answered Aug 23 '19 at 17:20

VirtualPhoton

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Perhaps you could provide more information about how this answers the OP's question? – Daddy Kropotkin Dec 20 '20 at 18:59

score 1 · Answer 13 · answered Oct 12 '18 at 18:15

Since time is not a dynamical variable in QM like $x$ or $p$. Therefore there is no spectral theorem in terms of time. The "operator" $i\hbar \partial_t$ is a just postulated method to drop the total energy as an average value. Surely there is a set of energies for every set of eigenstates, but those energies are determined by the Hamiltonian operator.

score 1 · Answer 14 · answered May 02 '13 at 12:57

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.

score 1 · Answer 15 · answered Apr 14 '22 at 10:49

Though already contained implicitly in other answers, I thought I'd just spell out my view explicitly here.

The state of a particle is characterized by a state vector. Observable quantities are represented by linear operators on the vectors, and the measurable values are the eigenvalues of the linear operators. The study of quantum mechanics is to determine how the state vector, and hence the measurable values, change over time. In particular, time is not an observable, i.e. you cannot speak of the time of a particle like its position or its momentum, because a state vector should exist for all time.

Now, when we say $\hat x = x$ and $\hat p = -i\hbar \partial_x$, we are talking about operations that turn vectors into other vectors. When we talk of $\partial_t$ in quantum mechanics, however, we are comparing the same vector to itself, just at different times. For this reason, a time derivative can't really be considered an operator at all.

Why can't $ i\hbar\frac{\partial}{\partial t}$ be considered the Hamiltonian operator?

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