## Motivation.

In my old classical mechanics notes it appears that I did covariant derivations of the Lorentz force equations a number of times, using different trial Lagrangians (relativistic and non-relativistic), and using both geometric algebra and tensor methods. However, none of these appear to have been done concisely, and a number not even coherently.

The following document has been drafted as replacement text for those incoherent classical mechanics notes. I’ll attempt to cover

• a lighting review of the geometric algebra STA (Space Time Algebra),
• relations between Dirac matrix algebra and STA,
• derivation of the relativistic form of the Euler-Lagrange equations from the covariant form of the action,
• relationship of the STA form of the Euler-Lagrange equations to their tensor equivalents,
• derivation of the Lorentz force equation from the STA Lorentz force Lagrangian,
• relationship of the STA Lorentz force equation to its equivalent in the tensor formalism,
• relationship of the STA Lorentz force equation to the traditional vector form.

Note that some of the prerequisite ideas and auxiliary details are presented as problems with solutions. If the reader has sufficient background to attempt those problems themselves, they are encouraged to do so.

The STA and geometric algebra ideas used here are not complete to learn from in isolation. The reader is referred to [1] for a more complete exposition of both STA and geometric algebra.

## Definition 1.1: Index conventions.

Latin indexes $$i, j, k, r, s, t, \cdots$$ are used to designate values in the range $$\setlr{ 1,2,3 }$$. Greek indexes are $$\alpha, \beta, \mu, \nu, \cdots$$ are used for indexes of spacetime quantities $$\setlr{0,1,2,3}$$.
The Einstein convention of implied summation for mixed upper and lower Greek indexes will be used, for example
\begin{equation*}
x^\alpha x_\alpha \equiv \sum_{\alpha = 0}^3 x^\alpha x_\alpha.
\end{equation*}

## Space Time Algebra (STA.)

In the geometric algebra literature, the Dirac algebra of quantum field theory has been rebranded Space Time Algebra (STA). The differences between STA and the Dirac theory that uses matrices ($$\gamma_\mu$$) are as follows

• STA completely omits any representation of the Dirac basis vectors $$\gamma_\mu$$. In particular, any possible matrix representation is irrelevant.
• STA provides a rich set of fundamental operations (grade selection, generalized dot and wedge products for multivector elements, rotation and reflection operations, …)
• Matrix trace, and commutator and anticommutator operations are nowhere to be found in STA, as geometrically grounded equivalents are available instead.
• The “slashed” quantities from Dirac theory, such as $$\gamma_\mu p^\mu$$ are nothing more than vectors in their entirety in STA (where the basis is no longer implicit, as is the case for coordinates.)

Our basis vectors have the following properties.

## Definition 1.2: Standard basis.

Let the four-vector standard basis be designated $$\setlr{\gamma_0, \gamma_1, \gamma_2, \gamma_3 }$$, where the basis vectors satisfy
\label{eqn:lorentzForceCovariant:1540}
\begin{aligned}
\gamma_0^2 &= -\gamma_i^2 = 1 \\
\gamma_\alpha \cdot \gamma_\beta &= 0, \forall \alpha \ne \beta.
\end{aligned}

## Problem: Commutator properties of the STA basis.

In Dirac theory, the commutator properties of the Dirac matrices is considered fundamental, namely
\begin{equation*}
\symmetric{\gamma_\mu}{\gamma_\nu} = 2 \eta_{\mu\nu}.
\end{equation*}

Show that this follows from the axiomatic assumptions of geometric algebra, and describe how the dot and wedge products are related to the anticommutator and commutator products of Dirac theory.

The anticommutator is defined as symmetric sum of products
\label{eqn:lorentzForceCovariant:1040}
\symmetric{\gamma_\mu}{\gamma_\nu}
\equiv
\gamma_\mu \gamma_\nu
+
\gamma_\nu \gamma_\mu,

but this is just twice the dot product in its geometric algebra form $$a b = (a b + ba)/2$$. Observe that the properties of the basis vectors defined in \ref{eqn:lorentzForceCovariant:1540} may be summarized as
\label{eqn:lorentzForceCovariant:1060}
\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu},

where $$\eta_{\mu\nu} = \text{diag}(+,-,-,-) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}$$ is the conventional metric tensor. This means
\label{eqn:lorentzForceCovariant:1080}
\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu} = 2 \symmetric{\gamma_\mu}{\gamma_\nu},

as claimed.

Similarly, observe that the commutator, defined as the antisymmetric sum of products
\label{eqn:lorentzForceCovariant:1100}
\antisymmetric{\gamma_\mu}{\gamma_\nu} \equiv
\gamma_\mu \gamma_\nu

\gamma_\nu \gamma_\mu,

is twice the wedge product $$a \wedge b = (a b – b a)/2$$. This provides geometric identifications for the respective anti-commutator and commutator products respectively
\label{eqn:lorentzForceCovariant:1120}
\begin{aligned}
\symmetric{\gamma_\mu}{\gamma_\nu} &= 2 \gamma_\mu \cdot \gamma_\nu \\
\antisymmetric{\gamma_\mu}{\gamma_\nu} &= 2 \gamma_\mu \wedge \gamma_\nu,
\end{aligned}

## Definition 1.3: Pseudoscalar.

The pseudoscalar for the space is denoted $$I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$$.

## Problem: Pseudoscalar.

Show that the STA pseudoscalar $$I$$ defined by \ref{eqn:lorentzForceCovariant:1540} satisfies
\begin{equation*}
\tilde{I} = I,
\end{equation*}
where the tilde operator designates reversion. Also show that $$I$$ has the properties of an imaginary number
\begin{equation*}
I^2 = -1.
\end{equation*}
Finally, show that, unlike the spatial pseudoscalar that commutes with all grades, $$I$$ anticommutes with any vector or trivector, and commutes with any bivector.

Since $$\gamma_\alpha \gamma_\beta = -\gamma_\beta \gamma_\alpha$$ for any $$\alpha \ne \beta$$, any permutation of the factors of $$I$$ changes the sign once. In particular
\label{eqn:lorentzForceCovariant:680}
\begin{aligned}
I &=
\gamma_0
\gamma_1
\gamma_2
\gamma_3 \\
&=

\gamma_1
\gamma_2
\gamma_3
\gamma_0 \\
&=

\gamma_2
\gamma_3
\gamma_1
\gamma_0 \\
&=
+
\gamma_3
\gamma_2
\gamma_1
\gamma_0
= \tilde{I}.
\end{aligned}

Using this, we have
\label{eqn:lorentzForceCovariant:700}
\begin{aligned}
I^2
&= I \tilde{I} \\
&=
(
\gamma_0
\gamma_1
\gamma_2
\gamma_3
)(
\gamma_3
\gamma_2
\gamma_1
\gamma_0
) \\
&=
\lr{\gamma_0}^2
\lr{\gamma_1}^2
\lr{\gamma_2}^2
\lr{\gamma_3}^2 \\
&=
(+1)
(-1)
(-1)
(-1) \\
&= -1.
\end{aligned}

To illustrate the anticommutation property with any vector basis element, consider the following two examples:
\label{eqn:lorentzForceCovariant:720}
\begin{aligned}
I \gamma_0 &=
\gamma_0
\gamma_1
\gamma_2
\gamma_3
\gamma_0 \\
&=

\gamma_0
\gamma_0
\gamma_1
\gamma_2
\gamma_3 \\
&=

\gamma_0 I,
\end{aligned}

\label{eqn:lorentzForceCovariant:740}
\begin{aligned}
I \gamma_2
&=
\gamma_0
\gamma_1
\gamma_2
\gamma_3
\gamma_2 \\
&=

\gamma_0
\gamma_1
\gamma_2
\gamma_2
\gamma_3 \\
&=

\gamma_2
\gamma_0
\gamma_1
\gamma_2
\gamma_3 \\
&= -\gamma_2 I.
\end{aligned}

A total of three sign swaps is required to “percolate” any given $$\gamma_\alpha$$ through the factors of $$I$$, resulting in an overall sign change of $$-1$$.

For any bivector basis element $$\alpha \ne \beta$$
\label{eqn:lorentzForceCovariant:760}
\begin{aligned}
I \gamma_\alpha \gamma_\beta
&=
-\gamma_\alpha I \gamma_\beta \\
&=
+\gamma_\alpha \gamma_\beta I.
\end{aligned}

Similarly for any trivector basis element $$\alpha \ne \beta \ne \sigma$$
\label{eqn:lorentzForceCovariant:780}
\begin{aligned}
I \gamma_\alpha \gamma_\beta \gamma_\sigma
&=
-\gamma_\alpha I \gamma_\beta \gamma_\sigma \\
&=
+\gamma_\alpha \gamma_\beta I \gamma_\sigma \\
&=
-\gamma_\alpha \gamma_\beta \gamma_\sigma I.
\end{aligned}

## Definition 1.4: Reciprocal basis.

The reciprocal basis $$\setlr{ \gamma^0, \gamma^1, \gamma^2, \gamma^3 }$$ is defined , such that the property $$\gamma^\alpha \cdot \gamma_\beta = {\delta^\alpha}_\beta$$ holds.

Observe that, $$\gamma^0 = \gamma_0$$ and $$\gamma^i = -\gamma_i$$.

## Theorem 1.1: Coordinates.

Coordinates are defined in terms of dot products with the standard basis, or reciprocal basis
\begin{equation*}
\begin{aligned}
x^\alpha &= x \cdot \gamma^\alpha \\
x_\alpha &= x \cdot \gamma_\alpha,
\end{aligned}
\end{equation*}

### Start proof:

Suppose that a coordinate representation of the following form is assumed
\label{eqn:lorentzForceCovariant:820}
x = x^\alpha \gamma_\alpha = x_\beta \gamma^\beta.

We wish to determine the representation of the $$x^\alpha$$ or $$x_\beta$$ coordinates in terms of $$x$$ and the basis elements. Taking the dot product with any standard basis element, we find
\label{eqn:lorentzForceCovariant:840}
\begin{aligned}
x \cdot \gamma_\mu
&= (x_\beta \gamma^\beta) \cdot \gamma_\mu \\
&= x_\beta {\delta^\beta}_\mu \\
&= x_\mu,
\end{aligned}

as claimed. Similarly, dotting with a reciprocal frame vector, we find
\label{eqn:lorentzForceCovariant:860}
\begin{aligned}
x \cdot \gamma^\mu
&= (x^\beta \gamma_\beta) \cdot \gamma^\mu \\
&= x^\beta {\delta_\beta}^\mu \\
&= x^\mu.
\end{aligned}

### End proof.

Observe that raising or lowering the index of a spatial index toggles the sign of a coordinate, but timelike indexes are left unchanged.
\label{eqn:lorentzForceCovariant:880}
\begin{aligned}
x^0 &= x_0 \\
x^i &= -x_i \\
\end{aligned}

\begin{equation*}
\grad = \gamma^\mu \partial_\mu = \gamma_\nu \partial^\nu,
\end{equation*}
where
\begin{equation*}
\partial_\mu = \PD{x^\mu}{},
\end{equation*}
and
\begin{equation*}
\partial^\mu = \PD{x_\mu}{}.
\end{equation*}

This definition of gradient is consistent with the Dirac gradient (sometimes denoted as a slashed $$\partial$$).

## Definition 1.6: Timelike and spacelike components of a four-vector.

Given a four vector $$x = \gamma_\mu x^\mu$$, that would be designated $$x^\mu = \setlr{ x^0, \Bx}$$ in conventional special relativity, we write
\begin{equation*}
x^0 = x \cdot \gamma_0,
\end{equation*}
and
\begin{equation*}
\Bx = x \wedge \gamma_0,
\end{equation*}
or
\begin{equation*}
x = (x^0 + \Bx) \gamma_0.
\end{equation*}

The spacetime split of a four-vector $$x$$ is relative to the frame. In the relativistic lingo, one would say that it is “observer dependent”, as the same operations with $${\gamma_0}’$$, the timelike basis vector for a different frame, would yield a different set of coordinates.

While the dot and wedge products above provide an effective mechanism to split a four vector into a set of timelike and spacelike quantities, the spatial component of a vector has a bivector representation in STA. Consider the following coordinate expansion of a spatial vector
\label{eqn:lorentzForceCovariant:1000}
\Bx =
x \wedge \gamma_0
=
\lr{ x^\mu \gamma_\mu } \wedge \gamma_0
=
\sum_{k = 1}^3 x^k \gamma_k \gamma_0.

## Definition 1.7: Spatial basis.

We designate
\label{eqn:lorentzForceCovariant:1560}
\Be_i = \gamma_i \gamma_0,

as the standard basis vectors for $$\mathbb{R}^3$$.

In the literature, this bivector representation of the spatial basis may be designated $$\sigma_i = \gamma_i \gamma_0$$, as these bivectors have the properties of the Pauli matrices $$\sigma_i$$. Because I intend to expand these notes to include purely non-relativistic applications, I won’t use the Pauli notation here.

## Problem: Orthonormality of the spatial basis.

Show that the spatial basis $$\setlr{ \Be_1, \Be_2, \Be_3 }$$, defined by \ref{eqn:lorentzForceCovariant:1560}, is orthonormal.

\label{eqn:lorentzForceCovariant:620}
\begin{aligned}
\Be_i \cdot \Be_j
&= \gpgradezero{ \gamma_i \gamma_0 \gamma_j \gamma_0 } \\
&= -\gpgradezero{ \gamma_i \gamma_j } \\
&= – \gamma_i \cdot \gamma_j.
\end{aligned}

This is zero for all $$i \ne j$$, and unity for any $$i = j$$.

## Problem: Spatial pseudoscalar.

Show that the STA pseudoscalar $$I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$$ equals the spatial pseudoscalar $$I = \Be_1 \Be_2 \Be_3$$.

The spatial pseudoscalar, expanded in terms of the STA basis vectors, is
\label{eqn:lorentzForceCovariant:1020}
\begin{aligned}
I
&= \Be_1 \Be_2 \Be_3 \\
&= \lr{ \gamma_1 \gamma_0 }
\lr{ \gamma_2 \gamma_0 }
\lr{ \gamma_3 \gamma_0 } \\
&= \lr{ \gamma_1 \gamma_0 } \gamma_2 \lr{ \gamma_0 \gamma_3 } \gamma_0 \\
&= \lr{ -\gamma_0 \gamma_1 } \gamma_2 \lr{ -\gamma_3 \gamma_0 } \gamma_0 \\
&= \gamma_0 \gamma_1 \gamma_2 \gamma_3 \lr{ \gamma_0 \gamma_0 } \\
&= \gamma_0 \gamma_1 \gamma_2 \gamma_3,
\end{aligned}

as claimed.

## Problem: Characteristics of the Pauli matrices.

The Pauli matrices obey the following anticommutation relations:
\label{eqn:lorentzForceCovariant:660}
\symmetric{ \sigma_a}{\sigma_b } = 2 \delta_{a b},

and commutation relations:
\label{eqn:lorentzForceCovariant:640}
\antisymmetric{ \sigma_a}{ \sigma_b } = 2 i \epsilon_{a b c}\,\sigma_c,

Show how these relate to the geometric algebra dot and wedge products, and determine the geometric algebra representation of the imaginary $$i$$ above.

## Euler-Lagrange equations.

I’ll start at ground zero, with the derivation of the relativistic form of the Euler-Lagrange equations from the action. A relativistic action for a single particle system has the form
\label{eqn:lorentzForceCovariant:20}
S = \int d\tau L(x, \dot{x}),

where $$x$$ is the spacetime coordinate, $$\dot{x} = dx/d\tau$$ is the four-velocity, and $$\tau$$ is proper time.

## Theorem 1.2: Relativistic Euler-Lagrange equations.

Let $$x \rightarrow x + \delta x$$ be any variation of the Lagrangian four-vector coordinates, where $$\delta x = 0$$ at the boundaries of the action integral. The variation of the action is
\label{eqn:lorentzForceCovariant:1580}
\delta S = \int d\tau \delta x \cdot \delta L(x, \dot{x}),

where
\label{eqn:lorentzForceCovariant:1600}

where $$\grad = \gamma^\mu \partial_\mu$$, and where we construct a similar velocity-gradient with respect to the proper-time derivatives of the coordinates $$\grad_v = \gamma^\mu \partial/\partial \dot{x}^\mu$$.The action is extremized when $$\delta S = 0$$, or when $$\delta L = 0$$. This latter condition is called the Euler-Lagrange equations.

### Start proof:

Let $$\epsilon = \delta x$$, and expand the Lagrangian in Taylor series to first order
\label{eqn:lorentzForceCovariant:60}
\begin{aligned}
S &\rightarrow S + \delta S \\
&= \int d\tau L( x + \epsilon, \dot{x} + \dot{\epsilon})
&=
\int d\tau \lr{
L(x, \dot{x}) + \epsilon \cdot \grad L + \dot{\epsilon} \cdot \grad_v L
}.
\end{aligned}

Subtracting off $$S$$ and integrating by parts, leaves
\label{eqn:lorentzForceCovariant:80}
\delta S =
\int d\tau \epsilon \cdot \lr{
}
+
\int d\tau \frac{d}{d\tau} (\grad_v L ) \cdot \epsilon.

The boundary integral
\label{eqn:lorentzForceCovariant:100}
\int d\tau \frac{d}{d\tau} (\grad_v L ) \cdot \epsilon
=
\evalbar{(\grad_v L ) \cdot \epsilon}{\Delta \tau} = 0,

is zero since the variation $$\epsilon$$ is required to vanish on the boundaries. So, if $$\delta S = 0$$, we must have
\label{eqn:lorentzForceCovariant:120}
0 =
\int d\tau \epsilon \cdot \lr{
},

for all variations $$\epsilon$$. Clearly, this requires that
\label{eqn:lorentzForceCovariant:140}

or
\label{eqn:lorentzForceCovariant:145}

which is the coordinate free statement of the Euler-Lagrange equations.

## Problem: Coordinate form of the Euler-Lagrange equations.

Working in coordinates, use the action argument show that the Euler-Lagrange equations have the form
\begin{equation*}
\PD{x^\mu}{L} = \frac{d}{d\tau} \PD{\dot{x}^\mu}{L}
\end{equation*}
Observe that this is identical to the statement of \ref{eqn:lorentzForceCovariant:1600} after contraction with $$\gamma^\mu$$.

In terms of coordinates, the first order Taylor expansion of the action is
\label{eqn:lorentzForceCovariant:180}
\begin{aligned}
S &\rightarrow S + \delta S \\
&= \int d\tau L( x^\alpha + \epsilon^\alpha, \dot{x}^\alpha + \dot{\epsilon}^\alpha) \\
&=
\int d\tau \lr{
L(x^\alpha, \dot{x}^\alpha) + \epsilon^\mu \PD{x^\mu}{L} + \dot{\epsilon}^\mu \PD{\dot{x}^\mu}{L}
}.
\end{aligned}

As before, we integrate by parts to separate out a pure boundary term
\label{eqn:lorentzForceCovariant:200}
\delta S =
\int d\tau \epsilon^\mu
\lr{
\PD{x^\mu}{L} – \frac{d}{d\tau} \PD{\dot{x}^\mu}{L}
}
+
\int d\tau \frac{d}{d\tau} \lr{
\epsilon^\mu \PD{\dot{x}^\mu}{L}
}.

The boundary term is killed since $$\epsilon^\mu = 0$$ at the end points of the action integral. We conclude that extremization of the action ($$\delta S = 0$$, for all $$\epsilon^\mu$$) requires
\label{eqn:lorentzForceCovariant:220}
\PD{x^\mu}{L} – \frac{d}{d\tau} \PD{\dot{x}^\mu}{L} = 0.

## Theorem 1.3: Lorentz force.

The relativistic Lagrangian for a charged particle is
\label{eqn:lorentzForceCovariant:1640}
L = \inv{2} m v^2 + q A \cdot v/c.

Application of the Euler-Lagrange equations to this Lagrangian yields the Lorentz-force equation
\label{eqn:lorentzForceCovariant:1660}
\frac{dp}{d\tau} = q F \cdot v/c,

where $$p = m v$$ is the proper momentum, $$F$$ is the Faraday bivector $$F = \grad \wedge A$$, and $$c$$ is the speed of light.

### Start proof:

To make life easier, let’s take advantage of the linearity of the Lagrangian, and break it into the free particle Lagrangian $$L_0 = (1/2) m v^2$$ and a potential term $$L_1 = q A \cdot v/c$$. For the free particle case we have
\label{eqn:lorentzForceCovariant:240}
\begin{aligned}
\delta L_0
&= – \frac{d}{d\tau} (m v) \\
&= – \frac{dp}{d\tau}.
\end{aligned}

For the potential contribution we have
\label{eqn:lorentzForceCovariant:260}
\begin{aligned}
\delta L_1
&= \frac{q}{c} \lr{ \grad (A \cdot v) – \frac{d}{d\tau} \lr{ \grad_v (A \cdot v)} } \\
&= \frac{q}{c} \lr{ \grad (A \cdot v) – \frac{dA}{d\tau} }.
\end{aligned}

The proper time derivative can be evaluated using the chain rule
\label{eqn:lorentzForceCovariant:280}
\frac{dA}{d\tau}
=
\frac{\partial x^\mu}{\partial \tau} \partial_\mu A

Putting all the pieces back together we have
\label{eqn:lorentzForceCovariant:300}
\begin{aligned}
0
&= \delta L \\
&=
-\frac{dp}{d\tau} + \frac{q}{c} \lr{ \grad (A \cdot v) – (v \cdot \grad) A } \\
&=
-\frac{dp}{d\tau} + \frac{q}{c} \lr{ \grad \wedge A } \cdot v.
\end{aligned}

## Problem: Gradient of a squared position vector.

Show that
\begin{equation*}
\grad (a \cdot x) = a,
\end{equation*}
and
\begin{equation*}
\end{equation*}
It should be clear that the same ideas can be used for the velocity gradient, where we obtain $$\grad_v (v^2) = 2 v$$, and $$\grad_v (A \cdot v) = A$$, as used in the derivation above.

The first identity follows easily by expansion in coordinates
\label{eqn:lorentzForceCovariant:320}
\begin{aligned}
&=
\gamma^\mu \partial_\mu a_\alpha x^\alpha \\
&=
\gamma^\mu a_\alpha \delta_\mu^\alpha \\
&=
\gamma^\mu a_\mu \\
&=
a.
\end{aligned}

The second identity follows by linearity of the gradient
\label{eqn:lorentzForceCovariant:340}
\begin{aligned}
&=
&=
\evalbar{\lr{\grad (x \cdot a)}}{a = x}
+
\evalbar{\lr{\grad (b \cdot x)}}{b = x} \\
&=
\evalbar{a}{a = x}
+
\evalbar{b}{b = x} \\
&=
2x.
\end{aligned}

It is desirable to put this relativistic Lorentz force equation into the usual vector and tensor forms for comparison.

## Theorem 1.4: Tensor form of the Lorentz force equation.

The tensor form of the Lorentz force equation is
\label{eqn:lorentzForceCovariant:1620}
\frac{dp^\mu}{d\tau} = \frac{q}{c} F^{\mu\nu} v_\nu,

where the antisymmetric Faraday tensor is defined as $$F^{\mu\nu} = \partial^\mu A^\nu – \partial^\nu A^\mu$$.

### Start proof:

We have only to dot both sides with $$\gamma^\mu$$. On the left we have
\label{eqn:lorentzForceCovariant:380}
\gamma^\mu \cdot \frac{dp}{d\tau}
=
\frac{dp^\mu}{d\tau}.

On the right, we have
\label{eqn:lorentzForceCovariant:400}
\begin{aligned}
\gamma^\mu \cdot \lr{ \frac{q}{c} F \cdot v }
&=
\frac{q}{c} (( \grad \wedge A ) \cdot v ) \cdot \gamma^\mu \\
&=
\frac{q}{c} ( \grad ( A \cdot v ) – (v \cdot \grad) A ) \cdot \gamma^\mu \\
&=
\frac{q}{c} \lr{ (\partial^\mu A^\nu) v_\nu – v_\nu \partial^\nu A^\mu } \\
&=
\frac{q}{c} F^{\mu\nu} v_\nu.
\end{aligned}

## Problem: Tensor expansion of $$F$$.

An alternate way to demonstrate \ref{eqn:lorentzForceCovariant:1620} is to first expand $$F = \grad \wedge A$$ in terms of coordinates, an expansion that can be expressed in terms of a second rank tensor antisymmetric tensor $$F^{\mu\nu}$$. Find that expansion, and re-evaluate the dot products of \ref{eqn:lorentzForceCovariant:400} using that.

\label{eqn:lorentzForceCovariant:900}
\begin{aligned}
F &=
&=
\lr{ \gamma_\mu \partial^\mu } \wedge \lr{ \gamma_\nu A^\nu } \\
&=
\lr{ \gamma_\mu \wedge \gamma_\nu } \partial^\mu A^\nu.
\end{aligned}

To this we can use the usual tensor trick (add self to self, change indexes, and divide by two), to give
\label{eqn:lorentzForceCovariant:920}
\begin{aligned}
F &=
\inv{2} \lr{
\lr{ \gamma_\mu \wedge \gamma_\nu } \partial^\mu A^\nu
+
\lr{ \gamma_\nu \wedge \gamma_\mu } \partial^\nu A^\mu
} \\
&=
\inv{2}
\lr{ \gamma_\mu \wedge \gamma_\nu } \lr{
\partial^\mu A^\nu

\partial^\nu A^\mu
},
\end{aligned}

which is just
\label{eqn:lorentzForceCovariant:940}
F =
\inv{2} \lr{ \gamma_\mu \wedge \gamma_\nu } F^{\mu\nu}.

Now, let’s expand $$(F \cdot v) \cdot \gamma^\mu$$ to compare to the earlier expansion in terms of $$\grad$$ and $$A$$.
\label{eqn:lorentzForceCovariant:960}
\begin{aligned}
(F \cdot v) \cdot \gamma^\mu
&=
\inv{2}
F^{\alpha\nu}
\lr{ \lr{ \gamma_\alpha \wedge \gamma_\nu } \cdot \lr{ \gamma^\beta v_\beta } } \cdot \gamma^\mu \\
&=
\inv{2}
F^{\alpha\nu} v_\beta
\lr{
{\delta_\nu}^\beta {\gamma_\alpha}^\mu

{\delta_\alpha}^\beta {\gamma_\nu}^\mu
} \\
&=
\inv{2}
\lr{
F^{\mu\beta} v_\beta

F^{\beta\mu} v_\beta
} \\
&=
F^{\mu\nu} v_\nu.
\end{aligned}

This alternate expansion illustrates some of the connectivity between the geometric algebra approach and the traditional tensor formalism.

## Problem: Lorentz force direct tensor derivation.

Instead of using the geometric algebra form of the Lorentz force equation as a stepping stone, we may derive the tensor form from the Lagrangian directly, provided the Lagrangian is put into tensor form
\begin{equation*}
L = \inv{2} m v^\mu v_\mu + q A^\mu v_\mu /c.
\end{equation*}
Evaluate the Euler-Lagrange equations in coordinate form and compare to \ref{eqn:lorentzForceCovariant:1620}.

Let $$\delta_\mu L = \gamma_\mu \cdot \delta L$$, so that we can write the Euler-Lagrange equations as
\label{eqn:lorentzForceCovariant:460}
0 = \delta_\mu L = \PD{x^\mu}{L} – \frac{d}{d\tau} \PD{\dot{x}^\mu}{L}.

Operating on the kinetic term of the Lagrangian, we have
\label{eqn:lorentzForceCovariant:480}
\delta_\mu L_0 = – \frac{d}{d\tau} m v_\mu.

For the potential term
\label{eqn:lorentzForceCovariant:500}
\begin{aligned}
\delta_\mu L_1
&=
\frac{q}{c} \lr{
v_\nu \PD{x^\mu}{A^\nu} – \frac{d}{d\tau} A_\mu
} \\
&=
\frac{q}{c} \lr{
v_\nu \PD{x^\mu}{A^\nu} – \frac{dx_\alpha}{d\tau} \PD{x_\alpha}{ A_\mu }
} \\
&=
\frac{q}{c} v^\nu \lr{
\partial_\mu A_\nu – \partial_\nu A_\mu
} \\
&=
\frac{q}{c} v^\nu F_{\mu\nu}.
\end{aligned}

Putting the pieces together gives
\label{eqn:lorentzForceCovariant:520}
\frac{d}{d\tau} (m v_\mu) = \frac{q}{c} v^\nu F_{\mu\nu},

which is identical\footnote{Some minor index raising and lowering gymnastics are required.} to the tensor form that we found by expanding the geometric algebra form of Maxwell’s equation in coordinates.

## Theorem 1.5: Vector Lorentz force equation.

Relative to a fixed observer’s frame, the Lorentz force equation of \ref{eqn:lorentzForceCovariant:1660} splits into a spatial rate of change of momentum, and (timelike component) rate of change of energy, as follows
\label{eqn:lorentzForceCovariant:1680}
\begin{aligned}
\ddt{(\gamma m \Bv)} &= q \lr{ \BE + \Bv \cross \BB } \\
\ddt{(\gamma m c^2)} &= q \Bv \cdot \BE,
\end{aligned}

where $$F = \BE + I c \BB$$, $$\gamma = 1/\sqrt{1 – \Bv^2/c^2 }$$.

### Start proof:

The first step is to eliminate the proper time dependencies in the Lorentz force equation. Consider first the coordinate representation of an arbitrary position four-vector $$x$$
\label{eqn:lorentzForceCovariant:1140}
x = c t \gamma_0 + x^k \gamma_k.

The corresponding four-vector velocity is
\label{eqn:lorentzForceCovariant:1160}
v = \ddtau{x} = c \ddtau{t} \gamma_0 + \ddtau{t} \ddt{x^k} \gamma_k.

By construction, $$v^2 = c^2$$ is a Lorentz invariant quantity (this is one of the relativistic postulates), so the LHS of \ref{eqn:lorentzForceCovariant:1160} must have the same square. That is
\label{eqn:lorentzForceCovariant:1240}
c^2 = \lr{ \ddtau{t} }^2 \lr{ c^2 – \Bv^2 },

where $$\Bv = v \wedge \gamma_0$$. This shows that we may make the identification
\label{eqn:lorentzForceCovariant:1260}
\gamma = \ddtau{t} = \inv{1 – \Bv^2/c^2 },

and
\label{eqn:lorentzForceCovariant:1280}
\ddtau{} = \ddtau{t} \ddt{} = \gamma \ddt{}.

We may now factor the four-velocity $$v$$ into its spacetime split
\label{eqn:lorentzForceCovariant:1300}
v = \gamma \lr{ c + \Bv } \gamma_0.

In particular the LHS of the Lorentz force equation can be rewritten as
\label{eqn:lorentzForceCovariant:1320}
\ddtau{p} = \gamma \ddt{}\lr{ \gamma \lr{ c + \Bv } } \gamma_0,

and the RHS of the Lorentz force equation can be rewritten as
\label{eqn:lorentzForceCovariant:1340}
\frac{q}{c} F \cdot v
=
\frac{\gamma q}{c} F \cdot \lr{ (c + \Bv) \gamma_0 }.

Equating timelike and spacelike components leaves us
\label{eqn:lorentzForceCovariant:1380}
\ddt{ (m \gamma c) } = \frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \cdot \gamma_0,

\label{eqn:lorentzForceCovariant:1400}
\ddt{ (m \gamma \Bv) } = \frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \wedge \gamma_0,

Evaluating these products requires some care, but is an essentially manual process. The reader is encouraged to do so once, but the end result may also be obtained easily using software (see lorentzForce.nb in [2]). One finds
\label{eqn:lorentzForceCovariant:1440}
F = \BE + I c \BB
=
E^1 \gamma_{10} +
+ E^2 \gamma_{20} +
+ E^3 \gamma_{30} +
– c B^1 \gamma_{23} +
– c B^2 \gamma_{31} +
– c B^3 \gamma_{12},

\label{eqn:lorentzForceCovariant:1460}
\frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \cdot \gamma_0
= \frac{q}{c} \BE \cdot \Bv,

\label{eqn:lorentzForceCovariant:1480}
\frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \wedge \gamma_0
= q \lr{ \BE + \Bv \cross \BB }.

## Problem: Algebraic spacetime split of the Lorentz force equation.

Derive the results of \ref{eqn:lorentzForceCovariant:1440} through \ref{eqn:lorentzForceCovariant:1480} algebraically.

## Problem: Spacetime split of the Lorentz force tensor equation.

Show that \ref{eqn:lorentzForceCovariant:1680} also follows from the tensor form of the Lorentz force equation (\ref{eqn:lorentzForceCovariant:1620}) provided we identify
\label{eqn:lorentzForceCovariant:1500}
F^{k0} = E^k,

and
\label{eqn:lorentzForceCovariant:1520}
F^{rs} = -\epsilon^{rst} B^t.

Also verify that the identifications of \ref{eqn:lorentzForceCovariant:1500} and \ref{eqn:lorentzForceCovariant:1520} is consistent with the geometric algebra Faraday bivector $$F = \BE + I c \BB$$, and the associated coordinate expansion of the field $$F = (1/2) (\gamma_\mu \wedge \gamma_\nu) F^{\mu\nu}$$.

# References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] Peeter Joot. Mathematica modules for Geometric Algebra’s GA(2,0), GA(3,0), and GA(1,3), 2017. URL https://github.com/peeterjoot/gapauli. [Online; accessed 24-Oct-2020].

## Explicit form of the square root of p . sigma.

December 10, 2018 phy2403 No comments , , , ,

With the help of Mathematica, a fairly compact form was found for the root of $$p \cdot \sigma$$
\label{eqn:DiracUVmatricesExplicit:121}
\sqrt{ p \cdot \sigma }
=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
\omega_\Bp- p^3 + \sqrt{ \omega_\Bp^2 – \Bp^2 } & – p^1 + i p^2 \\
– p^1 – i p^2 & \omega_\Bp+ p^3 + \sqrt{ \omega_\Bp^2 – \Bp^2 }
\end{bmatrix}.

A bit of examination shows that we can do much better. The leading scalar term can be simplified by squaring it
\label{eqn:squarerootpsigma:140}
\begin{aligned}
\lr{ \sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} } }^2
&=
\omega_\Bp- \Norm{\Bp} + \omega_\Bp+ \Norm{\Bp} + 2 \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
&=
2 \omega_\Bp + 2 m,
\end{aligned}

where the on-shell value of the energy $$\omega_\Bp^2 = m^2 + \Bp^2$$ has been inserted. Using that again in the matrix, we have
\label{eqn:squarerootpsigma:160}
\begin{aligned}
\sqrt{ p \cdot \sigma }
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\begin{bmatrix}
\omega_\Bp- p^3 + m & – p^1 + i p^2 \\
– p^1 – i p^2 & \omega_\Bp+ p^3 + m
\end{bmatrix} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0
-p^1 \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
-p^2 \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
-p^3 \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}
} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0
-p^1 \sigma^1
-p^2 \sigma^2
-p^3 \sigma^3
} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0 – \Bsigma \cdot \Bp
}.
\end{aligned}

We’ve now found a nice algebraic form for these matrix roots
\label{eqn:squarerootpsigma:180}
\boxed{
\begin{aligned}
\sqrt{p \cdot \sigma} &= \inv{\sqrt{ 2 \omega_\Bp + 2 m }} \lr{ m + p \cdot \sigma } \\
\sqrt{p \cdot \overline{\sigma}} &= \inv{\sqrt{ 2 \omega_\Bp + 2 m }} \lr{ m + p \cdot \overline{\sigma}}.
\end{aligned}}

As a check, let’s square one of these explicitly
\label{eqn:squarerootpsigma:101}
\begin{aligned}
\lr{ \sqrt{p \cdot \sigma} }^2
&= \inv{2 \omega_\Bp + 2 m }
\lr{ m^2 + (p \cdot \sigma)^2 + 2 m (p \cdot \sigma) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ m^2 + (\omega_\Bp^2 – 2 \omega_\Bp \Bsigma \cdot \Bp + \Bp^2) + 2 m (p \cdot \sigma) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ 2 \omega_\Bp^2 – 2 \omega_\Bp \Bsigma \cdot \Bp + 2 m (\omega_\Bp – \Bsigma \cdot \Bp) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ 2 \omega_\Bp \lr{ \omega_\Bp + m } – (2 \omega_\Bp + 2 m) \Bsigma \cdot \Bp } \\
&=
\omega_\Bp – \Bsigma \cdot \Bp \\
&=
p \cdot \sigma,
\end{aligned}

which validates the result.

## Explicit expansion of the Dirac u,v matrices

We found that the solution of the $$u(p), v(p)$$ matrices were
\label{eqn:DiracUVmatricesExplicit:20}
\begin{aligned}
u(p) &=
\begin{bmatrix}
\sqrt{p \cdot \sigma} \zeta \\
\sqrt{p \cdot \overline{\sigma}} \zeta \\
\end{bmatrix} \\
v(p) &=
\begin{bmatrix}
\sqrt{p \cdot \sigma} \eta \\
-\sqrt{p \cdot \overline{\sigma}} \eta \\
\end{bmatrix},
\end{aligned}

where
\label{eqn:DiracUVmatricesExplicit:40}
\begin{aligned}
p \cdot \sigma &= p_0 \sigma_0 – \Bsigma \cdot \Bp \\
p \cdot \overline{\sigma} &= p_0 \sigma_0 + \Bsigma \cdot \Bp.
\end{aligned}

It was pointed out that these square roots can be conceptualized as (in the right basis) as the diagonal matrices of the eigenvalue square roots.

It was also pointed out that we don’t tend to need the explicit form of these square roots.We saw that to be the case in all our calculations, where these always showed up in the end in quadratic combinations like $$\sqrt{ (p \cdot \sigma)^2 }, \sqrt{ (p \cdot \sigma)(p \cdot \overline{\sigma})}, \cdots$$, which nicely reduced each time without requiring the matrix roots.

I encountered a case where it would have been nice to have the explicit representation. In particular, I wanted to use Mathematica to symbolically expand $$\overline{\Psi} i \gamma^\mu \partial_\mu \Psi$$ in terms of $$a^s_\Bp, b^r_\Bp, \cdots$$ representation, to verify that the massless Dirac Lagrangian are in fact the energy and momentum operators (and to compare to the explicit form of the momentum operator found in eq. 3.105 [1]). For that mechanical task, I needed explicit representations of all the $$u^s(p), v^r(p)$$ matrices to plug in.

It happens that $$2 \times 2$$ matrices can be square-rooted symbolically (FIXME: link to squarerootOfFourSigmaDotP.nb notebook). In particular, the matrices $$p \cdot \sigma, p \cdot \overline{\sigma}$$ have nice simple eigenvalues $$\pm \Norm{\Bp} + \omega_\Bp$$. The corresponding unnormalized eigenvectors for $$p \cdot \sigma$$ are
\label{eqn:DiracUVmatricesExplicit:60}
\begin{aligned}
e_1 &=
\begin{bmatrix}
– p_x + i p_y \\
p_z + \Norm{\Bp}
\end{bmatrix} \\
e_1 &=
\begin{bmatrix}
– p_x + i p_y \\
p_z – \Norm{\Bp}
\end{bmatrix}.
\end{aligned}

This means that we can diagonalize $$p \cdot \sigma$$ as
\label{eqn:DiracUVmatricesExplicit:80}
p \cdot \sigma
= U
\begin{bmatrix}
\omega_\Bp+ \Norm{\Bp} & 0 \\
0 & \omega_\Bp- \Norm{\Bp}
\end{bmatrix}
U^\dagger,

where $$U$$ is the matrix of the normalized eigenvectors
\label{eqn:DiracUVmatricesExplicit:100}
U =
\begin{bmatrix}
e_1′ & e_2′
\end{bmatrix}
=
\inv{ \sqrt{ 2 \Bp^2 + 2 p_z \Norm{\Bp} } }
\begin{bmatrix}
-p_x + i p_y & -p_x + i p_y \\
p_z + \Norm{\Bp} & p_z – \Norm{\Bp}
\end{bmatrix}.

Letting Mathematica churn through the matrix products \ref{eqn:DiracUVmatricesExplicit:80} verifies the diagonalization, and for the roots, we find
\label{eqn:DiracUVmatricesExplicit:120}
\sqrt{ p \cdot \sigma }
=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
\omega_\Bp- p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } & – p_x + i p_y \\
– p_x – i p_y & \omega_\Bp+ p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 }
\end{bmatrix}.

Now we can plug in $$\zeta^{1\T} = (1,0), \zeta^{2\T} = (0,1), \eta^{1\T} = (1,0), \eta^{2\T} = (0,1)$$ to find the explicit form of our $$u$$’s and $$v$$’s
\label{eqn:DiracUVmatricesExplicit:140}
\begin{aligned}
u^1(p) &=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
\omega_\Bp- p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
– p_x – i p_y \\
\omega_\Bp+ p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
p_x + i p_y \\
\end{bmatrix} \\
u^2(p) &=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
– p_x + i p_y \\
\omega_\Bp+ p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
p_x – i p_y \\
\omega_\Bp- p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
\end{bmatrix} \\
v^1(p) &=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
\omega_\Bp- p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
– p_x – i p_y \\
-\omega_\Bp- p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
-p_x – i p_y \\
\end{bmatrix} \\
v^2(p) &=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
– p_x + i p_y \\
\omega_\Bp+ p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
-p_x + i p_y \\
-\omega_\Bp+ p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
\end{bmatrix}.
\end{aligned}

This is now a convenient form to try the next symbolic manipulation task. If nothing else this takes some of the mystery out of the original compact notation, since we see that the $$u,v$$’s are just $$4$$ element column vectors, and we know their explicit should we want them.

Also note that in class we made a note that we should take the positive roots of the eigenvalue diagonal matrix. It doesn’t look like that is really required. We need not even use the same sign for each root. Squaring the resulting matrix root in the end will recover the original $$p \cdot \sigma$$ matrix.

# References

[1] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

## Reflection using Pauli matrices.

In class yesterday (lecture 19, notes not yet posted) we used $$\Bsigma^\T = -\sigma_2 \Bsigma \sigma_2$$, which implicitly shows that $$(\Bsigma \cdot \Bx)^\T$$ is a reflection about the y-axis.
This form of reflection will be familiar to a student of geometric algebra (see [1] — a great book, one copy of which is in the physics library). I can’t recall any mention of the geometrical reflection identity from when I took QM. It’s a fun exercise to demonstrate the reflection identity when constrained to the Pauli matrix notation.

## Theorem: Reflection about a normal.

Given a unit vector $$\ncap \in \mathbb{R}^3$$ and a vector $$\Bx \in \mathbb{R}^3$$ the reflection of $$\Bx$$ about a plane with normal $$\ncap$$ can be represented in Pauli notation as
\begin{equation*}
-\Bsigma \cdot \ncap \Bsigma \cdot \Bx \Bsigma \cdot \ncap.
\end{equation*}

To prove this, first note that in standard vector notation, we can decompose a vector into its projective and rejective components
\label{eqn:reflection:20}
\Bx = (\Bx \cdot \ncap) \ncap + \lr{ \Bx – (\Bx \cdot \ncap) \ncap }.

A reflection about the plane normal to $$\ncap$$ just flips the component in the direction of $$\ncap$$, leaving the rest unchanged. That is
\label{eqn:reflection:40}
-(\Bx \cdot \ncap) \ncap + \lr{ \Bx – (\Bx \cdot \ncap) \ncap }
=
\Bx – 2 (\Bx \cdot \ncap) \ncap.

We may write this in $$\Bsigma$$ notation as
\label{eqn:reflection:60}
\Bsigma \cdot \Bx – 2 \Bx \cdot \ncap \Bsigma \cdot \ncap.

We also know that
\label{eqn:reflection:80}
\begin{aligned}
\Bsigma \cdot \Ba \Bsigma \cdot \Bb &= a \cdot b + i \Bsigma \cdot (\Ba \cross \Bb) \\
\Bsigma \cdot \Bb \Bsigma \cdot \Ba &= a \cdot b – i \Bsigma \cdot (\Ba \cross \Bb),
\end{aligned}

or
\label{eqn:reflection:100}
a \cdot b = \inv{2} \symmetric{\Bsigma \cdot \Ba}{\Bsigma \cdot \Bb},

where $$\symmetric{\Ba}{\Bb}$$ is the anticommutator of $$\Ba, \Bb$$.
Inserting \ref{eqn:reflection:100} into \ref{eqn:reflection:60} we find that the reflection is
\label{eqn:reflection:120}
\begin{aligned}
\Bsigma \cdot \Bx –
\symmetric{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap
&=
\Bsigma \cdot \Bx –
{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap

{\Bsigma \cdot \Bx}{\Bsigma \cdot \ncap}
\Bsigma \cdot \ncap \\
&=
\Bsigma \cdot \Bx –
{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap

{\Bsigma \cdot \Bx} \\
&=

{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap,
\end{aligned}

which completes the proof.

When we expand $$(\Bsigma \cdot \Bx)^\T$$ and find
\label{eqn:reflection:n}
(\Bsigma \cdot \Bx)^\T
=
\sigma^1 x^1 – \sigma^2 x^2 + \sigma^3 x^3,

it is clear that this coordinate expansion is a reflection about the y-axis. Knowing the reflection formula above provides a rationale for why we might want to write this in the compact form $$-\sigma^2 (\Bsigma \cdot \Bx) \sigma^2$$, which might not be obvious otherwise.

# References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

## Plane wave and spinor under time reversal

### Q: [1] pr 4.7

1. (a)
Find the time reversed form of a spinless plane wave state in three dimensions.

2. (b)
For the eigenspinor of $$\Bsigma \cdot \ncap$$ expressed in terms of polar and azimuthal angles $$\beta$$ and $$\gamma$$, show that $$-i \sigma_y \chi^\conj(\ncap)$$ has the reversed spin direction.

### A: part (a)

The Hamiltonian for a plane wave is

\label{eqn:timeReversalPlaneWaveAndSpinor:20}
H = \frac{\Bp^2}{2m} = i \PD{t}.

Under time reversal the momentum side transforms as

\label{eqn:timeReversalPlaneWaveAndSpinor:40}
\begin{aligned}
\Theta \frac{\Bp^2}{2m} \Theta^{-1}
&=
\frac{\lr{ \Theta \Bp \Theta^{-1}} \cdot \lr{ \Theta \Bp \Theta^{-1}} }{2m} \\
&=
\frac{(-\Bp) \cdot (-\Bp)}{2m} \\
&=
\frac{\Bp^2}{2m}.
\end{aligned}

The time derivative side of the equation is also time reversal invariant
\label{eqn:timeReversalPlaneWaveAndSpinor:60}
\begin{aligned}
\Theta i \PD{t}{} \Theta^{-1}
&=
\Theta i \Theta^{-1} \Theta \PD{t}{} \Theta^{-1} \\
&=
-i \PD{(-t)}{} \\
&=
i \PD{t}{}.
\end{aligned}

Solutions to this equation are linear combinations of

\label{eqn:timeReversalPlaneWaveAndSpinor:80}
\psi(\Bx, t) = e^{i \Bk \cdot \Bx – i E t/\Hbar},

where $$\Hbar^2 \Bk^2/2m = E$$, the energy of the particle. Under time reversal we have

\label{eqn:timeReversalPlaneWaveAndSpinor:100}
\begin{aligned}
\psi(\Bx, t)
\rightarrow e^{-i \Bk \cdot \Bx + i E (-t)/\Hbar}
&= \lr{ e^{i \Bk \cdot \Bx – i E (-t)/\Hbar} }^\conj \\
&=
\psi^\conj(\Bx, -t)
\end{aligned}

### A: part (b)

The text uses a requirement for time reversal of spin states to show that the Pauli matrix form of the time reversal operator is

\label{eqn:timeReversalPlaneWaveAndSpinor:120}
\Theta = -i \sigma_y K,

where $$K$$ is a complex conjugating operator. The form of the spin up state used in that demonstration was

\label{eqn:timeReversalPlaneWaveAndSpinor:140}
\begin{aligned}
\ket{\ncap ; +}
&= e^{-i S_z \beta/\Hbar} e^{-i S_y \gamma/\Hbar} \ket{+} \\
&= e^{-i \sigma_z \beta/2} e^{-i \sigma_y \gamma/2} \ket{+} \\
&= \lr{ \cos(\beta/2) – i \sigma_z \sin(\beta/2) }
\lr{ \cos(\gamma/2) – i \sigma_y \sin(\gamma/2) } \ket{+} \\
&= \lr{ \cos(\beta/2) – i \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \sin(\beta/2) }
\lr{ \cos(\gamma/2) – i \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \sin(\gamma/2) } \ket{+} \\
&=
\begin{bmatrix}
e^{-i\beta/2} & 0 \\
0 & e^{i \beta/2}
\end{bmatrix}
\begin{bmatrix}
\cos(\gamma/2) & -\sin(\gamma/2) \\
\sin(\gamma/2) & \cos(\gamma/2)
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\begin{bmatrix}
e^{-i\beta/2} & 0 \\
0 & e^{i \beta/2}
\end{bmatrix}
\begin{bmatrix}
\cos(\gamma/2) \\
\sin(\gamma/2) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\gamma/2)
e^{-i\beta/2}
\\
\sin(\gamma/2)
e^{i \beta/2}
\end{bmatrix}.
\end{aligned}

The state orthogonal to this one is claimed to be

\label{eqn:timeReversalPlaneWaveAndSpinor:180}
\begin{aligned}
\ket{\ncap ; -}
&= e^{-i S_z \beta/\Hbar} e^{-i S_y (\gamma + \pi)/\Hbar} \ket{+} \\
&= e^{-i \sigma_z \beta/2} e^{-i \sigma_y (\gamma + \pi)/2} \ket{+}.
\end{aligned}

We have

\label{eqn:timeReversalPlaneWaveAndSpinor:200}
\begin{aligned}
\cos((\gamma + \pi)/2)
&=
\textrm{Re} e^{i(\gamma + \pi)/2} \\
&=
\textrm{Re} i e^{i\gamma/2} \\
&=
-\sin(\gamma/2),
\end{aligned}

and
\label{eqn:timeReversalPlaneWaveAndSpinor:220}
\begin{aligned}
\sin((\gamma + \pi)/2)
&=
\textrm{Im} e^{i(\gamma + \pi)/2} \\
&=
\textrm{Im} i e^{i\gamma/2} \\
&=
\cos(\gamma/2),
\end{aligned}

so we should have

\label{eqn:timeReversalPlaneWaveAndSpinor:240}
\ket{\ncap ; -}
=
\begin{bmatrix}
-\sin(\gamma/2)
e^{-i\beta/2}
\\
\cos(\gamma/2)
e^{i \beta/2}
\end{bmatrix}.

This looks right, but we can sanity check orthogonality

\label{eqn:timeReversalPlaneWaveAndSpinor:260}
\begin{aligned}
\braket{\ncap ; -}{\ncap ; +}
&=
\begin{bmatrix}
-\sin(\gamma/2)
e^{i\beta/2}
&
\cos(\gamma/2)
e^{-i \beta/2}
\end{bmatrix}
\begin{bmatrix}
\cos(\gamma/2)
e^{-i\beta/2}
\\
\sin(\gamma/2)
e^{i \beta/2}
\end{bmatrix} \\
&=
0,
\end{aligned}

as expected.

The task at hand appears to be the operation on the column representation of $$\ket{\ncap; +}$$ using the Pauli representation of the time reversal operator. That is

\label{eqn:timeReversalPlaneWaveAndSpinor:160}
\begin{aligned}
\Theta \ket{\ncap ; +}
&=
-i \sigma_y K
\begin{bmatrix}
e^{-i\beta/2} \cos(\gamma/2) \\
e^{i \beta/2} \sin(\gamma/2)
\end{bmatrix} \\
&=
-i \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
e^{i\beta/2} \cos(\gamma/2) \\
e^{-i \beta/2} \sin(\gamma/2)
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
e^{i\beta/2} \cos(\gamma/2) \\
e^{-i \beta/2} \sin(\gamma/2)
\end{bmatrix} \\
&=
\begin{bmatrix}
-e^{-i \beta/2} \sin(\gamma/2) \\
e^{i\beta/2} \cos(\gamma/2) \\
\end{bmatrix} \\
&= \ket{\ncap ; -},
\end{aligned}

which is the result to be demononstrated.

# References

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

## Determining the rotation angle and normal for a rotation through Euler angles

November 2, 2015 phy1520 No comments , , ,

[1] pr. 3.9 poses the problem to determine the total rotation angle for a set of Euler rotations given by

\label{eqn:eulerAngleRotationAngleAndNormal:20}
\mathcal{D}^{1/2}(\alpha, \beta, \gamma)
=
\begin{bmatrix}
e^{-i(\alpha+\gamma)/2} \cos \frac{\beta}{2} & -e^{-i(\alpha-\gamma)/2} \sin \frac{\beta}{2} \\
e^{i(\alpha-\gamma)/2} \sin \frac{\beta}{2} & e^{i(\alpha+\gamma)/2} \cos \frac{\beta}{2}
\end{bmatrix}.

Compare this to the matrix for a rotation (again double sided) about a normal, given by

\label{eqn:eulerAngleRotationAngleAndNormal:40}
\mathcal{R}
= e^{-i \Bsigma \cdot \ncap \theta/2}
= \cos \frac{\theta}{2} I – i \Bsigma \cdot \ncap \sin \frac{\theta}{2}.

With $$\ncap = \lr{ \sin \Theta \cos\Phi, \sin \Theta \sin\Phi, \cos\Theta}$$, the normal direction in its Pauli basis is

\label{eqn:eulerAngleRotationAngleAndNormal:60}
\Bsigma \cdot \ncap
=
\begin{bmatrix}
\cos\Theta & \sin \Theta \cos\Phi – i \sin \Theta \sin\Phi \\
\sin \Theta \cos\Phi + i \sin \Theta \sin\Phi & -\cos\Theta
\end{bmatrix}
=
\begin{bmatrix}
\cos\Theta & \sin \Theta e^{-i \Phi} \\
\sin \Theta e^{i \Phi} & -\cos\Theta
\end{bmatrix},

so

\label{eqn:eulerAngleRotationAngleAndNormal:80}
\mathcal{R} =
\begin{bmatrix}
\cos \frac{\theta}{2} -i \sin \frac{\theta}{2} \cos\Theta & -i \sin \Theta e^{-i \Phi} \sin \frac{\theta}{2} \\
-i \sin \Theta e^{i \Phi} \sin \frac{\theta}{2} & \cos \frac{\theta}{2} +i \sin \frac{\theta}{2} \cos\Theta \\
\end{bmatrix}.

It’s not obvious how to put this into correspondence with the matrix for the Euler rotations. Doing so certainly doesn’t look fun. To solve this problem, let’s go the opposite direction, and put the matrix for the Euler rotations into the form of \ref{eqn:eulerAngleRotationAngleAndNormal:40}.

That is
\label{eqn:eulerAngleRotationAngleAndNormal:100}
\begin{aligned}
\mathcal{D}^{1/2}(\alpha, \beta, \gamma)
&=
\begin{bmatrix}
e^{-i(\alpha+\gamma)/2} \cos \frac{\beta}{2} & -e^{-i(\alpha-\gamma)/2} \sin \frac{\beta}{2} \\
e^{i(\alpha-\gamma)/2} \sin \frac{\beta}{2} & e^{i(\alpha+\gamma)/2} \cos \frac{\beta}{2}
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos\frac{\alpha+\gamma}{2} \cos \frac{\beta}{2} & – \cos\frac{\alpha-\gamma}{2} \sin \frac{\beta}{2} \\
\cos\frac{\alpha-\gamma}{2} \sin \frac{\beta}{2} & \cos\frac{\alpha+\gamma}{2} \cos \frac{\beta}{2}
\end{bmatrix} \\
i
\begin{bmatrix}
– \sin\frac{\alpha+\gamma}{2} \cos \frac{\beta}{2} & \sin\frac{\alpha-\gamma}{2} \sin \frac{\beta}{2} \\
\sin\frac{\alpha-\gamma}{2} \sin \frac{\beta}{2} & \sin\frac{\alpha+\gamma}{2} \cos \frac{\beta}{2}
\end{bmatrix} \\
&=
\cos\frac{\alpha+\gamma}{2} \cos \frac{\beta}{2}
+ i \sin\frac{\alpha-\gamma}{2} \sin \frac{\beta}{2} \sigma_x
– i \cos\frac{\alpha-\gamma}{2} \sin \frac{\beta}{2} \sigma_y
– i \sin\frac{\alpha+\gamma}{2} \cos \frac{\beta}{2} \sigma_z
\end{aligned},

This gives us

\label{eqn:eulerAngleRotationAngleAndNormal:120}
\begin{aligned}
\cos\frac{\theta}{2} &= \cos\frac{\alpha+\gamma}{2} \cos \frac{\beta}{2} \\
\ncap \sin\frac{\theta}{2} &= \lr{ -\sin\frac{\alpha-\gamma}{2} \sin \frac{\beta}{2}, \cos\frac{\alpha-\gamma}{2} \sin \frac{\beta}{2}, \sin\frac{\alpha+\gamma}{2} \cos \frac{\beta}{2} }.
\end{aligned}

The angle is

\label{eqn:eulerAngleRotationAngleAndNormal:140}
\theta
= 2 \textrm{arctan} \frac{
\sqrt{\sin^2\frac{\beta}{2} + \sin^2\frac{\alpha+\gamma}{2} \cos^2\frac{\beta}{2}
}
}{\cos\frac{\alpha+\gamma}{2} \cos \frac{\beta}{2}},

or
\label{eqn:eulerAngleRotationAngleAndNormal:180}
\boxed{
\theta = 2 \textrm{arctan} \frac{
\sqrt{\tan^2\frac{\beta}{2} + \sin^2\frac{\alpha+\gamma}{2}
}
}{\cos\frac{\alpha+\gamma}{2}
},
}

and the normal direction is
\label{eqn:eulerAngleRotationAngleAndNormal:160}
\boxed{
\ncap
=
\inv{\sqrt{1 – \cos^2\frac{\alpha+\gamma}{2} \cos^2\frac{\beta}{2} }}
\lr{ -\sin\frac{\alpha-\gamma}{2} \sin \frac{\beta}{2}, \cos\frac{\alpha-\gamma}{2} \sin \frac{\beta}{2}, \sin\frac{\alpha+\gamma}{2} \cos \frac{\beta}{2} }.
}

# References

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

## (a)

Determine the matrix representation of $$\ket{\alpha}\bra{\beta}$$ given a complete set of eigenvectors $$\ket{a^r}$$.

## (b)

Verify with $$\ket{\alpha} = \ket{s_z = \Hbar/2}, \ket{s_x = \Hbar/2}$$.

## (a)

Forming the matrix element

\label{eqn:moreBraKetProblems:20}
\begin{aligned}
\bra{a^r} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^s}
&=
\braket{a^r}{\alpha}\braket{\beta}{a^s} \\
&=
\braket{a^r}{\alpha}
\braket{a^s}{\beta}^\conj,
\end{aligned}

the matrix representation is seen to be

\label{eqn:moreBraKetProblems:40}
\ket{\alpha}\bra{\beta}
\sim
\begin{bmatrix}
\bra{a^1} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^1} & \bra{a^1} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^2} & \cdots \\
\bra{a^2} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^1} & \bra{a^2} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^2} & \cdots \\
\vdots & \vdots & \ddots \\
\end{bmatrix}
=
\begin{bmatrix}
\braket{a^1}{\alpha} \braket{a^1}{\beta}^\conj & \braket{a^1}{\alpha} \braket{a^2}{\beta}^\conj & \cdots \\
\braket{a^2}{\alpha} \braket{a^1}{\beta}^\conj & \braket{a^2}{\alpha} \braket{a^2}{\beta}^\conj & \cdots \\
\vdots & \vdots & \ddots \\
\end{bmatrix}.

## (b)

First compute the spin-z representation of $$\ket{s_x = \Hbar/2 }$$.

\label{eqn:moreBraKetProblems:60}
\begin{aligned}
\lr{ S_x – \Hbar/2 I }
\begin{bmatrix}
a \\
b
\end{bmatrix}
&=
\lr{
\begin{bmatrix}
0 & \Hbar/2 \\
\Hbar/2 & 0 \\
\end{bmatrix}

\begin{bmatrix}
\Hbar/2 & 0 \\
0 & \Hbar/2 \\
\end{bmatrix}
} \\
&=
\begin{bmatrix}
a \\
b
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
-1 & 1 \\
1 & -1 \\
\end{bmatrix}
\begin{bmatrix}
a \\
b
\end{bmatrix},
\end{aligned}

so $$\ket{s_x = \Hbar/2 } \propto (1,1)$$.

Normalized we have

\label{eqn:moreBraKetProblems:80}
\begin{aligned}
\ket{\alpha} &= \ket{s_z = \Hbar/2 } =
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
\ket{\beta} &= \ket{s_z = \Hbar/2 }
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1
\end{bmatrix}.
\end{aligned}

Using \ref{eqn:moreBraKetProblems:40} the matrix representation is

\label{eqn:moreBraKetProblems:100}
\ket{\alpha}\bra{\beta}
\sim
\begin{bmatrix}
(1) (1/\sqrt{2})^\conj & (1) (1/\sqrt{2})^\conj \\
(0) (1/\sqrt{2})^\conj & (0) (1/\sqrt{2})^\conj \\
\end{bmatrix}
=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix}.

This can be confirmed with direct computation
\label{eqn:moreBraKetProblems:120}
\begin{aligned}
\ket{\alpha}\bra{\beta}
&=
\begin{bmatrix}
1 \\
0
\end{bmatrix}
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1
\end{bmatrix} \\
&=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix}.
\end{aligned}

## Question: eigenvalue of sum of kets ([1] pr. 1.6)

Given eigenkets $$\ket{i}, \ket{j}$$ of an operator $$A$$, what are the conditions that $$\ket{i} + \ket{j}$$ is also an eigenvector?

Let $$A \ket{i} = i \ket{i}, A \ket{j} = j \ket{j}$$, and suppose that the sum is an eigenket. Then there must be a value $$a$$ such that

\label{eqn:moreBraKetProblems:140}
A \lr{ \ket{i} + \ket{j} } = a \lr{ \ket{i} + \ket{j} },

so

\label{eqn:moreBraKetProblems:160}
i \ket{i} + j \ket{j} = a \lr{ \ket{i} + \ket{j} }.

Operating with $$\bra{i}, \bra{j}$$ respectively, gives

\label{eqn:moreBraKetProblems:180}
\begin{aligned}
i &= a \\
j &= a,
\end{aligned}

so for the sum to be an eigenket, both of the corresponding energy eigenvalues must be identical (i.e. linear combinations of degenerate eigenkets are also eigenkets).

## Question: Null operator ([1] pr. 1.7)

Given eigenkets $$\ket{a’}$$ of operator $$A$$

## (a)

show that

\label{eqn:moreBraKetProblems:200}
\prod_{a’} \lr{ A – a’ }

is the null operator.

## (b)

\label{eqn:moreBraKetProblems:220}
\prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”}

## (c)

Illustrate using $$S_z$$ for a spin 1/2 system.

## (a)

Application of $$\ket{a}$$, the eigenket of $$A$$ with eigenvalue $$a$$ to any term $$A – a’$$ scales $$\ket{a}$$ by $$a – a’$$, so the product operating on $$\ket{a}$$ is

\label{eqn:moreBraKetProblems:240}
\prod_{a’} \lr{ A – a’ } \ket{a} = \prod_{a’} \lr{ a – a’ } \ket{a}.

Since $$\ket{a}$$ is one of the $$\setlr{\ket{a’}}$$ eigenkets of $$A$$, one of these terms must be zero.

## (b)

Again, consider the action of the operator on $$\ket{a}$$,

\label{eqn:moreBraKetProblems:260}
\prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”} \ket{a}
=
\prod_{a” \ne a’} \frac{\lr{ a – a” }}{a’ – a”} \ket{a}.

If $$\ket{a} = \ket{a’}$$, then $$\prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”} \ket{a} = \ket{a}$$, whereas if it does not, then it equals one of the $$a”$$ energy eigenvalues. This is a representation of the Kronecker delta function

\label{eqn:moreBraKetProblems:300}
\prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”} \ket{a} \equiv \delta_{a’, a} \ket{a}

## (c)

For operator $$S_z$$ the eigenvalues are $$\setlr{ \Hbar/2, -\Hbar/2 }$$, so the null operator must be

\label{eqn:moreBraKetProblems:280}
\begin{aligned}
\prod_{a’} \lr{ A – a’ }
&=
\lr{ \frac{\Hbar}{2} }^2 \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} – \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \\
&=
\begin{bmatrix}
0 & 0 \\
0 & -2
\end{bmatrix}
\begin{bmatrix}
2 & 0 \\
0 & 0 \\
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 0 \\
0 & 0 \\
\end{bmatrix}
\end{aligned}

For the delta representation, consider the $$\ket{\pm}$$ states and their eigenvalue. The delta operators are

\label{eqn:moreBraKetProblems:320}
\begin{aligned}
\prod_{a” \ne \Hbar/2} \frac{\lr{ A – a” }}{\Hbar/2 – a”}
&=
\frac{S_z – (-\Hbar/2) I}{\Hbar/2 – (-\Hbar/2)} \\
&=
\inv{2} \lr{ \sigma_z + I } \\
&=
\inv{2} \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \\
&=
\inv{2}
\begin{bmatrix}
2 & 0 \\
0 & 0
\end{bmatrix}
\\
&=
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}.
\end{aligned}

\label{eqn:moreBraKetProblems:340}
\begin{aligned}
\prod_{a” \ne -\Hbar/2} \frac{\lr{ A – a” }}{-\Hbar/2 – a”}
&=
\frac{S_z – (\Hbar/2) I}{-\Hbar/2 – \Hbar/2} \\
&=
\inv{2} \lr{ \sigma_z – I } \\
&=
\inv{2} \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} – \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \\
&=
\inv{2}
\begin{bmatrix}
0 & 0 \\
0 & -2
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}.
\end{aligned}

These clearly have the expected delta function property acting on kets $$\ket{+} = (1,0), \ket{-} = (0, 1)$$.

## Question: Spin half general normal ([1] pr. 1.9)

Construct $$\ket{\BS \cdot \ncap ; + }$$, where $$\ncap = ( \cos\alpha \sin\beta, \sin\alpha \sin\beta, \cos\beta )$$ such that

\label{eqn:moreBraKetProblems:360}
\BS \cdot \ncap \ket{\BS \cdot \ncap ; + } =
\frac{\Hbar}{2} \ket{\BS \cdot \ncap ; + },

Solve this as an eigenvalue problem.

The spin operator for this direction is

\label{eqn:moreBraKetProblems:380}
\begin{aligned}
\BS \cdot \ncap
&= \frac{\Hbar}{2} \Bsigma \cdot \ncap \\
&= \frac{\Hbar}{2}
\lr{
\cos\alpha \sin\beta \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} + \sin\alpha \sin\beta \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} + \cos\beta \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}
} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\beta &
e^{-i\alpha}
\sin\beta
\\
e^{i\alpha}
\sin\beta
& -\cos\beta
\end{bmatrix}.
\end{aligned}

Observed that this is traceless and has a $$-\Hbar/2$$ determinant like any of the $$x,y,z$$ spin operators.

Assuming that this has an $$\Hbar/2$$ eigenvalue (to be verified later), the eigenvalue problem is

\label{eqn:moreBraKetProblems:400}
\begin{aligned}
0
&=
\BS \cdot \ncap – \Hbar/2 I \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\beta -1 &
e^{-i\alpha}
\sin\beta
\\
e^{i\alpha}
\sin\beta
& -\cos\beta -1
\end{bmatrix} \\
&=
\Hbar
\begin{bmatrix}
– \sin^2 \frac{\beta}{2} &
e^{-i\alpha}
\sin\frac{\beta}{2} \cos\frac{\beta}{2}
\\
e^{i\alpha}
\sin\frac{\beta}{2} \cos\frac{\beta}{2}
& -\cos^2 \frac{\beta}{2}
\end{bmatrix}
\end{aligned}

This has a zero determinant as expected, and the eigenvector $$(a,b)$$ will satisfy

\label{eqn:moreBraKetProblems:420}
\begin{aligned}
0
&= – \sin^2 \frac{\beta}{2} a +
e^{-i\alpha}
\sin\frac{\beta}{2} \cos\frac{\beta}{2}
b \\
&= \sin\frac{\beta}{2} \lr{ – \sin \frac{\beta}{2} a +
e^{-i\alpha} b
\cos\frac{\beta}{2}
}
\end{aligned}

\label{eqn:moreBraKetProblems:440}
\begin{bmatrix}
a \\
b
\end{bmatrix}
\propto
\begin{bmatrix}
\cos\frac{\beta}{2} \\
e^{i\alpha}
\sin\frac{\beta}{2}
\end{bmatrix}.

This is appropriately normalized, so the ket for $$\BS \cdot \ncap$$ is

\label{eqn:moreBraKetProblems:460}
\ket{ \BS \cdot \ncap ; + } =
\cos\frac{\beta}{2} \ket{+} +
e^{i\alpha}
\sin\frac{\beta}{2}
\ket{-}.

Note that the other eigenvalue is

\label{eqn:moreBraKetProblems:480}
\ket{ \BS \cdot \ncap ; – } =
-\sin\frac{\beta}{2} \ket{+} +
e^{i\alpha}
\cos\frac{\beta}{2}
\ket{-}.

It is straightforward to show that these are orthogonal and that this has the $$-\Hbar/2$$ eigenvalue.

## Question: Two state Hamiltonian ([1] pr. 1.10)

Solve the eigenproblem for

\label{eqn:moreBraKetProblems:500}
H = a \biglr{
\ket{1}\bra{1}
-\ket{2}\bra{2}
+\ket{1}\bra{2}
+\ket{2}\bra{1}
}

In matrix form the Hamiltonian is

\label{eqn:moreBraKetProblems:520}
H = a
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}.

The eigenvalue problem is

\label{eqn:moreBraKetProblems:540}
\begin{aligned}
0
&= \Abs{ H – \lambda I } \\
&= (a – \lambda)(-a – \lambda) – a^2 \\
&= (-a + \lambda)(a + \lambda) – a^2 \\
&= \lambda^2 – a^2 – a^2,
\end{aligned}

or

\label{eqn:moreBraKetProblems:560}
\lambda = \pm \sqrt{2} a.

An eigenket proportional to $$(\alpha,\beta)$$ must satisfy

\label{eqn:moreBraKetProblems:580}
0
= ( 1 \mp \sqrt{2} ) \alpha + \beta,

so

\label{eqn:moreBraKetProblems:600}
\ket{\pm} \propto
\begin{bmatrix}
-1 \\
1 \mp \sqrt{2}
\end{bmatrix},

or

\label{eqn:moreBraKetProblems:620}
\begin{aligned}
\ket{\pm}
&=
\inv{2(2 – \sqrt{2})}
\begin{bmatrix}
-1 \\
1 \mp \sqrt{2}
\end{bmatrix} \\
&=
\frac{2 + \sqrt{2}}{4}
\begin{bmatrix}
-1 \\
1 \mp \sqrt{2}
\end{bmatrix}.
\end{aligned}

That is
\label{eqn:moreBraKetProblems:640}
\ket{\pm} =
\frac{2 + \sqrt{2}}{4} \lr{
-\ket{1} + (1 \mp \sqrt{2}) \ket{2}
}.

## Question: Spin half probability and dispersion ([1] pr. 1.12)

A spin $$1/2$$ system $$\BS \cdot \ncap$$, with $$\ncap = \sin \gamma \xcap + \cos\gamma \zcap$$, is in state with eigenvalue $$\Hbar/2$$.

## (a)

If $$S_x$$ is measured. What is the probability of getting $$+ \Hbar/2$$?

## (b)

Evaluate the dispersion in $$S_x$$, that is,

\label{eqn:moreBraKetProblems:660}
\expectation{\lr{ S_x – \expectation{S_x}}^2}.

## (a)

In matrix form the spin operator for the system is

\label{eqn:moreBraKetProblems:680}
\begin{aligned}
\BS \cdot \ncap
&= \frac{\Hbar}{2} \lr{ \cos\gamma \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} + \sin\gamma \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}} \\
&= \frac{\Hbar}{2}
\begin{bmatrix}
\cos\gamma & \sin\gamma \\
\sin\gamma & -\cos\gamma \\
\end{bmatrix}
\end{aligned}

An eigenket $$\ket{\BS \cdot \ncap ; + } = (a,b)$$ must satisfy

\label{eqn:moreBraKetProblems:700}
\begin{aligned}
0
&= \lr{ \cos \gamma – 1 } a + \sin\gamma b \\
&= \lr{ -2 \sin^2 \frac{\gamma}{2} } a + 2 \sin\frac{\gamma}{2} \cos\frac{\gamma}{2} b \\
&= -\sin \frac{\gamma}{2} a + \cos\frac{\gamma}{2} b,
\end{aligned}

so the eigenstate is
\label{eqn:moreBraKetProblems:720}
\ket{\BS \cdot \ncap ; + }
=
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix}.

Pick $$\ket{S_x ; \pm } = \inv{\sqrt{2}} \begin{bmatrix} 1 \\ \pm 1 \end{bmatrix}$$ as the basis for the $$S_x$$ operator. Then, for the probability that the system will end up in the $$+ \Hbar/2$$ state of $$S_x$$, we have

\label{eqn:moreBraKetProblems:740}
\begin{aligned}
P
&= \Abs{\braket{ S_x ; + }{ \BS \cdot \ncap ; + } }^2 \\
&= \Abs{ \inv{\sqrt{2} }
{
\begin{bmatrix}
1 \\
1
\end{bmatrix}}^\dagger
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix}
}^2 \\
&=\inv{2}
\Abs{
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix}
}^2 \\
&=
\inv{2}
\lr{
\cos\frac{\gamma}{2} +
\sin\frac{\gamma}{2}
}^2 \\
&=
\inv{2}
\lr{ 1 + 2 \cos\frac{\gamma}{2} \sin\frac{\gamma}{2} } \\
&=
\inv{2}
\lr{ 1 + \sin\gamma }.
\end{aligned}

This is a reasonable seeming result, with $$P \in [0, 1]$$. Some special values also further validate this

\label{eqn:moreBraKetProblems:760}
\begin{aligned}
\gamma &= 0, \ket{\BS \cdot \ncap ; + } =
\begin{bmatrix}
1 \\
0
\end{bmatrix}
=
\ket{S_z ; +}
=
\inv{\sqrt{2}} \ket{S_x;+}
+\inv{\sqrt{2}} \ket{S_x;-}
\\
\gamma &= \pi/2, \ket{\BS \cdot \ncap ; + } =
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
=
\ket{S_x ; +}
\\
\gamma &= \pi, \ket{\BS \cdot \ncap ; + } =
\begin{bmatrix}
0 \\
1
\end{bmatrix}
=
\ket{S_z ; -}
=
\inv{\sqrt{2}} \ket{S_x;+}
-\inv{\sqrt{2}} \ket{S_x;-},
\end{aligned}

where we see that the probabilites are in proportion to the projection of the initial state onto the measured state $$\ket{S_x ; +}$$.

## (b)

The $$S_x$$ expectation is

\label{eqn:moreBraKetProblems:780}
\begin{aligned}
\expectation{S_x}
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\frac{\gamma}{2} & \sin\frac{\gamma}{2}
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\frac{\gamma}{2} & \sin\frac{\gamma}{2}
\end{bmatrix}
\begin{bmatrix}
\sin\frac{\gamma}{2} \\
\cos\frac{\gamma}{2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2} 2 \sin\frac{\gamma}{2} \cos\frac{\gamma}{2} \\
&=
\frac{\Hbar}{2} \sin\gamma.
\end{aligned}

Note that $$S_x^2 = (\Hbar/2)^2I$$, so

\label{eqn:moreBraKetProblems:800}
\begin{aligned}
\expectation{S_x^2}
&=
\lr{\frac{\Hbar}{2}}^2
\begin{bmatrix}
\cos\frac{\gamma}{2} & \sin\frac{\gamma}{2}
\end{bmatrix}
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix} \\
&=
\lr{ \frac{\Hbar}{2} }^2
\cos^2\frac{\gamma}{2} + \sin^2 \frac{\gamma}{2} \\
&=
\lr{ \frac{\Hbar}{2} }^2.
\end{aligned}

The dispersion is

\label{eqn:moreBraKetProblems:820}
\begin{aligned}
\expectation{\lr{ S_x – \expectation{S_x}}^2}
&=
\expectation{S_x^2} – \expectation{S_x}^2 \\
&=
\lr{ \frac{\Hbar}{2} }^2
\lr{1 – \sin^2 \gamma} \\
&=
\lr{ \frac{\Hbar}{2} }^2
\cos^2 \gamma.
\end{aligned}

At $$\gamma = \pi/2$$ the dispersion is 0, which is expected since $$\ket{\BS \cdot \ncap ; + } = \ket{ S_x ; + }$$ at that point. Similarily, the dispersion is maximized at $$\gamma = 0,\pi$$ where the $$\ket{\BS \cdot \ncap ; + }$$ component in the $$\ket{S_x ; + }$$ direction is minimized.

# References

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

## Pauli matrix problems

July 21, 2015 phy1520 No comments ,

### Q: [1] problem 1.2.

Given an arbitrary $$2 \times 2$$ matrix $$X = a_0 + \Bsigma \cdot \Ba$$,
show the relationships between $$a_\mu$$ and $$\textrm{tr}(X), \textrm{tr}(\sigma_k X)$$, and $$X_{ij}$$.

### A.

Observe that each of the Pauli matrices $$\sigma_k$$ are traceless

\label{eqn:pauliProblems:20}
\begin{aligned}
\sigma_x &= \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \\
\sigma_y &= \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \\
\sigma_z &= \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\
\end{aligned},

so $$\textrm{tr}(X) = 2 a_0$$. Note that $$\textrm{tr}(\sigma_k \sigma_m) = 2 \delta_{k m}$$, so $$\textrm{tr}(\sigma_k X) = 2 a_k$$.

Notationally, it would seem to make sense to define $$\sigma_0 \equiv I$$, so that $$\textrm{tr}(\sigma_\mu X) = a_\mu$$. I don’t know if that is common practice.

For the opposite relations, given

\label{eqn:pauliProblems:40}
\begin{aligned}
X
&= a_0 + \Bsigma \cdot \Ba \\
&= \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} a_0 + \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} a_1 + \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} a_2 + \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} a_3 \\
&=
\begin{bmatrix}
a_0 + a_3 & a_1 – i a_2 \\
a_1 + i a_2 & a_0 – a_3
\end{bmatrix} \\
&=
\begin{bmatrix}
X_{11} & X_{12} \\
X_{21} & X_{22} \\
\end{bmatrix},
\end{aligned}

so

\label{eqn:pauliProblems:80}
\begin{aligned}
a_0 &= \inv{2} \lr{ X_{11} + X_{22} } \\
a_1 &= \inv{2} \lr{ X_{12} + X_{21} } \\
a_2 &= \inv{2 i} \lr{ X_{21} – X_{12} } \\
a_3 &= \inv{2} \lr{ X_{11} – X_{22} }
\end{aligned}.

### Q: [1] problem 1.3.

Determine the structure and determinant of the transformation

\label{eqn:pauliProblems:100}
\Bsigma \cdot \Ba \rightarrow
\Bsigma \cdot \Ba’ =
\exp\lr{ i \Bsigma \cdot \ncap \phi/2}
\Bsigma \cdot \Ba
\exp\lr{ -i \Bsigma \cdot \ncap \phi/2}.

### A.

Knowing Geometric Algebra, this is recognized as a rotation transformation. In GA, $$i$$ is treated as a pseudoscalar (which commutes with all grades in \R{3}), and the expression can be reduced to one involving dot and wedge products. Let’s see how can this be reduced using only the Pauli matrix toolbox.

First, consider the determinant of one of the exponentials. Showing that one such exponential has unit determinant is sufficient. The matrix representation of the unit normal is

\label{eqn:pauliProblems:120}
\begin{aligned}
\Bsigma \cdot \ncap
&= n_x \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
+ n_y \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
+ n_z \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\
&=
\begin{bmatrix}
n_z & n_x – i n_y \\
n_x + i n_y & -n_z
\end{bmatrix}.
\end{aligned}

This is expected to have a unit square, and does

\label{eqn:pauliProblems:140}
\begin{aligned}
\lr{ \Bsigma \cdot \ncap }^2
&=
\begin{bmatrix}
n_z & n_x – i n_y \\
n_x + i n_y & -n_z
\end{bmatrix}
\begin{bmatrix}
n_z & n_x – i n_y \\
n_x + i n_y & -n_z
\end{bmatrix} \\
&=
\lr{ n_x^2 + n_y^2 + n_z^2 }
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \\
&=
1.
\end{aligned}

This allows for a cosine and sine expansion of the exponential, as in

\label{eqn:pauliProblems:160}
\begin{aligned}
\exp\lr{ i \Bsigma \cdot \ncap \theta}
&=
\cos\theta + i \Bsigma \cdot \ncap \sin\theta \\
&=
\cos\theta
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
+
i \sin\theta
\begin{bmatrix}
n_z & n_x – i n_y \\
n_x + i n_y & -n_z
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos\theta + i n_z \sin\theta & \lr{ n_x – i n_y } i \sin\theta \\
\lr{ n_x + i n_y } i \sin\theta & \cos\theta – i n_z \sin\theta \\
\end{bmatrix}.
\end{aligned}

This has determinant

\label{eqn:pauliProblems:180}
\begin{aligned}
\Abs{\exp\lr{ i \Bsigma \cdot \ncap \theta} }
&=
\cos^2\theta + n_z^2 \sin^2\theta

\lr{ -n_x^2 + -n_y^2 } \sin^2\theta \\
&=
\cos^2\theta + \lr{ n_x^2 + n_y^2 + n_z^2 } \sin^2\theta \\
&= 1,
\end{aligned}

as expected.

Next step is to show that this transformation is a rotation, and determine the sense of the rotation. Let $$C = \cos\phi/2, S = \sin\phi/2$$, so that

\label{eqn:pauliProblems:200}
\begin{aligned}
\Bsigma \cdot \Ba’
&=
\exp\lr{ i \Bsigma \cdot \ncap \phi/2}
\Bsigma \cdot \Ba
\exp\lr{ -i \Bsigma \cdot \ncap \phi/2} \\
&=
\lr{ C + i \Bsigma \cdot \ncap S }
\Bsigma \cdot \Ba
\lr{ C – i \Bsigma \cdot \ncap S } \\
&=
\lr{ C + i \Bsigma \cdot \ncap S }
\lr{ C \Bsigma \cdot \Ba – i \Bsigma \cdot \Ba \Bsigma \cdot \ncap S } \\
&=
C^2 \Bsigma \cdot \Ba + \Bsigma \cdot \ncap \Bsigma \cdot \Ba \Bsigma \cdot \ncap S^2
+ i \lr{
-\Bsigma \cdot \Ba \Bsigma \cdot \ncap
+ \Bsigma \cdot \ncap \Bsigma \cdot \Ba
} S C \\
&=
\inv{2} \lr{ 1 + \cos\phi}
\Bsigma \cdot \Ba
+ \Bsigma \cdot \ncap \Bsigma \cdot \Ba \Bsigma \cdot \ncap \inv{2} \lr{ 1 – \cos\phi}
+ i
\antisymmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba }
\inv{2} \sin\phi \\
&=
\inv{2}
\Bsigma \cdot \ncap
\symmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba }
+ \inv{2}
\Bsigma \cdot \ncap
\antisymmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba } \cos\phi
+
\inv{2}
i
\antisymmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba }
\sin\phi.
\end{aligned}

Observe that the angle dependent portion can be written in a compact exponential form

\label{eqn:pauliProblems:220}
\begin{aligned}
\Bsigma \cdot \Ba’
&=
\inv{2}
\Bsigma \cdot \ncap
\symmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba }
+
\lr{
\cos\phi
+
i
\Bsigma \cdot \ncap
\sin\phi
}
\inv{2}
\Bsigma \cdot \ncap
\antisymmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba } \\
&=
\inv{2}
\Bsigma \cdot \ncap
\symmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba }
+
\exp\lr{ i \Bsigma \cdot \ncap \phi }
\inv{2}
\Bsigma \cdot \ncap
\antisymmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba }.
\end{aligned}

The anticommutator and commutator products with the unit normal can be identified as projections and rejections respectively. Consider the symmetric product first

\label{eqn:pauliProblems:240}
\begin{aligned}
\inv{2}
\symmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba } \\
&=
\inv{2}
\sum n_r a_s \lr{ \sigma_r \sigma_s + \sigma_s \sigma_r } \\
&=
\inv{2}
\sum_{r \ne s} n_r a_s \lr{ \sigma_r \sigma_s + \sigma_s \sigma_r }
+
\inv{2}
\sum_{r } n_r a_r 2 \\
&= 2 \ncap \cdot \Ba.
\end{aligned}

This shows that
\label{eqn:pauliProblems:260}
\inv{2}
\Bsigma \cdot \ncap
\symmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba }
=
\lr{ \ncap \cdot \Ba } \Bsigma \cdot \ncap,

which is the projection of $$\Ba$$ in the direction of the normal $$\ncap$$. To show that the commutator term is the rejection, consider the sum of the two

\label{eqn:pauliProblems:280}
\begin{aligned}
\inv{2}
\Bsigma \cdot \ncap
\symmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba }
+
\inv{2}
\Bsigma \cdot \ncap
\antisymmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba }
&=
\Bsigma \cdot \ncap
\Bsigma \cdot \ncap \Bsigma \cdot \Ba \\
&=
\Bsigma \cdot \Ba,
\end{aligned}

so we must have

\label{eqn:pauliProblems:300}
\Bsigma \cdot \Ba – \lr{ \ncap \cdot \Ba } \Bsigma \cdot \ncap
=
\inv{2}
\Bsigma \cdot \ncap
\antisymmetric{
\Bsigma \cdot \ncap }{\Bsigma \cdot \Ba }.

This is the component of $$\Ba$$ that has the projection in the $$\ncap$$ direction removed. Looking back to \ref{eqn:pauliProblems:220}, the transformation leaves components of the vector that are colinear with the unit normal unchanged, and applies an exponential operation to the component that lies in what is presumed to be the rotation plane. To verify that this latter portion of the transformation is a rotation, and to determine the sense of the rotation, let’s expand the factor of the sine of \ref{eqn:pauliProblems:200}.

That is

\label{eqn:pauliProblems:320}
\begin{aligned}
\frac{i}{2} \antisymmetric{ \Bsigma \cdot \ncap }{\Bsigma \cdot \Ba }
&=
\frac{i}{2} \sum n_r a_s \antisymmetric{ \sigma_r }{\sigma_s } \\
&=
\frac{i}{2} \sum n_r a_s 2 i \epsilon_{r s t} \sigma_t \\
&=
– \sum \sigma_t n_r a_s \epsilon_{r s t} \\
&=
-\Bsigma \cdot \lr{ \ncap \cross \Ba } \\
&=
\Bsigma \cdot \lr{ \Ba \cross \ncap }.
\end{aligned}

Since $$\Ba \cross \ncap = \lr{ \Ba – \ncap (\ncap \cdot \Ba) } \cross \ncap$$, this vector is seen to lie in the plane normal to $$\ncap$$, but perpendicular to the rejection of $$\ncap$$ from $$\Ba$$. That completes the demonstration that this is a rotation transformation.

To understand the sense of this rotation, consider $$\ncap = \zcap, \Ba = \xcap$$, so

\label{eqn:pauliProblems:340}
\Bsigma \cdot \lr{ \Ba \cross \ncap }
=
\Bsigma \cdot \lr{ \xcap \cross \zcap }
=
-\Bsigma \cdot \ycap,

and
\label{eqn:pauliProblems:360}
\Bsigma \cdot \Ba’
=
\xcap \cos\phi – \ycap \sin\phi,

showing that this rotation transformation has a clockwise sense.

# References

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

## Motivation

The conventional form for the Pauli matrices is

\label{eqn:pauliMatrixXYgeometry:20}
\begin{aligned}
\sigma_x &=
\begin{bmatrix}
0 & 1 \\
1 & 0 \\
\end{bmatrix} \\
\sigma_y &=
\begin{bmatrix}
0 & -i \\
i & 0 \\
\end{bmatrix} \\
\sigma_z &=
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\end{aligned}.

In [1] these forms are derived based on the commutation relations

\label{eqn:pauliMatrixXYgeometry:40}
\antisymmetric{\sigma_r}{\sigma_s} = 2 i \epsilon_{r s t} \sigma_t,

by defining raising and lowering operators $$\sigma_{\pm} = \sigma_x \pm i \sigma_y$$ and figuring out what form the matrix must take. I noticed an interesting geometrical relation hiding in that derivation if $$\sigma_{+}$$ is not assumed to be real.

## Derivation

For completeness, I’ll repeat the argument of [1], which builds on the commutation relations of the raising and lowering operators. Those are

\label{eqn:pauliMatrixXYgeometry:60}
\begin{aligned}
\antisymmetric{\sigma_z}{\sigma_{\pm}}
&=
\sigma_z \lr{ \sigma_x \pm i \sigma_y }
-\lr{ \sigma_x \pm i \sigma_y } \sigma_z \\
&=
\antisymmetric{\sigma_z}{\sigma_x} \pm i \antisymmetric{\sigma_z}{\sigma_y} \\
&=
2 i \sigma_y \pm i (-2 i) \sigma_x \\
&= \pm 2 \lr{ \sigma_x \pm i \sigma_y } \\
&= \pm 2 \sigma_{\pm},
\end{aligned}

and

\label{eqn:pauliMatrixXYgeometry:80}
\begin{aligned}
\antisymmetric{\sigma_{+}}{\sigma_{-}}
&=
\lr{ \sigma_x + i \sigma_y } \lr{ \sigma_x – i \sigma_y }
-\lr{ \sigma_x – i \sigma_y } \lr{ \sigma_x + i \sigma_y } \\
&=
-i \sigma_x \sigma_y + i \sigma_y \sigma_x
– i \sigma_x \sigma_y + i \sigma_y \sigma_x \\
&= 2 i \antisymmetric{ \sigma_y }{\sigma_x} \\
&= 2 i (-2i) \sigma_z \\
&= 4 \sigma_z
\end{aligned}

From these a matrix representation containing unknown values can be assumed. Let

\label{eqn:pauliMatrixXYgeometry:100}
\sigma_{+} =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}.

The commutator with $$\sigma_z$$ can be computed

\label{eqn:pauliMatrixXYgeometry:120}
\begin{aligned}
\antisymmetric{\sigma_z}{\sigma_{+}}
&=
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}

\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\\
&=
\begin{bmatrix}
a & b \\
-c & -d
\end{bmatrix}

\begin{bmatrix}
a & -b \\
c & -d
\end{bmatrix} \\
&=
2
\begin{bmatrix}
0 & b \\
-c & 0
\end{bmatrix}
\end{aligned}

Now compare this with \ref{eqn:pauliMatrixXYgeometry:60}

\label{eqn:pauliMatrixXYgeometry:140}
2
\begin{bmatrix}
0 & b \\
-c & 0
\end{bmatrix}
=
2 \sigma_{+}
=
2
\begin{bmatrix}
a & b \\
d & d
\end{bmatrix}.

This shows that $$a = 0$$, and $$d = 0$$. Similarly the $$\sigma_z$$ commutator with the lowering operator is

\label{eqn:pauliMatrixXYgeometry:160}
\begin{aligned}
\antisymmetric{\sigma_z}{\sigma_{-}}
&=
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\begin{bmatrix}
0 & -c^\conj \\
b^\conj & 0
\end{bmatrix}

\begin{bmatrix}
0 & -c^\conj \\
b^\conj & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\\
&=
\begin{bmatrix}
0 & -c^\conj \\
-b^\conj & 0
\end{bmatrix}

\begin{bmatrix}
0 & c^\conj \\
b^\conj & 0
\end{bmatrix} \\
&=
-2
\begin{bmatrix}
0 & c^\conj \\
b^\conj & 0
\end{bmatrix}
\end{aligned}

Again comparing to \ref{eqn:pauliMatrixXYgeometry:60}, we have
\label{eqn:pauliMatrixXYgeometry:180}
-2
\begin{bmatrix}
0 & c^\conj \\
b^\conj & 0
\end{bmatrix}
= – 2 \sigma_{-}
= – 2
\begin{bmatrix}
0 & -c^\conj \\
b^\conj & 0
\end{bmatrix},

so $$c = 0$$. Computing the commutator of the raising and lowering operators fixes $$b$$

\label{eqn:pauliMatrixXYgeometry:200}
\begin{aligned}
\antisymmetric{\sigma_{+}}{\sigma_{-}}
&=
\begin{bmatrix}
0 & b \\
0 & 0 \\
\end{bmatrix}
\begin{bmatrix}
0 & 0 \\
b^\conj & 0 \\
\end{bmatrix}

\begin{bmatrix}
0 & 0 \\
b^\conj & 0 \\
\end{bmatrix}
\begin{bmatrix}
0 & b \\
0 & 0 \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\Abs{b}^2 & 0 \\
0 & 0
\end{bmatrix}

\begin{bmatrix}
0 & 0
0 & -\Abs{b}^2 \\
\end{bmatrix} \\
&=
\Abs{b}^2
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\\
&=
\Abs{b}^2 \sigma_z.
\end{aligned}

From \ref{eqn:pauliMatrixXYgeometry:80} it must be that $$\Abs{b}^2 = 4$$, so the most general form of the raising operator is

\label{eqn:pauliMatrixXYgeometry:220}
\sigma_{+}
=
2
\begin{bmatrix}
0 & e^{i \phi} \\
0 & 0
\end{bmatrix}.

## Observation

The conventional choice is to set $$\phi = 0$$, but I found it interesting to see the form of $$\sigma_x, \sigma_y$$ without that choice. That is

\label{eqn:pauliMatrixXYgeometry:240}
\begin{aligned}
\sigma_x
&= \inv{2} \lr{ \sigma_{+} + \sigma_{-} } \\
&=
\begin{bmatrix}
0 & e^{i \phi} \\
e^{-i \phi} & 0 \\
\end{bmatrix}
\end{aligned}

\label{eqn:pauliMatrixXYgeometry:260}
\begin{aligned}
\sigma_y
&= \inv{2 i} \lr{ \sigma_{+} – \sigma_{-} } \\
&=
\begin{bmatrix}
0 & -i e^{i \phi} \\
-i e^{-i \phi} & 0 \\
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & e^{i (\phi – \pi/2) } \\
e^{-i (\phi – \pi/2)} & 0 \\
\end{bmatrix}.
\end{aligned}

Notice that the Pauli matrices $$\sigma_x$$ and $$\sigma_y$$ actually both have the same form as $$\sigma_x$$, but the phase of the complex argument of each differs by $$90^\circ$$. That $$90^\circ$$ separation isn’t obvious in the standard form \ref{eqn:pauliMatrixXYgeometry:20}.

It’s a small detail, but I thought it was kind of cool that the orthogonality of these matrix unit vector representations is built directly into the structure of their matrix representations.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.