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Why is the "canonical momentum" for the Dirac equation not defined in terms of the "gauge covariant derivative"?

+ 4 like - 0 dislike
136 views

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.

This post imported from StackExchange Physics at 2015-03-30 13:52 (UTC), posted by SE-user rubenvb
asked Dec 3, 2012 in Theoretical Physics by rubenvb (30 points) [ no revision ]

2 Answers

+ 6 like - 0 dislike

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.

This post imported from StackExchange Physics at 2015-03-30 13:52 (UTC), posted by SE-user Qmechanic
answered Dec 3, 2012 by Qmechanic (2,840 points) [ no revision ]
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!

This post imported from StackExchange Physics at 2015-03-30 13:52 (UTC), posted by SE-user rubenvb
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.

This post imported from StackExchange Physics at 2015-03-30 13:52 (UTC), posted by SE-user Qmechanic
+ 1 like - 0 dislike

(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 $(x^0,x^1,...,x^{n-1})$. When there is no electromagnetic field then the action of particle is given as

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

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 $-iA_{\mu}dx^{\mu}$ then we can add one more Lorentz invariant term to this action to generalize it as :-

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

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

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

So

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

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

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

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 chosing 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 -

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

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

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

Or taking $-i\hbar$ common we get

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

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

This post imported from StackExchange Physics at 2015-03-30 13:52 (UTC), posted by SE-user user10001
answered Dec 3, 2012 by user10001 (635 points) [ no revision ]

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