bra-ket manipulation problems

July 22, 2015 phy1520 , , , , , , , ,

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Some bra-ket manipulation problems.([1] pr. 1.4)

Using braket logic expand

(a)

\begin{equation}\label{eqn:braketManip:20}
\textrm{tr}{X Y}
\end{equation}

(b)

\begin{equation}\label{eqn:braketManip:40}
(X Y)^\dagger
\end{equation}

(c)

\begin{equation}\label{eqn:braketManip:60}
e^{i f(A)},
\end{equation}

where \( A \) is Hermitian with a complete set of eigenvalues.

(d)

\begin{equation}\label{eqn:braketManip:80}
\sum_{a’} \Psi_{a’}(\Bx’)^\conj \Psi_{a’}(\Bx”),
\end{equation}

where \( \Psi_{a’}(\Bx”) = \braket{\Bx’}{a’} \).

Answers

(a)

\begin{equation}\label{eqn:braketManip:100}
\begin{aligned}
\textrm{tr}{X Y}
&= \sum_a \bra{a} X Y \ket{a} \\
&= \sum_{a,b} \bra{a} X \ket{b}\bra{b} Y \ket{a} \\
&= \sum_{a,b}
\bra{b} Y \ket{a}
\bra{a} X \ket{b} \\
&= \sum_{a,b}
\bra{b} Y
X \ket{b} \\
&= \textrm{tr}{ Y X }.
\end{aligned}
\end{equation}

(b)

\begin{equation}\label{eqn:braketManip:120}
\begin{aligned}
\bra{a} \lr{ X Y}^\dagger \ket{b}
&=
\lr{ \bra{b} X Y \ket{a} }^\conj \\
&=
\sum_c \lr{ \bra{b} X \ket{c}\bra{c} Y \ket{a} }^\conj \\
&=
\sum_c \lr{ \bra{b} X \ket{c} }^\conj \lr{ \bra{c} Y \ket{a} }^\conj \\
&=
\sum_c
\lr{ \bra{c} Y \ket{a} }^\conj
\lr{ \bra{b} X \ket{c} }^\conj \\
&=
\sum_c
\bra{a} Y^\dagger \ket{c}
\bra{c} X^\dagger \ket{b} \\
&=
\bra{a} Y^\dagger
X^\dagger \ket{b},
\end{aligned}
\end{equation}

so \( \lr{ X Y }^\dagger = Y^\dagger X^\dagger \).

(c)

Let’s presume that the function \( f \) has a Taylor series representation

\begin{equation}\label{eqn:braketManip:140}
f(A) = \sum_r b_r A^r.
\end{equation}

If the eigenvalues of \( A \) are given by

\begin{equation}\label{eqn:braketManip:160}
A \ket{a_s} = a_s \ket{a_s},
\end{equation}

this operator can be expanded like

\begin{equation}\label{eqn:braketManip:180}
\begin{aligned}
A
&= \sum_{a_s} A \ket{a_s} \bra{a_s} \\
&= \sum_{a_s} a_s \ket{a_s} \bra{a_s},
\end{aligned}
\end{equation}

To compute powers of this operator, consider first the square

\begin{equation}\label{eqn:braketManip:200}
\begin{aligned}
A^2 =
&=
\sum_{a_s} a_s \ket{a_s} \bra{a_s}
\sum_{a_r} a_r \ket{a_r} \bra{a_r} \\
&=
\sum_{a_s, a_r} a_s a_r \ket{a_s} \bra{a_s} \ket{a_r} \bra{a_r} \\
&=
\sum_{a_s, a_r} a_s a_r \ket{a_s} \delta_{s r} \bra{a_r} \\
&=
\sum_{a_s} a_s^2 \ket{a_s} \bra{a_s}.
\end{aligned}
\end{equation}

The pattern for higher powers will clearly just be

\begin{equation}\label{eqn:braketManip:220}
A^k =
\sum_{a_s} a_s^k \ket{a_s} \bra{a_s},
\end{equation}

so the expansion of \( f(A) \) will be

\begin{equation}\label{eqn:braketManip:240}
\begin{aligned}
f(A)
&= \sum_r b_r A^r \\
&= \sum_r b_r
\sum_{a_s} a_s^r \ket{a_s} \bra{a_s} \\
&=
\sum_{a_s} \lr{ \sum_r b_r a_s^r } \ket{a_s} \bra{a_s} \\
&=
\sum_{a_s} f(a_s) \ket{a_s} \bra{a_s}.
\end{aligned}
\end{equation}

The exponential expansion is

\begin{equation}\label{eqn:braketManip:260}
\begin{aligned}
e^{i f(A)}
&=
\sum_t \frac{i^t}{t!} f^t(A) \\
&=
\sum_t \frac{i^t}{t!}
\lr{ \sum_{a_s} f(a_s) \ket{a_s} \bra{a_s} }^t \\
&=
\sum_t \frac{i^t}{t!}
\sum_{a_s} f^t(a_s) \ket{a_s} \bra{a_s} \\
&=
\sum_{a_s}
e^{i f(a_s) }
\ket{a_s} \bra{a_s}.
\end{aligned}
\end{equation}

(d)

\begin{equation}\label{eqn:braketManip:n}
\begin{aligned}
\sum_{a’} \Psi_{a’}(\Bx’)^\conj \Psi_{a’}(\Bx”)
&=
\sum_{a’}
\braket{\Bx’}{a’}^\conj
\braket{\Bx”}{a’} \\
&=
\sum_{a’}
\braket{a’}{\Bx’}
\braket{\Bx”}{a’} \\
&=
\sum_{a’}
\braket{\Bx”}{a’}
\braket{a’}{\Bx’} \\
&=
\braket{\Bx”}{\Bx’} \\
&= \delta_{\Bx” – \Bx’}.
\end{aligned}
\end{equation}

References

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

Pauli matrix problems

July 21, 2015 phy1520 ,

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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

\begin{equation}\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},
\end{equation}

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

\begin{equation}\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}
\end{equation}

so

\begin{equation}\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}.
\end{equation}

Q: [1] problem 1.3.

Determine the structure and determinant of the transformation

\begin{equation}\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}.
\end{equation}

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

\begin{equation}\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}
\end{equation}

This is expected to have a unit square, and does

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

This has determinant

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

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

\begin{equation}\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}
\end{equation}

so we must have

\begin{equation}\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 }.
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

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

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

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.

An observation about the geometry of Pauli x,y matrices

July 19, 2015 phy1520 , , , ,

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Motivation

The conventional form for the Pauli matrices is

\begin{equation}\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}.
\end{equation}

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

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

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

\begin{equation}\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}
\end{equation}

and

\begin{equation}\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}
\end{equation}

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

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

The commutator with \( \sigma_z \) can be computed

\begin{equation}\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}
\end{equation}

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

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

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

\begin{equation}\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}
\end{equation}

Again comparing to \ref{eqn:pauliMatrixXYgeometry:60}, we have
\begin{equation}\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},
\end{equation}

so \( c = 0 \). Computing the commutator of the raising and lowering operators fixes \( b \)

\begin{equation}\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}
\end{equation}

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

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

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

\begin{equation}\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}
\end{equation}

\begin{equation}\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}
\end{equation}

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.

Schwartz inequality in bra-ket notation

July 6, 2015 phy1520 , , , , , ,

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

Motivation

In [2] the Schwartz inequality

\begin{equation}\label{eqn:qmSchwartz:20}
\boxed{
\braket{a}{a}
\braket{b}{b}
\ge \Abs{\braket{a}{b}}^2,
}
\end{equation}

is used in the derivation of the uncertainty relation. The proof of the Schwartz inequality uses a sneaky substitution that doesn’t seem obvious, and is even less obvious since there is a typo in the value to be substituted. Let’s understand where that sneakiness is coming from.

Without being sneaky

My ancient first year linear algebra text [1] contains a non-sneaky proof, but it only works for real vector spaces. Recast in bra-ket notation, this method examines the bounds of the norms of sums and differences of unit states (i.e. \( \braket{a}{a} = \braket{b}{b} = 1 \).)

\begin{equation}\label{eqn:qmSchwartz:40}
\braket{a – b}{a – b}
= \braket{a}{a} + \braket{b}{b} – \braket{a}{b} – \braket{b}{a}
= 2 – 2 \textrm{Re} \braket{a}{b}
\ge 0,
\end{equation}

so
\begin{equation}\label{eqn:qmSchwartz:60}
1 \ge \textrm{Re} \braket{a}{b}.
\end{equation}

Similarily

\begin{equation}\label{eqn:qmSchwartz:80}
\braket{a + b}{a + b}
= \braket{a}{a} + \braket{b}{b} + \braket{a}{b} + \braket{b}{a}
= 2 + 2 \textrm{Re} \braket{a}{b}
\ge 0,
\end{equation}

so
\begin{equation}\label{eqn:qmSchwartz:100}
\textrm{Re} \braket{a}{b} \ge -1.
\end{equation}

This means that for normalized state vectors

\begin{equation}\label{eqn:qmSchwartz:120}
-1 \le \textrm{Re} \braket{a}{b} \le 1,
\end{equation}

or
\begin{equation}\label{eqn:qmSchwartz:140}
\Abs{\textrm{Re} \braket{a}{b}} \le 1.
\end{equation}

Writing out the unit vectors explicitly, that last inequality is

\begin{equation}\label{eqn:qmSchwartz:180}
\Abs{ \textrm{Re} \braket{ \frac{a}{\sqrt{\braket{a}{a}}} }{ \frac{b}{\sqrt{\braket{b}{b}}} } } \le 1,
\end{equation}

squaring and rearranging gives

\begin{equation}\label{eqn:qmSchwartz:200}
\Abs{\textrm{Re} \braket{a}{b}}^2 \le
\braket{a}{a}
\braket{b}{b}.
\end{equation}

This is similar to, but not identical to the Schwartz inequality. Since \( \Abs{\textrm{Re} \braket{a}{b}} \le \Abs{\braket{a}{b}} \) the Schwartz inequality cannot be demonstrated with this argument. This first year algebra method works nicely for demonstrating the inequality for real vector spaces, so a different argument is required for a complex vector space (i.e. quantum mechanics state space.)

Arguing with projected and rejected components

Notice that the equality condition in the inequality holds when the vectors are colinear, and the largest inequality holds when the vectors are normal to each other. Given those geometrical observations, it seems reasonable to examine the norms of projected or rejected components of a vector. To do so in bra-ket notation, the correct form of a projection operation must be determined. Care is required to get the ordering of the bra-kets right when expressing such a projection.

Suppose we wish to calculation the rejection of \( \ket{a} \) from \( \ket{b} \), that is \( \ket{b – \alpha a}\), such that

\begin{equation}\label{eqn:qmSchwartz:220}
0
= \braket{a}{b – \alpha a}
= \braket{a}{b} – \alpha \braket{a}{a},
\end{equation}

or
\begin{equation}\label{eqn:qmSchwartz:240}
\alpha =
\frac{\braket{a}{b} }{ \braket{a}{a} }.
\end{equation}

Therefore, the projection of \( \ket{b} \) on \( \ket{a} \) is

\begin{equation}\label{eqn:qmSchwartz:260}
\textrm{Proj}_{\ket{a}} \ket{b}
= \frac{\braket{a}{b} }{ \braket{a}{a} } \ket{a}
= \frac{\braket{b}{a}^\conj }{ \braket{a}{a} } \ket{a}.
\end{equation}

The conventional way to write this in QM is in the operator form

\begin{equation}\label{eqn:qmSchwartz:300}
\textrm{Proj}_{\ket{a}} \ket{b}
= \frac{\ket{a}\bra{a}}{\braket{a}{a}} \ket{b}.
\end{equation}

In this form the rejection of \( \ket{a} \) from \( \ket{b} \) can be expressed as

\begin{equation}\label{eqn:qmSchwartz:280}
\textrm{Rej}_{\ket{a}} \ket{b} = \ket{b} – \frac{\ket{a}\bra{a}}{\braket{a}{a}} \ket{b}.
\end{equation}

This state vector is normal to \( \ket{a} \) as desired

\begin{equation}\label{eqn:qmSchwartz:320}
\braket{a}{b – \frac{\braket{a}{b} }{ \braket{a}{a} } a }
=
\braket{a}{ b} – \frac{ \braket{a}{b} }{ \braket{a}{a} } \braket{a}{a}
=
\braket{a}{ b} – \braket{a}{b}
= 0.
\end{equation}

How about it’s length? That is

\begin{equation}\label{eqn:qmSchwartz:340}
\begin{aligned}
\braket{b – \frac{\braket{a}{b} }{ \braket{a}{a} } a}{b – \frac{\braket{a}{b} }{ \braket{a}{a} } a }
&=
\braket{b}{b} – 2 \frac{\Abs{\braket{a}{b}}^2}{\braket{a}{a}} +\frac{\Abs{\braket{a}{b}}^2 }{ \braket{a}{a}^2 } \braket{a}{a} \\
&=
\braket{b}{b} – \frac{\Abs{\braket{a}{b}}^2}{\braket{a}{a}}.
\end{aligned}
\end{equation}

Observe that this must be greater to or equal to zero, so

\begin{equation}\label{eqn:qmSchwartz:360}
\braket{b}{b} \ge \frac{ \Abs{ \braket{a}{b} }^2 }{ \braket{a}{a} }.
\end{equation}

Rearranging this gives \ref{eqn:qmSchwartz:20} as desired. The Schwartz proof in [2] obscures the geometry involved and starts with

\begin{equation}\label{eqn:qmSchwartz:380}
\braket{b + \lambda a}{b + \lambda a} \ge 0,
\end{equation}

where the “proof” is nothing more than a statement that one can “pick” \( \lambda = -\braket{b}{a}/\braket{a}{a} \). The Pythagorean context of the Schwartz inequality is not mentioned, and without thinking about it, one is left wondering what sort of magic hat that \( \lambda \) selection came from.

References

[1] W Keith Nicholson. Elementary linear algebra, with applications. PWS-Kent Publishing Company, 1990.

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

Impedance transformation

May 16, 2015 ece1229 , , , ,

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

In our final problem set we used the impedance transformation for calculations related to a microslot antenna. This transformation wasn’t familiar to me, and is apparently covered in the third year ECE fields class. I found a derivation of this in [1], but the idea is really simple and follows from the reflection coefficient calculation for a normal reflection configuration.

Consider a normal field reflection between two interfaces, as sketched in fig. 1.

normalTransmissionFig1

fig. 1. Normal reflection and transmission between two media.

The fields are

\begin{equation}\label{eqn:impedanceTransformation:40}
\BE^\textrm{i} = \xcap E_0 e^{-j k_1 z}
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:60}
\BH^\textrm{i} = \ycap \frac{E_0}{\eta_1} e^{-j k_1 z}
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:80}
\BE^\textrm{r} = \xcap \Gamma E_0 e^{j k_1 z}
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:100}
\BH^\textrm{r} = -\ycap \Gamma \frac{E_0}{\eta_1} e^{j k_1 z}
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:120}
\BE^\textrm{t} = \xcap E_0 T e^{-j k_2 z}
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:140}
\BH^\textrm{t} = \ycap \frac{E_0}{\eta_1} T e^{-j k_2 z}.
\end{equation}

The field orientations have been picked so that the tangential component of the electric field is \( \xcap \) oriented for all of the incident, reflected, and transmitted components. Requiring equality of the tangential field components at the interface gives

\begin{equation}\label{eqn:impedanceTransformation:180}
1 + \Gamma = T
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:200}
\inv{\eta_1} – \frac{\Gamma}{\eta_1} = \frac{T}{\eta_2}.
\end{equation}

Solving for the transmission coefficient gives

\begin{equation}\label{eqn:impedanceTransformation:220}
\begin{aligned}
T
&= \frac{2}{ 1 + \frac{\eta_1}{\eta_2} } \\
&= \frac{2 \eta_2}{ \eta_2 + \eta_1 },
\end{aligned}
\end{equation}

and for the reflection coefficient

\begin{equation}\label{eqn:impedanceTransformation:240}
\begin{aligned}
\Gamma
&= T – 1 \\
&= \frac{2 \eta_2 – \eta_1 – \eta_2}{ \eta_2 + \eta_1 } \\
&= \frac{\eta_2 – \eta_1 }{ \eta_2 + \eta_1 }.
\end{aligned}
\end{equation}

The total fields in medium 1 at the point \( z = -l \) are

\begin{equation}\label{eqn:impedanceTransformation:280}
\BE^\textrm{i} + \BE^\textrm{r}
=
\xcap E_0 \lr{ e^{ -j k_1 (-l)} + \Gamma e^{ j k_1 (-l) } }
\end{equation}
\begin{equation}\label{eqn:impedanceTransformation:300}
\BH^\textrm{i} + \BH^\textrm{r}
=
\ycap \frac{E_0}{\eta_1} \lr{ e^{ -j k_1 (-l)} – \Gamma e^{ j k_1 (-l) }}.
\end{equation}

The ratio of the electric field strength to the magnetic field strength is defined as the input impedance

\begin{equation}\label{eqn:impedanceTransformation:320}
Z_{\textrm{in}} \equiv \evalbar{\frac{E^\textrm{i} + E^\textrm{r}}{H^\textrm{i} + H^\textrm{r}}}{ z = -l}.
\end{equation}

That is

\begin{equation}\label{eqn:impedanceTransformation:340}
\begin{aligned}
Z_{\textrm{in}}
&=
\eta_1 \frac{
e^{ j k_1 l} + \Gamma e^{ -j k_1 l }
}{
e^{ j k_1 l} – \Gamma e^{ -j k_1 l }
} \\
&=
\eta_1 \frac{
\lr{ \eta_1 + \eta_2} e^{ j k_1 l} + \lr{ \eta_2 – \eta_1} e^{ -j k_1 l }
}{
\lr{ \eta_1 + \eta_2} e^{ j k_1 l} – \lr{ \eta_2 – \eta_1} e^{ -j k_1 l }
} \\
&=
\eta_1 \frac{
\eta_2 \cos( k_1 l ) + \eta_1 j \sin( k_1 l)
}{
\eta_2 j \sin( k_1 l ) + \eta_1 \cos( k_1 l)
},
\end{aligned}
\end{equation}

or
\begin{equation}\label{eqn:impedanceTransformation:360}
\boxed{
Z_{\textrm{in}}
=
\eta_1 \frac{
\eta_2 + j \eta_1 \tan( k_1 l)
}{
\eta_1 + j \eta_2 \tan( k_1 l )
}.
}
\end{equation}

References

[1] Constantine A Balanis. Advanced engineering electromagnetics, chapter {Reflection and transmission}. Wiley New York, 1989.