## Commutators of angular momentum and a central force Hamiltonian

September 30, 2015 phy1520 , , , ,

[Click here for a PDF of this post with nicer formatting]

In problem 1.17 of [1] we are to show that non-commuting operators that both commute with the Hamiltonian, have, in general, degenerate energy eigenvalues. It suggests considering $$L_x, L_z$$ and a central force Hamiltonian $$H = \Bp^2/2m + V(r)$$ as examples.

Let’s just demonstrate these commutators act as expected in these cases.

With $$\BL = \Bx \cross \Bp$$, we have

\label{eqn:angularMomentumAndCentralForceCommutators:20}
\begin{aligned}
L_x &= y p_z – z p_y \\
L_y &= z p_x – x p_z \\
L_z &= x p_y – y p_x.
\end{aligned}

The $$L_x, L_z$$ commutator is

\label{eqn:angularMomentumAndCentralForceCommutators:40}
\begin{aligned}
\antisymmetric{L_x}{L_z}
&=
\antisymmetric{y p_z – z p_y }{x p_y – y p_x} \\
&=
\antisymmetric{y p_z}{x p_y}
-\antisymmetric{y p_z}{y p_x}
-\antisymmetric{z p_y }{x p_y}
+\antisymmetric{z p_y }{y p_x} \\
&=
x p_z \antisymmetric{y}{p_y}
+ z p_x \antisymmetric{p_y }{y} \\
&=
i \Hbar \lr{ x p_z – z p_x } \\
&=
– i \Hbar L_y
\end{aligned}

cyclicly permuting the indexes shows that no pairs of different $$\BL$$ components commute. For $$L_y, L_x$$ that is

\label{eqn:angularMomentumAndCentralForceCommutators:60}
\begin{aligned}
\antisymmetric{L_y}{L_x}
&=
\antisymmetric{z p_x – x p_z }{y p_z – z p_y} \\
&=
\antisymmetric{z p_x}{y p_z}
-\antisymmetric{z p_x}{z p_y}
-\antisymmetric{x p_z }{y p_z}
+\antisymmetric{x p_z }{z p_y} \\
&=
y p_x \antisymmetric{z}{p_z}
+ x p_y \antisymmetric{p_z }{z} \\
&=
i \Hbar \lr{ y p_x – x p_y } \\
&=
– i \Hbar L_z,
\end{aligned}

and for $$L_z, L_y$$

\label{eqn:angularMomentumAndCentralForceCommutators:80}
\begin{aligned}
\antisymmetric{L_z}{L_y}
&=
\antisymmetric{x p_y – y p_x }{z p_x – x p_z} \\
&=
\antisymmetric{x p_y}{z p_x}
-\antisymmetric{x p_y}{x p_z}
-\antisymmetric{y p_x }{z p_x}
+\antisymmetric{y p_x }{x p_z} \\
&=
z p_y \antisymmetric{x}{p_x}
+ y p_z \antisymmetric{p_x }{x} \\
&=
i \Hbar \lr{ z p_y – y p_z } \\
&=
– i \Hbar L_x.
\end{aligned}

If these angular momentum components are also shown to commute with themselves (which they do), the commutator relations above can be summarized as

\label{eqn:angularMomentumAndCentralForceCommutators:100}
\antisymmetric{L_a}{L_b} = i \Hbar \epsilon_{a b c} L_c.

In the example to consider, we’ll have to consider the commutators with $$\Bp^2$$ and $$V(r)$$. Picking any one component of $$\BL$$ is sufficent due to the symmetries of the problem. For example

\label{eqn:angularMomentumAndCentralForceCommutators:120}
\begin{aligned}
\antisymmetric{L_x}{\Bp^2}
&=
\antisymmetric{y p_z – z p_y}{p_x^2 + p_y^2 + p_z^2} \\
&=
\antisymmetric{y p_z}{{p_x^2} + p_y^2 + {p_z^2}}
-\antisymmetric{z p_y}{{p_x^2} + {p_y^2} + p_z^2} \\
&=
p_z \antisymmetric{y}{p_y^2}
-p_y \antisymmetric{z}{p_z^2} \\
&=
p_z 2 i \Hbar p_y
2 i \Hbar p_y
-p_y 2 i \Hbar p_z \\
&=
0.
\end{aligned}

How about the commutator of $$\BL$$ with the potential? It is sufficient to consider one component again, for example

\label{eqn:angularMomentumAndCentralForceCommutators:140}
\begin{aligned}
\antisymmetric{L_x}{V}
&=
\antisymmetric{y p_z – z p_y}{V} \\
&=
y \antisymmetric{p_z}{V} – z \antisymmetric{p_y}{V} \\
&=
-i \Hbar y \PD{z}{V(r)} + i \Hbar z \PD{y}{V(r)} \\
&=
-i \Hbar y \PD{r}{V}\PD{z}{r} + i \Hbar z \PD{r}{V}\PD{y}{r} \\
&=
-i \Hbar y \PD{r}{V} \frac{z}{r} + i \Hbar z \PD{r}{V}\frac{y}{r} \\
&=
0.
\end{aligned}

We’ve shown that all the components of $$\BL$$ commute with a central force Hamiltonian, and each different component of $$\BL$$ do not commute.

The next step will be figuring out how to use this to show that there are energy degeneracies.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## PHY1520H Graduate Quantum Mechanics. Lecture 4: Quantum Harmonic oscillator and coherent states. Taught by Prof. Arun Paramekanti

September 29, 2015 phy1520 , , , ,

[Click here for a PDF of this post with nicer formatting]

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough. This lecture reviewed a lot of quantum harmonic oscillator theory, and wouldn’t make sense without having seen raising and lowering operators (ladder operators), number operators, and the like.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 2 content.

### Classical Harmonic Oscillator

Recall the classical Harmonic oscillator equations in their Hamiltonian form

\label{eqn:qmLecture4:40}
\ddt{x} = \frac{p}{m}

\label{eqn:qmLecture4:60}
\ddt{p} = -k x.

With

\label{eqn:qmLecture4:140}
\begin{aligned}
x(t = 0) &= x_0 \\
p(t = 0) &= p_0 \\
k &= m \omega^2,
\end{aligned}

the solutions are ellipses in phase space

\label{eqn:qmLecture4:100}
x(t) = x_0 \cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t)

\label{eqn:qmLecture4:120}
p(t) = p_0 \cos(\omega t) – m \omega x_0 \sin(\omega t).

After a suitable scaling of the variables, these elliptical orbits can be transformed into circular trajectories.

### Quantum Harmonic Oscillator

\label{eqn:qmLecture4:160}
\hat{H} = \frac{\hat{p}^2}{2 m} + \inv{2} k \hat{x}^2

Set

\label{eqn:qmLecture4:200}
\hat{X} = \sqrt{\frac{m \omega}{\Hbar}} \hat{x}

\label{eqn:qmLecture4:220}
\hat{P} = \sqrt{\inv{m \omega \Hbar}} \hat{p}

The commutators after this change of variables goes from

\label{eqn:qmLecture4:240}
\antisymmetric{ \hat{x}}{\hat{p}} = i \Hbar,

to
\label{eqn:qmLecture4:260}
\antisymmetric{ \hat{X}}{\hat{P}} = i.

The Hamiltonian takes the form

\label{eqn:qmLecture4:280}
\begin{aligned}
\hat{H}
&= \frac{\Hbar \omega}{2} \lr{ \hat{X}^2 + \hat{P}^2 } \\
&= \Hbar \omega \lr{ \lr{ \frac{\hat{X} -i \hat{P}}{\sqrt{2}} } \lr{ \frac{\hat{X} +i \hat{P}}{\sqrt{2}}} + \inv{2} }.
\end{aligned}

Define ladder operators (raising and lowering operators respectively)

\label{eqn:qmLecture4:320}
\hat{a}^\dagger = \frac{\hat{X} -i \hat{P}}{\sqrt{2}}

\label{eqn:qmLecture4:340}
\hat{a} = \frac{\hat{X} +i \hat{P}}{\sqrt{2}}

so

\label{eqn:qmLecture4:360}
\hat{H} = \Hbar \omega \lr{ \hat{a}^\dagger \hat{a} + \inv{2} }.

We can show

\label{eqn:qmLecture4:380}
\antisymmetric{\hat{a}}{\hat{a}^\dagger} = 1,

and

\label{eqn:qmLecture4:400}
N \ket{n} \equiv \hat{a}^\dagger a = n \ket{n},

where $$n \ge 0$$ is an integer. Recall that

\label{eqn:qmLecture4:420}
\hat{a} \ket{0} = 0,

and

\label{eqn:qmLecture4:440}
\bra{X} X + i P \ket{0} = 0.

With

\label{eqn:qmLecture4:460}
\braket{x}{0} = \Psi_0(x),

we can show

\label{eqn:qmLecture4:480}
\inv{\sqrt{2}} \lr{ X + \PD{X}{} } \Psi_0(X) = 0.

Also recall that

\label{eqn:qmLecture4:520}
\hat{a} \ket{n} = \sqrt{n} \ket{n-1}

\label{eqn:qmLecture4:540}
\hat{a}^\dagger \ket{n} = \sqrt{n + 1} \ket{n+1}

### Coherent states

Coherent states for the quantum harmonic oscillator are the eigenkets for the creation and annihilation operators

\label{eqn:qmLecture4:580}
\hat{a} \ket{z} = z \ket{z}

\label{eqn:qmLecture4:600}
\hat{a}^\dagger \ket{\tilde{z}} = \tilde{z} \ket{\tilde{z}} ,

where

\label{eqn:qmLecture4:620}
\ket{z} = \sum_{n = 0}^\infty c_n \ket{n},

and $$z$$ is allowed to be a complex number.

Looking for such a state, we compute

\label{eqn:qmLecture4:640}
\begin{aligned}
\hat{a} \ket{z}
&= \sum_{n=1}^\infty c_n \hat{a} \ket{n} \\
&= \sum_{n=1}^\infty c_n \sqrt{n} \ket{n-1}
\end{aligned}

compare this to

\label{eqn:qmLecture4:660}
\begin{aligned}
z \ket{z}
&=
z \sum_{n=0}^\infty c_n \ket{n} \\
&=
\sum_{n=1}^\infty c_n \sqrt{n} \ket{n-1} \\
&=
\sum_{n=0}^\infty c_{n+1} \sqrt{n+1} \ket{n},
\end{aligned}

so

\label{eqn:qmLecture4:680}
c_{n+1} \sqrt{n+1} = z c_n

This gives

\label{eqn:qmLecture4:700}
c_{n+1} = \frac{z c_n}{\sqrt{n+1}}

\label{eqn:qmLecture4:720}
\begin{aligned}
c_1 &= c_0 z \\
c_2 &= \frac{z c_1}{\sqrt{2}} = \frac{z^2 c_0}{\sqrt{2}} \\
\vdots &
\end{aligned}

or

\label{eqn:qmLecture4:740}
c_n = \frac{z^n}{\sqrt{n!}}.

So the desired state is

\label{eqn:qmLecture4:760}
\ket{z} = c_0 \sum_{n=0}^\infty \frac{z^n}{\sqrt{n!}} \ket{n}.

Also recall that

\label{eqn:qmLecture4:780}
\ket{n} = \frac{\lr{ \hat{a}^\dagger }^n}{\sqrt{n!}} \ket{0},

which gives

\label{eqn:qmLecture4:800}
\begin{aligned}
\ket{z}
&= c_0 \sum_{n=0}^\infty \frac{\lr{z \hat{a}^\dagger}^n }{n!} \ket{0} \\
&= c_0 e^{z \hat{a}^\dagger} \ket{0}.
\end{aligned}

The normalization is

\label{eqn:qmLecture4:820}
c_0 = e^{-\Abs{z}^2/2}.

While we have $$\braket{n_1}{n_2} = \delta_{n_1, n_2}$$, these $$\ket{z}$$ states are not orthonormal. Figuring out that this overlap

\label{eqn:qmLecture4:840}
\braket{z_1}{z_2} \ne 0,

will be left for homework.

### Dynamics

We don’t know much about these coherent states. For example does a coherent state at time zero evolve to a coherent state?

\label{eqn:qmLecture4:860}
\ket{z} \stackrel{?}{\rightarrow} \ket{z(t)}

It turns out that these questions are best tackled in the Heisenberg picture, considering

\label{eqn:qmLecture4:880}
e^{-i \hat{H} t/\Hbar } \ket{z}.

For example, what is the average of the position operator

\label{eqn:qmLecture4:900}
\bra{z} e^{i \hat{H} t/\Hbar } \hat{x} e^{-i \hat{H} t/\Hbar } \ket{z}
=
\sum_{n, n’ = 0}^\infty
\bra{n} c_n^\conj e^{i E_n t/\Hbar}
\lr{ a + a^\dagger} \sqrt{ \frac{\Hbar}{m \omega} }
c_{n’} e^{i E_{n’} t/\Hbar}
\ket{n}.

This is very messy to attempt. Instead if we know how the operator evolves we can calculate

\label{eqn:qmLecture4:920}
\bra{z} \hat{x}_{\textrm{H}}(t) \ket{z},

that is

\label{eqn:qmLecture4:940}
\expectation{\hat{x}}(t) = \bra{z} \hat{x}_{\textrm{H}}(t) \ket{z},

and for momentum

\label{eqn:qmLecture4:960}
\expectation{\hat{p}}(t) = \bra{z} \hat{p}_{\textrm{H}}(t) \ket{z}.

The question to ask is what are the expansions of

\label{eqn:qmLecture4:1000}
\hat{a}_{\textrm{H}}(t) = e^{i \hat{H} t/\Hbar} \hat{a} e^{-i \hat{H} t/\Hbar}.

\label{eqn:qmLecture4:1020}
\hat{a}^\dagger_{\textrm{H}}(t) = e^{i \hat{H} t/\Hbar} \hat{a}^\dagger e^{-i \hat{H} t/\Hbar}.

The question to ask is how do these operators ask on the basis states

\label{eqn:qmLecture4:1040}
\begin{aligned}
\hat{a}_{\textrm{H}}(t) \ket{n}
&= e^{i \hat{H} t/\Hbar} \hat{a} e^{-i \hat{H} t/\Hbar} \ket{n} \\
&= e^{i \hat{H} t/\Hbar} \hat{a} e^{-i t \omega (n + 1/2)} \ket{n} \\
&=
e^{-i t \omega (n + 1/2)}
e^{i \hat{H} t/\Hbar}
\sqrt{n} \ket{n-1} \\
&=
\sqrt{n}
e^{-i t \omega (n + 1/2)}
e^{i t \omega (n – 1/2)}
\ket{n-1} \\
&=
\sqrt{n} e^{-i \omega t} \ket{n-1} \\
&=
e^{-i \omega t} \ket{n}.
\end{aligned}

So we have found

\label{eqn:qmLecture4:1060}
\begin{aligned}
\hat{a}_{\textrm{H}}(t) &= a e^{-i\omega t} \\
\hat{a}^\dagger_{\textrm{H}}(t) &= a^\dagger e^{i\omega t}
\end{aligned}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Can anticommuting operators have a simulaneous eigenket?

September 28, 2015 phy1520 , ,

[Click here for a PDF of this post with nicer formatting]

## Question: Can anticommuting operators have a simulaneous eigenket? ([1] pr. 1.16)

Two Hermitian operators anticommute

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:20}
\symmetric{A}{B} = A B + B A = 0.

Is it possible to have a simultaneous eigenket of $$A$$ and $$B$$? Prove or illustrate your assertion.

## Answer

Suppose that such a simultaneous non-zero eigenket $$\ket{\alpha}$$ exists, then

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:40}
A \ket{\alpha} = a \ket{\alpha},

and

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:60}
B \ket{\alpha} = b \ket{\alpha}

This gives

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:80}
\lr{ A B + B A } \ket{\alpha}
=
\lr{A b + B a} \ket{\alpha}
= 2 a b \ket{\alpha}.

If this is zero, one of the operators must have a zero eigenvalue. Knowing that we can construct an example of such operators. In matrix form, let

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:120}
A =
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & a \\
\end{bmatrix}

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:140}
B =
\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & b \\
\end{bmatrix}.

These are both Hermitian, and anticommute provided at least one of $$a, b$$ is zero. These have a common eigenket

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:160}
\ket{\alpha} =
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}.

A zero eigenvalue of one of the commuting operators may not be a sufficient condition for such anticommutation.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Grade 11 physics handout: “The Big Five … in Physics”?

September 26, 2015 math and physics play ,

[Click here for a PDF of this post with nicer formatting]

## Motivation

Check out fig. 1, a handout given to my daughter has a handout from her grade 11 physics class, titled “The Big Five”, covering some dynamics equations.

fig. 1. The Big Five… in Physics

I found this handout disorienting. Part of that disorientation is because of the weird African animal theme which I couldn’t see the rationale for. Aurora showed me that the second equation can be outlined with an elephant, apparently justifying the animal theme. The equations themselves are not in a form that I would have expected, and have a lot of redundancy built in. The assumptions required for these equations to be valid are also not stated. Those equations are

\label{eqn:theBigFivePhysics:40}
v_2 = v_1 + a \Delta t

\label{eqn:theBigFivePhysics:60}
\Delta d = \inv{2} \lr{ v_1 + v_2 } \Delta t

\label{eqn:theBigFivePhysics:80}
\Delta d = v_1 \Delta t + \inv{2} a \lr{\Delta t}^2

\label{eqn:theBigFivePhysics:100}
\Delta d = v_2 \Delta t – \inv{2} a \lr{\Delta t}^2

\label{eqn:theBigFivePhysics:120}
v_2^2 = v_1^2 + 2 a \Delta d.

## Reverse engineering “the big five”.

### Difference of velocity

The first equation \ref{eqn:theBigFivePhysics:40} is just a discrete version of the definition of scalar acceleration

\label{eqn:theBigFivePhysics:140}
a = \frac{dv}{dt}.

The approximation of that is

\label{eqn:theBigFivePhysics:160}
a = \frac{\Delta v}{\Delta t},

or

\label{eqn:theBigFivePhysics:180}
\Delta v = v_2 – v_1 = a \Delta t.

### Constant acceleration

Next in the list, it’s clear that the equation(s) for $$\Delta d$$ is really based on an assumption of constant acceleration. In fact, all four of the next equations are nothing more than variations of

\label{eqn:theBigFivePhysics:200}
v = a t.

To arrive at fig. 2.

fig. 2. Displacement as area under the curve.

Should we wish to integrate, it’s the second simplest integral we could possibly do

\label{eqn:theBigFivePhysics:220}
\begin{aligned}
\Delta x
&= \int_{t_1}^{t_2} v(t) dt \\
&= \int_{t_1}^{t_2} a t dt \\
&= \inv{2} a \lr{ t_2^2 – t_1^2 } \\
&= \inv{2} a \Delta t \lr{ t_2 + t_1 }.
\end{aligned}

This looks a little different than what’s on the formula cheet, but since (for constant acceleration) we have

\label{eqn:theBigFivePhysics:240}
t = \frac{v}{a},

this can be written as

\label{eqn:theBigFivePhysics:260}
\begin{aligned}
\Delta x
&= \inv{2} a \Delta t \lr{ \frac{v_2}{a} + \frac{v_1}{a} } \\
&= \inv{2} \Delta t \lr{ v_1 + v_2 },
\end{aligned}

as found on the formula sheet (except for them using $$\Delta d$$ for the difference in position .)

Each of the next equations follow from straight algebra

\label{eqn:theBigFivePhysics:280}
\begin{aligned}
\Delta x – v_1 \Delta t
&= \inv{2} \Delta t \lr{ v_1 + v_2 } – v_1 \Delta t \\
&= \inv{2} \Delta t \lr{ -v_1 + v_2 } \\
&= \inv{2} \lr{\Delta t}^2 \frac{\Delta v}{\Delta t} \\
&= \inv{2} a \lr{\Delta t}^2,
\end{aligned}

and

\label{eqn:theBigFivePhysics:300}
\begin{aligned}
\Delta x – v_2 \Delta t
&= \inv{2} \Delta t \lr{ v_1 + v_2 } – v_2 \Delta t \\
&= \inv{2} \Delta t \lr{ v_1 – v_2 } \\
&= -\inv{2} \lr{\Delta t}^2 \frac{\Delta v}{\Delta t} \\
&= -\inv{2} a \lr{\Delta t}^2,
\end{aligned}

and finally

\label{eqn:theBigFivePhysics:320}
\begin{aligned}
\Delta x
&= \inv{2} a \lr{ t_2^2 – t_1^2 } \\
&= \inv{2 a } \lr{ v_2^2 – v_1^2 }.
\end{aligned}

### A better set of equations.

If I had to write these “big five” equation, I’d be more inclined to write them as

\label{eqn:theBigFivePhysics:360}
a = \frac{\Delta v}{\Delta t}

\label{eqn:theBigFivePhysics:380}
v = a t = \frac{\Delta x}{\Delta t}

\label{eqn:theBigFivePhysics:400}
\Delta x = \int_{t_1}^{t_2} v dt = \inv{2} a \lr{ t_2^2 – t_1^2 }

\label{eqn:theBigFivePhysics:420}
t_1 = \frac{v_1}{a}

\label{eqn:theBigFivePhysics:440}
t_2 = \frac{v_2}{a}.

Anything more than that is just algebra. The last two could be omitted since they really follow from \ref{eqn:theBigFivePhysics:380}. For high school where calculus isn’t known, I’d swap out \ref{eqn:theBigFivePhysics:400} for \ref{eqn:theBigFivePhysics:60} which can be derived graphically by understanding that the distance is the area under the velocity curve.

I’d also leave out all mentions of big African animals, which is, just plain weird!

## Lagrangian for magnetic portion of Lorentz force

September 26, 2015 phy1520 , , , ,

[Click here for a PDF of this post with nicer formatting]

In [1] it is claimed in an Aharonov-Bohm discussion that a Lagrangian modification to include electromagnetism is

\label{eqn:magneticLorentzForceLagrangian:20}
\LL \rightarrow \LL + \frac{e}{c} \Bv \cdot \BA.

That can’t be the full Lagrangian since there is no $$\phi$$ term, so what exactly do we get?

If you have somehow, like I did, forgot the exact form of the Euler-Lagrange equations (i.e. where do the dots go), then the derivation of those equations can come to your rescue. The starting point is the action

\label{eqn:magneticLorentzForceLagrangian:40}
S = \int \LL(x, \xdot, t) dt,

where the end points of the integral are fixed, and we assume we have no variation at the end points. The variational calculation is

\label{eqn:magneticLorentzForceLagrangian:60}
\begin{aligned}
\delta S
&= \int \delta \LL(x, \xdot, t) dt \\
&= \int \lr{ \PD{x}{\LL} \delta x + \PD{\xdot}{\LL} \delta \xdot } dt \\
&= \int \lr{ \PD{x}{\LL} \delta x + \PD{\xdot}{\LL} \delta \ddt{x} } dt \\
&= \int \lr{ \PD{x}{\LL} – \ddt{}\lr{\PD{\xdot}{\LL}} } \delta x dt
+ \delta x \PD{\xdot}{\LL}.
\end{aligned}

The boundary term is killed after evaluation at the end points where the variation is zero. For the result to hold for all variations $$\delta x$$, we must have

\label{eqn:magneticLorentzForceLagrangian:80}
\boxed{
\PD{x}{\LL} = \ddt{}\lr{\PD{\xdot}{\LL}}.
}

Now lets apply this to the Lagrangian at hand. For the position derivative we have

\label{eqn:magneticLorentzForceLagrangian:100}
\PD{x_i}{\LL}
=
\frac{e}{c} v_j \PD{x_i}{A_j}.

For the canonical momentum term, assuming $$\BA = \BA(\Bx)$$ we have

\label{eqn:magneticLorentzForceLagrangian:120}
\begin{aligned}
\ddt{} \PD{\xdot_i}{\LL}
&=
\ddt{}
\lr{ m \xdot_i
+
\frac{e}{c} A_i
} \\
&=
m \ddot{x}_i
+
\frac{e}{c}
\ddt{A_i} \\
&=
m \ddot{x}_i
+
\frac{e}{c}
\PD{x_j}{A_i} \frac{dx_j}{dt}.
\end{aligned}

Assembling the results, we’ve got

\label{eqn:magneticLorentzForceLagrangian:140}
\begin{aligned}
0
&=
\ddt{} \PD{\xdot_i}{\LL}

\PD{x_i}{\LL} \\
&=
m \ddot{x}_i
+
\frac{e}{c}
\PD{x_j}{A_i} \frac{dx_j}{dt}

\frac{e}{c} v_j \PD{x_i}{A_j},
\end{aligned}

or
\label{eqn:magneticLorentzForceLagrangian:160}
\begin{aligned}
m \ddot{x}_i
&=
\frac{e}{c} v_j \PD{x_i}{A_j}

\frac{e}{c}
\PD{x_j}{A_i} v_j \\
&=
\frac{e}{c} v_j
\lr{
\PD{x_i}{A_j}

\PD{x_j}{A_i}
} \\
&=
\frac{e}{c} v_j B_k \epsilon_{i j k}.
\end{aligned}

In vector form that is

\label{eqn:magneticLorentzForceLagrangian:180}
m \ddot{\Bx}
=
\frac{e}{c} \Bv \cross \BB.

So, we get the magnetic term of the Lorentz force. Also note that this shows the Lagrangian (and the end result), was not in SI units. The $$1/c$$ term would have to be dropped for SI.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.