# Does a gauge transformation leave physics invariant?

It is usually said that gauge invariance is captured by the invariance of the electric and magnetic fields on the following transformation of the potentials :

$$\phi' = \phi - \frac{\partial{\chi}}{\partial{t}} \hspace{5ex} A'= A + \nabla \chi$$

In quantum mechanics, considering an hamiltonian on the form

$$H = \frac{1}{2m} (P- qA)^2 + q \phi ,$$

the effect of the gauge transformation can be canceled by transforming the wave function into

$$\psi' = e^{i\frac{q}{\hbar}\chi} \psi .$$

Am I right so far?

If I am then because the gauge transformation is not supposed to change anything, it means that every expectation can be calculated equivalently using $\psi'$ or $\psi$, which does not seem not true.

An example could be a particle in a $E=B=0$ region. So I can choose $\phi = A = 0$, to make the hamiltonian

$$H = \frac{1}{2m} P^2$$

But one of my friends could have selected $\phi =0$, $A = a$, which is supposed to be equivalent to my choice since they are linked by a gauge transformation ($\chi = ax$). Then, the hamiltonian would be

$$H = \frac{1}{2m} (P- q a)^2$$

The gauge transformed wave function he can take to make his hamiltonian similar to mine is $\psi' = e^{i\frac{q}{\hbar} a x} \psi$, but this function is not equivalent.

The expectation value of momentum for me is:

$$\langle P\rangle_{\psi} = -i\hbar \int \psi^* \partial_{x} \psi dx$$

For my friend it is:

$$\langle P\rangle_{\psi'} = -i\hbar \int e^{-i\frac{q}{\hbar} a x} \psi^* \partial_{x} (e^{i\frac{q}{\hbar} a x} \psi) dx = q a \int \psi^* \psi dx -i\hbar \int \psi^* \partial_{x} \psi dx$$

What is it that I don't understand?

• what do you think "equivalent" means in this context? May 18, 2017 at 16:06
• Try to give an example of expectation values that do not match.
– noah
May 18, 2017 at 16:08
• Multiplication of a wavefunction by an overall phase as you would in a 'gauge transform' makes no difference to calculations since it gets cancelled by its complex conjugate whenever you calculate an expectation value. May 18, 2017 at 16:11
• Also note that the momentum that appears in the Hamiltonian is canonical momentum $\partial L / \partial \dot x$, not mechanical momentum $mv$. May 18, 2017 at 16:23
• I added the example of the expectation value of momentum, but it makes me think that maybe the correct momentum I should use for the expectation value is p - qa? If this is my mistake, can someone answer by giving the general rule for how observables should transform? May 18, 2017 at 18:55

The statement

because the gauge transformation is not supposed to change anything, it means that every expectation can be calculated equivalently using $\psi'$ or $\psi$

isn't particularly correct. All physically measurable expectation values can be calculated correctly in any arbitrary gauge, but the operator representation of some operators can change from one gauge to another.

The particular sticking point here is the difference between kinetic momentum and canonical momentum. To be clear,

• kinetic momentum is always equal to $m\mathbf v$, it is physically measurable, and it must therefore be gauge invariant, whereas
• canonical momentum is the quantity $\mathbf p$ such that $[\hat x_i,\hat p_j] = i\hbar \delta_{ij}$ (so therefore its position representation is always the gradient, i.e. $\langle \mathbf x|\hat p_j = -i\hbar \frac{\partial}{\partial x_j}\langle \mathbf x|$), and it is not gauge invariant. As an example, in the presence of an electromagnetic field, the canonical momentum typically differs from the kinetic momentum by a multiple of the vector potential $\mathbf A$.

When you change gauges, it's important to take care to transform all the relevant observables in the process; failure to do this is probably the most common reason for confusion regarding spurious gauge dependences in results.

It's also important to note that none of this is specific to quantum mechanics, and you get exactly the same issues and concepts from the beginnings of lagrangian mechanics, from gauges and gauge changes to the mismatch between canonical and kinetic momentum. For details, see your favourite analytical mechanics textbook.

• Makes sense. Is there a way to make this whole gauge transformation thing more "quantum mechanical like" (by that I mean not having to refer to lagrangian mechanics) by introducing some kind of unitary operator associated with a gauge transformation? May 18, 2017 at 21:09
• Yes - and you've already done it, it's the factor of $\hat U=e^{i\frac{q}{\hbar}\chi(\hat{\mathbf r} )}$. However, trying to decouple QM entirely and forgetting its basis and analogues in lagrangian mechanics is a thoroughly bad idea (and misleading statements of the firm "X is this wonky new feature that's exclusively quantum mechanical" are particularly pernicious when they are false, as is the case for gauge transformations). What you're asking for is simply the formally correct way to do this in QM. May 18, 2017 at 21:53
• Thanks, you are right. I guess the next step for me is to get a more general understanding of how symetries work in QM. May 18, 2017 at 22:10

You are right. The momentum operatur is not gauge invariant. If you demand that measurable quantities are gauge invariant (it is a sensible requirement), the momentum.is not a measurable quantity. Velocity is though. The only book I could find mentioning this is Ballentines. Ballentine 'Quantum Mechanics' states in Chapter 11: "Since p, like A, is changed by a gauge transformation, it too lacks a direct physical significance." (then some calculations on the velocity operator follow)