Heisenberg uncertainty principle and the canonical momentum operator

[joneall]

I'm arriving late for this discussion, but it is just in time as I am now reading about the subject in Sakurai and Napolitano. Just to get things straighter, let me quote more from them, p. 128.

"In general, the canonical momentum p is not a gauge-invariant quantity; its numerical value depends on the particular gauge used even when we are referring to the same physical situation. In contrast, the kinematic [or mechanical] momentum $\Pi$, or $mdx/dt$ that traces the trajectory of the particle [!] is a gauge-invariant quantity... Because $p$ and $mdx/dt$ are related..., p must change to compensate for the change in A..."

Here, the relation he is referring to is $\Pi = m\frac{dx}{dt} = p - \frac{eA}{c}$.

Two pages later, they say:

"To summarize, when vector potentials in different gauges are used for the same physical situation,the corresponding state kets (or wave functions) must necessarily be different. However, only a simple change is needed; we can go from a gauge specified by $A$ to another specified by $A + \nabla\Lambda$ by merely multiplying the old ket (the old wave function) by $exp[ie\Lambda(x)/\hbar c]exp[ie\Lambda(x^{\prime})/\hbar c]$. The canonical momentum, defined as the generator of translation, is manifestly gauge dependent in the sense that its expectation value depends on the particular gauge chosen, while the kinematic [or mechanical] momentum and the probability flux are gauge invariant."

Let's see if I understand this correctly. The case considered is that of a charged particle in a magnetic field (ignoring eventual magnetic effects due to the charge).

-- That the mechanical momentum traces the trajectory is just because it is proportional to the acceleration.

-- The canonical momentum is the generator of translation means just that. A translation $\delta x$ multiplies by $exp(i\delta x p/\hbar)$ where the p in question is the non-invariant canonical momentum.

-- The non-invariance of the canonical momentum in this case, plus its being the generator of translation, means simply that translation symmetry does not hold and the canonical momentum is not conserved (which we already knew). But is $mv$ conserved?

-- This still leaves me wondering which one is replaced by $-i \hbar \partial_x$, or. more importantly, why.

 

[pines-demon]

That the mechanical momentum traces the trajectory is just because it is proportional to the acceleration.

Well this makes sense, it is the one related to the position of the particle, at least classically.

-- The canonical momentum is the generator of translation means just that. A translation $\delta x$ multiplies by $exp(i\delta x p/\hbar)$ where the p in question is the non-invariant canonical momentum.

Yes.

-- The non-invariance of the canonical momentum in this case, plus its being the generator of translation, means simply that translation symmetry does not hold and the canonical momentum is not conserved (which we already knew). But is $mv$ conserved?

I guess what is conserved is not $m\mathbf v$ but the kinetic energy $mv^2/2$? You do not need quantum mechanics to see that.

-- This still leaves me wondering which one is replaced by $-i \hbar \partial_x$, or. more importantly, why.

The answer to this is clear, it is the canonical momentum that gets transformed into $-i\hbar \nabla$, that's how canonical quantization works and you can check this by just looking examples of calculations, like for the Landau levels. If you use other approaches (like Ballentine's book), the one that becomes $-i\hbar \nabla$ is the one associated to translation invariance, again the canonical momentum.

 

[div_grad]

The Schrödinger formulation of QM is based on the Hamiltonian; and since the Hamiltonian is always given in terms of the canonical momentum, then in the appropriate Schrödinger equation it is precisely the canonical momentum that is to be replaced with $-\mathrm{i}\hbar\nabla$. It only just so happens that for a free particle this canonical momentum $\mathbf{p}_{\rm{can}}$ is the same as the kinetic (i.e., physical/observable/gauge-independent) momentum $\mathbf{p}_{\rm{kin}}=m\mathbf{v}$. In the presence of external vector potential $\mathbf{A}$ one has $\mathbf{p}_{\rm{can}} = \mathbf{p}_{\rm{kin}} + q\mathbf{A}$. Thus, the kinetic energy is written as
$$
\begin{align}
E_{\rm{kin}} &= \frac{m\mathbf{v}^2}{2} = \frac{\mathbf{p}^2_{\rm{kin}}}{2m} = \frac{\left(\mathbf{p}_{\rm{can}} - q\mathbf{A}\right)^2}{2m} =\\
&=\frac{\mathbf{p}^2_{\rm{can}}}{2m} - \frac{q}{2m}\left(\mathbf{p}_{\rm{can}}\cdot\mathbf{A} + \mathbf{A}\cdot\mathbf{p}_{\rm{can}}\right) + \frac{q^2}{2m}|\mathbf{A}|^2
\end{align}
$$
from which you immediately see that the $\mathbf{p}^2_{\rm{can}}/2m$ term, on its own, is not the kinetic energy of the particle. But since this term enters the Hamiltonian, and is replaced in the Schrödinger equation by $-\mathrm{i}\hbar\nabla$, then it takes the same differential form $-\hbar^2 \nabla^2 / 2m$ as the actual free-particle kinetic energy $\mathbf{p}^2_{\rm{kin}}/2m$ (because for a free particle, $\mathbf{p}_{\rm{can}} = \mathbf{p}_{\rm{kin}}$). So if one works solely in the position representation, then it is easy to make a mistake and wrongly interpret the unobservable gauge-dependent $\mathbf{p}^2_{\rm{can}}/2m$ term as the particle's kinetic energy, which has consequences for calculating matrix elements in different electromagnetic gauges.