The canonical momentum is always used to add an EM field to the Schrödinger/Pauli/Dirac equations. Why does one not use the gauge covariant derivative? As far as I can see, the difference is a factor i in front of the vector potential. I know I'm combining two seemingly unrelated things, but they seem very similar, an the covariant form seems much "better" with respect to the inherent gauge freedom in the EM field. I can also see that with the canonical momentum form, the equations remain unchanged after an EM and a QM (phase) gauge transformation. Suffice to say my field theory knowledge is not that impressive.


2 Answers 2


The identification goes as follows:

$$ \text{Kin. Mom.}~=~ \text{Can. Mom.} ~-~\text{Charge} \times \text{Gauge Pot.} $$

$$ \updownarrow $$

$$ m\hat{v}_{\mu} ~=~ \hat{p}_{\mu} - qA_{\mu}(\hat{x})$$

$$ \updownarrow $$

$$ \frac{\hbar}{i} D_{\mu} ~=~ \frac{\hbar}{i}\partial_{\mu} - qA_{\mu}(x) $$

$$ \updownarrow $$

$$ D_{\mu} ~=~ \partial_{\mu} -\frac{i}{\hbar} qA_{\mu}(x) $$

$$ \updownarrow $$

$$ \text{Cov. Der.}~=~ \text{Par. Der.} ~-~\frac{i}{\hbar}\text{Charge} \times \text{Gauge Pot.} $$

The imaginary unit $i$ is needed, e.g. because the derivative is an anti-hermitian operator (recall the usual integration-by-part proof), while the momentum is required to be a hermitian operator in quantum mechanics.

  • $\begingroup$ Wow, I knew this. I wasn't thinking straight today, gotta remember to not post questions when I have a fever. Thanks for the clear explanation! $\endgroup$
    – rubenvb
    Commented Dec 3, 2012 at 18:19
  • 1
    $\begingroup$ Comment to the answer (v1): we focus on spatial directions $\mu=1,2,3$, and assume the sign convention $(-,+,+,+)$. See also this Phys.SE post. Note that the Hamiltonian and the wavefunction are gauge-covariant rather than gauge-invariant objects. $\endgroup$
    – Qmechanic
    Commented Feb 26, 2013 at 21:47

(Qmechanic has already given the answer. However, since I spent some time writing the answer below so I am anyway posting it.)

Consider charged particle with charge $q$, and (nonzero) rest mass $m$ moving in a spacetime with coordinates $(x^0,x^1,\ldots ,x^{n-1})$. When there is no electromagnetic field then the action of particle is given as

$$\tag {1} S=-mc\int \sqrt {\eta_{\mu\nu}\dot x^\mu(\lambda)\dot x^\nu(\lambda)}\,\mathrm d\lambda$$

Where $\lambda$ is the parameter along the trajectory $x^{\mu}(\lambda)$ of the particle and $\dot x^{\mu}$ means $\partial x^{\mu}(\lambda)/\partial\lambda$. Note that the action is Lorentz invariant. When there is a $U(1)$ gauge field $-iA_{\mu}\,\mathrm dx^{\mu}$ then we can add one more Lorentz invariant term to this action to generalize it as :-

$$\tag{2}S=-mc\int \sqrt {\eta_{\mu\nu}\dot x^\mu (\lambda)\dot x^\nu (\lambda)}\,\mathrm d\lambda-(q/c)\int \eta_{\mu\nu}A^{\mu}\dot x^{\mu}(\lambda)\,\mathrm d\lambda$$

Now in order to proceed its convenient to work in a particular inertial frame, and look at things from the viewpoint of inertial observer corresponding to that frame. In such a frame we can take $x^0(\lambda)/c=t=\lambda$. Above integral was from some point $\lambda_{0}$ to some $\lambda_{1}$. Now it becomes an integral from $t_0=x^0(\lambda_0)/c$ to $t_1=x^0(\lambda_1)/c$ and can be written as

$$\tag{3} S=-mc^2\int \sqrt {1-v^2/c^2}dt-\int \left(qA^{0}-\frac {q}{c} \sum_{i}v^{i}.A^{i}\right) \mathrm dt$$


$$\tag{4} L=-mc^2\sqrt {1-v^2/c^2}- \left(qA^{0}- \frac {q}{c} \sum_{i}v^{i}.A^{i}\right)$$

canonical momentum corresponding to $x^i$ can now be obtained as partial derivative of $L$ wrt $v^i=\mathrm dx^i/\mathrm dt$ and is given as :-

$$\tag{5} \pi_i=mv^i/\sqrt {1-v^2/c^2}+\frac{q}{c}A^{i}$$

Thus, as Qmechanic has answered, canonical momentum corresponding to $i$ th coordinate is physical momentum along that coordinate plus a contribution from gauge potential.

Even without choosing a particular inertial frame we could find the canonical momentum $\pi_{\mu}$ corresponding to $x^\mu$ by taking the derivative of $L$ in its covariant form wrt $\dot x^\mu$. This would give -

$$\tag{6}\pi_{\mu}=-mc\eta_{\mu\nu}\dot x^{\nu}/\sqrt {\eta_{\alpha\beta}\dot x^\alpha\dot x^\beta}-\frac{q}{c}A_{\mu}$$

Now in classical mechanics above equation is nothing but a map from velocity space to phase space. Its only when we move to QM that we represent canonical momentums as derivatives wrt spatial coordinates. Again for convenience lets work in a particular inertial frame. Here momentum conjugate to $x^i$ is $\pi_i$ as given by the equation [5]. So as usual in QM we quantize by requiring the corresponding operators to satisfy

$$\tag{7}[ X^i,\Pi_j]=i\delta^i_{j}\hbar$$

We can represent this algebra on Hilbert space of functions on space $R^{n-1}$ (note that spacetime is $R^n$) by defining $X^i$ to be the operator which acts as multiplication by $x^i$, and $\Pi_i$ to be the operator which acts as the derivative $-i\hbar\partial/\partial x^i$. From equation [5] we see that

$$\tag{8}\textrm{mechanical momentum operator}=-i\hbar\partial/\partial x^i-\frac{q}{c}A^{i}$$

Or taking $-i\hbar$ common we get

$$\tag{9}\textrm{mechanical momentum operator}=-i\hbar\left(\partial/\partial x^i-i\frac{q}{c\hbar}A^{i}\right)=-i\hbar D_i$$

Where $D_i=\partial/\partial x^i-i\displaystyle\frac{q}{c\hbar}A^{i}$


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