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I am getting a little confused by the notion of potentials. In classical mechanics we are never worried about the value of a potential, we are always concerned mainly in the way a potential varies, or potential difference between points. We can always set our potential to have any value anywhere we want (most of the times).

On the other hand in quantum mechanics we explicitly use the value of the potential $V(\boldsymbol{r})$. Is there a difference in quantum mechanics between the potentials $V_1 (x) = V(x)$ and $V_2(x) = V(x) + C$?

If we include both of these potentials into Schrodinger's equation there is a difference between the solutions we get.

Qmechanic
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Daniel Duque
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    It's exactly like in classical mechanics. If you want to shift the potential, you need to shift the total energy too. Then the Schrodinger equation comes out exactly the same. – knzhou Oct 25 '17 at 15:47

2 Answers2

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You're correct that the solutions don't come out precisely the same. However, if we look at Schrodinger's equation with the added constant, $$ i\hbar \frac{\partial}{\partial t}|\psi\rangle = (\hat H+C)|\psi\rangle $$

we notice that if we let

$$|\psi\rangle = e^{-iCt/\hbar}|\psi'\rangle $$

then the Schrodinger equation becomes $$i\hbar \frac{\partial}{\partial t}|\psi\rangle = e^{-iCt/\hbar}\cdot i\hbar \frac{\partial}{\partial t} |\psi'\rangle +Ce^{-iCt/\hbar}|\psi'\rangle= (\hat H+C)e^{-iCt/\hbar}|\psi'\rangle$$

Which simplifies to

$$i \hbar \frac{\partial}{\partial t}|\psi'\rangle = \hat H |\psi'\rangle$$

In other words, shifting the energy by some constant $C$ is equivalent to multiplying all of the states by the phase factor $e^{-iCt/\hbar}$. Because this unitary factor is applied to every state, it cancels out in all physical calculations (e.g. expectation values) and so does not impact the predictions of the theory.

Classical physics is insensitive to shifts in potential energy because measurable physical quantities depend only on derivatives of the potential. Similarly, quantum mechanics is insensitive to shifts in potential energy because measurable physical quantities depend only on inner products between states, and the transformation induced by the potential shift ($|\psi\rangle \rightarrow e^{-iCt/\hbar}|\psi\rangle$) is unitary, meaning that it leaves the inner products unchanged. Both of these are particular examples of the more general notion of gauge invariance.

J. Murray
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Potentials in QM and classical mechanics are exactly the same. In fact, one reason why classical mechanics in Lagrangian or Hamiltonian formalism (both based on potential) is preferred over the Newtonian formalism (which brings in the concept of force) is because the same governing equations generalize to a majority of problems and also paves an easy transition to QM.

Let's see the spectrum of the two Hamiltonian operators, consider $H = T + V$ with eigenvalues $\{E_n\}$ and eigenstates $\{ \psi_n \}$ $$ H \psi_n = E_n \psi_n$$ Now consider the action of the new Hamiltonian (with shifted potential) $H' = H + C$ on the $\psi_n$ $$H'\psi_n = (H + C)\psi_n = E_n \psi_n + C\psi_n = (E_n + C)\psi_n$$ which is equivalent to saying that the eigenvalues, which are the possible energy measurements, corresponding to the new reference for potential are $$E_n' = E_n + C$$ which is what you'd expect in classical scenario.

the boy who believed
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