I understand that charge and electric potential are conjugate observables in QM.

See https://en.wikipedia.org/wiki/Conjugate_variables

The quantum mechanical operator for charge, q, is simply equal to q. I am guessing that the quantum mechanical operator for electrical potential (also called emf, in this context) must have a derivative in it, but am not sure and can't seem to Google it.

So, my question is what is the operator for electric potential? For instance, for linear momentum, it is $-i\hbar\nabla$

  • 2
    $\begingroup$ In non-relativistic QM the charge is not a dynamic variable and the potential is simply the potential, i.e. in the electrostatic case it's the electric potential times the charge. In quantum field theory a classical quantity like the electrostatic potential doesn't exist, one can only quantize the Maxwell equations as a whole. $\endgroup$
    – CuriousOne
    Feb 12, 2016 at 5:52

1 Answer 1


For a non-varying charge, your electric potential, $V$, is a function of the radial distance $r$. In Quantum Mechanics, the equivalence theorem permits us to derive a classical expression for a given physical observable, then convert it to an operator expression. We know a classical formula for the electric potential $V(r)$. To obtain a quantum mechanical operator for this potential, we simply convert $r$ to an operator, that is $r\to\hat{r}$. The operator for electric potential is then just $\hat{V} = V(\hat{r})$.

For the momentum operator, however, we can directly derive a quantum mechanical expression. Consider the free particle wave function:

$Ψ(x,t) = Ae^{i(kx - ωt)}$ where $A$ is a normalisation constant.

Using these facts: $E = hf = ħω$ ; $p = \frac{h}{λ} =\frac{2πħ}{λ}=ħk$, we can rewrite $Ψ(x,t)$ as:

$Ψ(x,t) = Ae^{i\frac{(px - Et)}{ħ}}$

Now take the derivative of $Ψ$ with respect to $x$ (just to see if we can come up with something):

$\frac{∂Ψ}{∂x}=\frac{ip}{ħ}Ae^{i\frac{(px - Et)}{ħ}}=\frac{ip}{ħ}Ψ$

Now it's getting interesting. We rearrange the above expression into:


Now this is an eigenvalue equation. We have an operator $\frac{ħ}{i}\frac{∂}{∂x}$ acting on $Ψ$ with eigenvalues $p$ which are momentum values. From this, we extract directly that the momentum operator must be:


For simplicity,I have derived this expression in one dimension, but, of course, the partial derivative can be extended to 3D space as well. The general expression for the momentum operator is then:

$\hat{P}=-iħ(\frac{∂}{∂x}+\frac{∂}{∂y}+\frac{∂}{∂z}) = -iħ∇$.

  • $\begingroup$ @ 2goodforthis and Everett You: Thank you for that answer! I am just wondering, the QM operator for momentum does not seem to follow that rule, as far as I can see. It uses a $\nabla$ operator, not a derivative with respect to time, as in the classical case. There is also an $i$ and $\hbar$ in that formula. $\endgroup$
    – David
    Feb 12, 2016 at 19:42
  • $\begingroup$ @David: We only need to use the equivalence theorem if we can't otherwise directly derive a quantum mechanical expression for a given operator. In the case of momentum, however, it turns out that we can. I have added the derivation of the momentum operator to my answer to your original question. $\endgroup$
    – 2good4this
    Feb 15, 2016 at 17:54

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