Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the Schrödinger equation in the position representation $$ i\hbar\frac{\partial}{\partial t}\Psi(x,t) ~=~[\frac{-\hbar^2}{2m}\nabla^2+V(x,t)]\Psi(x,t), $$ is the potential $V(x,t)$ an operator acting on $\Psi$ or merely a scalar?

share|cite|improve this question
Designing by $\hat V$ the operator, you have : $\hat V \Psi(x,t) = V(x,t)\Psi(x,t)$, where $V(x,t)$ is a scalar quantity. – Trimok Jul 21 '14 at 9:39
Echoing @Trimok's comment: A multiplication operator is also an operator. – Qmechanic Jul 21 '14 at 10:26
up vote 3 down vote accepted

In the representation given by you, the $V(x,t)$ is a scalar function depending on $x$ and $t$. However, you already have choses a basis (the $x$-basis). In general, the $\hat{V}$ or rather $\hat{H}$ in the Schödinger Equation is an operator. This operator can then be evaluated in a basis of your choice. If you are familliar to the dirac notation one can write $\langle x |\hat{V}|x\rangle=V(x)$. Correspondingly, you choose the same basis for your state $|\psi\rangle$ with $\psi(x)=\langle x|\psi\rangle$.

share|cite|improve this answer
Is V(x,t) always a real number or sometimes imaginary? – elflyao Jul 21 '14 at 10:36
The potential is normally a real number, as it is the expectation value of an "observable", so a value which you can meassure in some way. – Hagadol Jul 21 '14 at 10:39

The potential is, without any ambiguity, an operator. There is no other way, because it changes the spatial dependence of the wavefunction: $$ \Psi(x,t)\text{ changes to }V(x,t)\Psi(x,t). $$ It's unclear to me what you mean by "scalar", though, because this word can have multiple meanings. For example, it is indeed a scalar operator - as opposed to vector operators like the position or momentum - but it is not a scalar in the Hilbert-space sense.

share|cite|improve this answer

As usually, physicists are being a bit sloopy with notation here. Mathematically, there's a big difference between a function (such as $V$) and a function's value for some arguments (such as $V(x,t)$). The latter is indeed simply a number, scalar if you like. However, in physics, functions are often used just as "families of values" and values with some free variables in them are used as functions.

Now, actually it doesn't matter that much here1, because both scalars and functions can be seen as operators that simply perform a pointwise multiplication. So basically the answer to your question is

$V(x,t)$ is an operator, though it is merely a scalar.

However, to speak a bit more rigorously again, $V(x,t)$ does not operate on the system's Hilbert space but merely on the complex numbers. I'd therefore consider it rather unnecessary complicating to call it an operator; it's just a number.

This destiction doesn't matter while you're working in a fixed basis (position basis here), but it becomes a problem when you consider the states as abstract, basis-invariant entities. Often it's more useful to work in the momentum basis. In the function-with-arguments writing, this would be expressed as simply re-labeling the arguments: $V(k,t)$. But this is just physicists' shorthand again: really, to go to momentum space you need to Fourier transform the potential.

The way to get around all this confusion is to work without any explicit basis as much as possible. So the "preferred" way of writing the Schrödinger equation is $$ i\hslash \partial_t | \psi(t) \rangle = (\tfrac{-\hslash^2}{2m}\nabla^2 + \hat{V})|\psi(t)\rangle $$ Here it's clear that $\hat{V}$ is an operator which acts directly on the Hilbert space. No matter what basis you like to express your states in, $V$ will have to follow the convention.

1What's much more critical is actually the nabla operator: this is not simply position-wise multiplication. It is actually momentum-wise multiplication though!

share|cite|improve this answer

To answer the question I think you meant to ask, $V(x,t)$ is a real-valued function. But you can also think of it as an operator. An operator is anything that maps functions to other functions, and multiplication by a fixed function is one way to do that.

To be a little more pedantic with the notation, if you use $\hat{V}$ to represent the operator, then the action of this operator is defined as

$$\hat{V}f(x,t) = V(x,t)f(x,t)$$

for any $f(x,t)$ for which this equation makes sense.

The peculiar feature of these multiplicative operators is that they are diagonal in the position basis, so that $\langle x\rvert\hat{V}\lvert y\rangle \propto \delta(x - y)$. This means that, in a rough sense, a multiplicative operator like $\hat{V}$ doesn't do anything "weird" or "interesting" to the position-space function it's acting on - unlike a derivative operator, for example. (Of course things are different if you work in another basis.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.