In the Schrödinger equation we can see an operator associated with the position This operator is used in the expression for the kinetic energy $T$, being part of the quantum mechanical Hamiltonian.
Why isn't an operator for $V$ to be seen in the equation? $V$ being the potential energy in the Hamiltonian?
The potential energy depends both on $x$ and $t$: $V=V(x,t)$ if we consider only one spatial dimension.
It's been said in answer to this question that we have already chosen a basis and that $V$ is a scalar.
But isn't $x$ a scalar too? Why can't we explicitly state how $V(x,t)$ looks like and plug the ensuing function of $x$ and $t$ in the SE?

The Schrödinger equation:

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So, we can approach a problem with a specific $V(x,t)$.

So the question: Can $V(x,t)$ be considered an operator? Or maybe even put better, why not? In the case of the hydrogen atom, isn't $V(\vec{r},t)$ considered an operator?

Is it because for every situation $V(x,t)$ is different, while $T$ always has the same form?

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    $\begingroup$ Have you take a course in quantum mechanics? It looks like gross misunderstanding of QM... $\endgroup$
    – Roger V.
    Sep 3, 2020 at 9:49
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    $\begingroup$ The potential energy of the Hamiltonian in QM is an operator. There is no position operator explicit in the shrodinger equation. $\endgroup$
    – joseph h
    Sep 3, 2020 at 9:56
  • $\begingroup$ @Vadim What is my misunderstanding? $\endgroup$ Sep 3, 2020 at 10:41
  • $\begingroup$ I'm not sure what you're asking. The $\hat{V}$ in the SE is an operator, which can usually be thought of as some function in the position operator $\hat{x}$ - does that help? $\endgroup$
    – jacob1729
    Sep 3, 2020 at 10:42
  • $\begingroup$ @jacob1729 So $V$ is an operator? $\endgroup$ Sep 3, 2020 at 10:43

3 Answers 3


The Schroedinger equation is an equation in terms of state vectors in a Hilbert space and reads:

$$ i\partial_t |\psi\rangle = \hat{H} |\psi\rangle$$

For some hermitian operator $\hat{H}:\mathcal{H}\to\mathcal{H}$. It is usually possible to split a Hamiltonian into a 'kinetic' part and a 'potential' part:


These are both operators as their sum is an operator. For instance, for a harmonic oscillator, $\hat{V}$ is the operator $\hat{x}^2$ (which, with the usual caveats about this not being rigorous, is the hermitian operator whose eigenfunctions are the delta functions $\delta(x-a)$ and has eigenvalues $a^2$).


Yes, the potential energy is an operator. Just because it simply acts as scalar multiplication does not mean it is not an operator. Recall that an operator is a map from functions to functions and $\psi \mapsto V(x)\psi$ is taking a function to a function.

I think the root of your misunderstanding is in thinking that $\hat{p}=-i\hbar\partial_x$ and $\hat{x}=x$ are somehow fundamental. But they are not.

We can write the Hamiltonian you are using as simply

$$ \hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x}) $$

At this point, we have not said anything about what $\hat{p}$ and $\hat{x}$ look like. The only thing we actually require is that they satisfy their commutation relation $$ [\hat{x}, \hat{p}] = i\hbar $$ One choice which will satisfy this is the usual one which you have used already to write down your Schrodinger equation ($\hat{p} = -i\hbar\partial_x$ and $\hat{x}=x$). But this is not the only choice. We could just as well choose \begin{align} \hat{p}=p && \hat{x}=i\hbar\partial_p \end{align} and this is just as valid. But I suspect that you would then say the kinetic energy looks like it is not an operator. Take for example the harmonic oscillator Hamiltonian in this representation: $$ \hat{H} = \frac{p^2}{2m} - \frac{m\omega^2 \hbar^2}{2}\frac{\partial^2}{\partial p^2} $$ This acts on the momentum space wavefunction $\tilde{\psi}(p)$, but now the kinetic energy only acts by scalar multiplication instead of the potential energy.

  • $\begingroup$ That's the way I like it. BUT, the thing is, in the equation $V$ is stated just as $V$ and not as an operator because $V$ is situation-dependent, contrary to $T$. $\endgroup$ Sep 3, 2020 at 13:24
  • $\begingroup$ V just stands for whatever potential operator you put in for it. It is always an operator. $\endgroup$
    – jgw
    Sep 3, 2020 at 13:33
  • $\begingroup$ Of course. $V$ represents an operator, but this operator is not explicitly written out in the SE, as is the case with $T$. $\endgroup$ Sep 3, 2020 at 13:38
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    $\begingroup$ $T$ could strictly speaking be different too. Think about rigid bodies, or particles in magnetic fields. The division is ultimately arbitrary. $\endgroup$
    – jacob1729
    Sep 3, 2020 at 15:29

In the space of functions, you’re probably used to constants (constant functions) always outputting some real/complex number (more generally, an element of the ground field), regardless of the function’s input(s). Functions take elements in the ground field (which tend to be real/complex numbers) to elements in the ground field.

Example: $f(t) =$ constant real/complex number for any (admissible) input $t$.

In the space of operators (sending functions to functions), the “constants” are no longer numbers. The “constants” are functions. So a constant operator will send the input function(s) to (the function(s) given by the operator) “acting on” the input function(s). The potential energy and position operators are constants in the space of operators. That is, they don’t care which function you act on with them, respectively (to be a little more rigorous, I am of course assuming the input function is in the domain of the operator).

Example: $\hat{V}(\psi(x)) = V(x)\psi(x)$ for any (admissible) input $\psi(x)$. $V(x)$ may be a constant function itself, or not.

Example: $\hat{x}\psi(x) = x\psi(x)$.

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    $\begingroup$ Mile Grazie, Antonino! $\endgroup$ Sep 3, 2020 at 17:22

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