Is the potential in Schrödinger equation an operator?

Question

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

Answer 1

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 $\psi(x)=\langle x|\psi\rangle$.

Answer 2

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.

Answer 3

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.)

Answer 4

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.

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

Answer 5

Surely it is, but in the following sense: for every $t_0 \geq 0$ the function $V(x,t_0)$ is an operator in $L^2$ with complete set of eigenvalues. Bear in mind that the potential must be physical, e.g. not too singular.