phy1520

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

November 2, 2015 phy1520 , , ,

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

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

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

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

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

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

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

so

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

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
\begin{equation}\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} \\
&\quad +
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},
\end{equation}

This gives us

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

The angle is

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

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

and the normal direction is
\begin{equation}\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} }.
}
\end{equation}

References

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

Ensembles for spin one half

November 1, 2015 phy1520 , , , ,

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

Mixed ensemble averages

In [1], Sakurai leaves it to the reader to verify that knowledge of the three ensemble averages [S_x], [S_y],[S_z] is sufficient to reconstruct the density operator for a spin one half system.

I’ll do this in two parts, the first using a spin-up/down ensemble to see what form this has, then the general case. The general case is a bit messy algebraically. After first attempting it the hard way, I did the grunt work portion of that calculation in Mathematica, but then realized it’s not so bad to do it manually.

Consider first an ensemble with density operator

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:20}
\rho =
w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-},
\end{equation}

where these are the \( \BS \cdot (\pm \zcap) \) eigenstates. The traces are

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:40}
\begin{aligned}
\textrm{Tr}( \rho \sigma_x )
&=
\bra{+} \rho \sigma_x \ket{+}
+
\bra{-} \rho \sigma_x \ket{-} \\
&=
\bra{+} \rho \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \ket{+}
+
\bra{-} \rho \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \ket{-} \\
&=
\bra{+} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{-}
+
\bra{-} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{+} \\
&=
\bra{+} w_{-} \ket{-}
+
\bra{-} w_{+} \ket{+} \\
&=
0,
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:60}
\begin{aligned}
\textrm{Tr}( \rho \sigma_y )
&=
\bra{+} \rho \sigma_y \ket{+}
+
\bra{-} \rho \sigma_y \ket{-} \\
&=
\bra{+} \rho \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \ket{+}
+
\bra{-} \rho \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \ket{-} \\
&=
i \bra{+} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{-}
–
i \bra{-} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{+} \\
&=
i \bra{+} w_{-} \ket{-}
–
i \bra{-} w_{+} \ket{+} \\
&=
0,
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:ensemblesForSpinOneHalf:100}
\begin{aligned}
\textrm{Tr}( \rho \sigma_z )
&=
\bra{+} \rho \sigma_z \ket{+}
+
\bra{-} \rho \sigma_z \ket{-} \\
&=
\bra{+} \rho \ket{+}
–
\bra{-} \rho \ket{-} \\
&=
\bra{+} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{+}
–
\bra{-} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{-} \\
&=
\bra{+} w_{+} \ket{+}
–
\bra{-} w_{-} \ket{-} \\
&=
w_{+} – w_{-}.
\end{aligned}
\end{equation}

Since \( w_{+} + w_{-} = 1 \), this gives

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:80}
\boxed{
\begin{aligned}
w_{+} &= \frac{1 + \textrm{Tr}( \rho \sigma_z )}{2} \\
w_{-} &= \frac{1 – \textrm{Tr}( \rho \sigma_z )}{2}
\end{aligned}
}
\end{equation}

Attempting to do a similar set of trace expansions this way for a more general spin basis turns out to be a really bad idea and horribly messy. So much so that I resorted to \href{https://raw.githubusercontent.com/peeterjoot/mathematica/master/phy1520/spinOneHalfSymbolicManipulation.nb}{Mathematica to do this symbolic work}. However, it’s not so bad if the trace is done completely in matrix form.

Using the basis

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:120}
\begin{aligned}
\ket{\BS \cdot \ncap ; + } &=
\begin{bmatrix}
\cos(\theta/2) \\
\sin(\theta/2) e^{i \phi}
\end{bmatrix} \\
\ket{\BS \cdot \ncap ; – } &=
\begin{bmatrix}
\sin(\theta/2) e^{-i \phi} \\
-\cos(\theta/2) \\
\end{bmatrix},
\end{aligned}
\end{equation}

the projector matrices are

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:140}
\begin{aligned}
\ket{\BS \cdot \ncap ; + } \bra{\BS \cdot \ncap ; + }
&=
\begin{bmatrix}
\cos(\theta/2) \\
\sin(\theta/2) e^{i \phi}
\end{bmatrix}
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos^2(\theta/2) & \cos(\theta/2) \sin(\theta/2) e^{-i \phi} \\
\sin(\theta/2) \cos(\theta/2) e^{i \phi} & \sin^2(\theta/2)
\end{bmatrix},
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:ensemblesForSpinOneHalf:160}
\begin{aligned}
\ket{\BS \cdot \ncap ; – } \bra{\BS \cdot \ncap ; – }
&=
\begin{bmatrix}
\sin(\theta/2) e^{-i \phi} \\
-\cos(\theta/2) \\
\end{bmatrix}
\begin{bmatrix}
\sin(\theta/2) e^{i \phi} & -\cos(\theta/2) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\sin^2(\theta/2) & -\cos(\theta/2) \sin(\theta/2) e^{-i \phi} \\
-\cos(\theta/2) \sin(\theta/2) e^{i \phi} & \cos^2(\theta/2)
\end{bmatrix}
\end{aligned}
\end{equation}

With \( C = \cos(\theta/2), S = \sin(\theta/2) \), a general density operator in this basis has the form

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:180}
\begin{aligned}
\rho
&=
w_{+}
\begin{bmatrix}
C^2 & C S e^{-i \phi} \\
S C e^{i \phi} & S^2
\end{bmatrix}
+
w_{-}
\begin{bmatrix}
S^2 & -C S e^{-i \phi} \\
-C S e^{i \phi} & C^2
\end{bmatrix} \\
&=
\begin{bmatrix}
w_{+} C^2 + w_{-} S^2 & (w_{+} – w_{-})C S e^{-i \phi} \\
(w_{+} -w_{-} ) S C e^{i \phi} & w_{+} S^2 + w_{-} C^2
\end{bmatrix}.
\end{aligned}
\end{equation}

The products with the Pauli matrices are

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:200}
\begin{aligned}
\rho \sigma_x
&=
\begin{bmatrix}
w_{+} C^2 + w_{-} S^2 & (w_{+} – w_{-})C S e^{-i \phi} \\
(w_{+} -w_{-} ) S C e^{i \phi} & w_{+} S^2 + w_{-} C^2
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \\
&=
\begin{bmatrix}
(w_{+} – w_{-})C S e^{-i \phi} & w_{+} C^2 + w_{-} S^2 \\
w_{+} S^2 + w_{-} C^2 & (w_{+} -w_{-} ) S C e^{i \phi} \\
\end{bmatrix}
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:220}
\begin{aligned}
\rho \sigma_y
&=
\begin{bmatrix}
w_{+} C^2 + w_{-} S^2 & (w_{+} – w_{-})C S e^{-i \phi} \\
(w_{+} -w_{-} ) S C e^{i \phi} & w_{+} S^2 + w_{-} C^2
\end{bmatrix}
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \\
&=
i
\begin{bmatrix}
(w_{+} – w_{-})C S e^{-i \phi} & -w_{+} C^2 – w_{-} S^2 \\
w_{+} S^2 + w_{-} C^2 & -(w_{+} -w_{-} ) S C e^{i \phi} \\
\end{bmatrix}
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:240}
\begin{aligned}
\rho \sigma_z
&=
\begin{bmatrix}
w_{+} C^2 + w_{-} S^2 & (w_{+} – w_{-})C S e^{-i \phi} \\
(w_{+} -w_{-} ) S C e^{i \phi} & w_{+} S^2 + w_{-} C^2
\end{bmatrix}
\begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\
&=
\begin{bmatrix}
w_{+} C^2 + w_{-} S^2 & -(w_{+} – w_{-})C S e^{-i \phi} \\
(w_{+} -w_{-} ) S C e^{i \phi} & – (w_{+} S^2 + w_{-} C^2)
\end{bmatrix}
\end{aligned}
\end{equation}

The respective traces can be read right off the matrices
\begin{equation}\label{eqn:ensemblesForSpinOneHalf:260}
\begin{aligned}
\textrm{Tr}( \rho \sigma_x ) &= (w_{+} – w_{-}) \sin\theta \cos\phi \\
\textrm{Tr}( \rho \sigma_y ) &= (w_{+} – w_{-}) \sin\theta \sin\phi \\
\textrm{Tr}( \rho \sigma_z ) &= (w_{+} – w_{-}) \cos\theta \\
\end{aligned}.
\end{equation}

This gives

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:280}
(w_{+} – w_{-}) \ncap = \lr{ \textrm{Tr}( \rho \sigma_x ), \textrm{Tr}( \rho \sigma_y ), \textrm{Tr}( \rho \sigma_z ) },
\end{equation}

or

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:281}
\boxed{
w_{\pm} = \frac{1 \pm \sqrt{ \textrm{Tr}^2( \rho \sigma_x ) + \textrm{Tr}^2( \rho \sigma_y ) + \textrm{Tr}^2( \rho \sigma_z )} }{2} .
}
\end{equation}

So, as claimed, it’s possible to completely describe the ensemble weight factors using the ensemble averages of \( [S_x], [S_y], [S_z] \). I used the Pauli matrices instead, but the difference is just an \( \Hbar/2 \) scaling adjustment.

Pure ensemble

It turns out that doing the above is also pr. 3.10(b). Part (a) of that problem is to show how the expectation values \( \expectation{S_x}, \expectation{S_y},\expectation{S_x} \) fully determine the spin orientation for a pure ensemble.

Suppose that the system is in the state \( \ket{\BS \cdot \ncap ; + } \) as defined above, then the expectation values of \( \sigma_x, \sigma_y, \sigma_z \) with respect to this state are

\begin{equation}\label{eqn:ensemblesForSpinOneHalf:300}
\begin{aligned}
\expectation{\sigma_x}
&=
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
\cos(\theta/2) \\
\sin(\theta/2) e^{i \phi}
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix}
\sin(\theta/2) e^{i \phi} \\
\cos(\theta/2) \\
\end{bmatrix} \\
&=
\sin\theta \cos\phi,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:ensemblesForSpinOneHalf:340}
\begin{aligned}
\expectation{\sigma_y}
&=
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
\cos(\theta/2) \\
\sin(\theta/2) e^{i \phi}
\end{bmatrix} \\
&=
i
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix}
-\sin(\theta/2) e^{i \phi} \\
\cos(\theta/2) \\
\end{bmatrix} \\
&=
\sin\theta \sin\phi,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:ensemblesForSpinOneHalf:360}
\begin{aligned}
\expectation{\sigma_z}
&=
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}
\begin{bmatrix}
\cos(\theta/2) \\
\sin(\theta/2) e^{i \phi}
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix}
\cos(\theta/2) \\
-\sin(\theta/2) e^{i \phi}
\end{bmatrix} \\
&=
\cos\theta.
\end{aligned}
\end{equation}

So we have
\begin{equation}\label{eqn:ensemblesForSpinOneHalf:380}
\boxed{
\ncap = \lr{ \expectation{\sigma_x}, \expectation{\sigma_y}, \expectation{\sigma_z} }.
}
\end{equation}

The spin direction is completely determined by this vector of expectation values (or equivalently, the expectation values of \( S_x, S_y, S_z \)).

References

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

Unimodular transformation

October 31, 2015 phy1520 ,

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

Q: Show that

This is ([1] pr. 3.3)

Given the matrix

\begin{equation}\label{eqn:unimodularAndRotation:20}
U =
\frac
{a_0 + i \sigma \cdot \Ba}
{a_0 – i \sigma \cdot \Ba},
\end{equation}

where \( a_0, \Ba \) are real valued constant and vector respectively.

  • Show that this is a unimodular and unitary transformation.
  • A unitary transformation can represent an arbitary rotation. Determine the rotation angle and direction in terms of \( a_0, \Ba \).

A: unimodular

Let’s call these factors \( A_{\pm} \), which expand to

\begin{equation}\label{eqn:unimodularAndRotation:40}
\begin{aligned}
A_{\pm}
&=
a_0 \pm i \sigma \cdot \Ba \\
&=
\begin{bmatrix}
a_0 \pm i a_z & \pm \lr{ a_y + i a_x} \\
\mp (a_y – i a_x) & a_0 \mp i a_z \\
\end{bmatrix},
\end{aligned}
\end{equation}

or with \( z = a_0 + i a_z \), and \( w = a_y + i a_x \), these are

\begin{equation}\label{eqn:unimodularAndRotation:120}
A_{+}
=
\begin{bmatrix}
z & w \\
-w^\conj & z^\conj
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:unimodularAndRotation:180}
A_{-}
=
\begin{bmatrix}
z^\conj & -w \\
w^\conj & z
\end{bmatrix}.
\end{equation}

These both have a determinant of
\begin{equation}\label{eqn:unimodularAndRotation:60}
\begin{aligned}
\Abs{z}^2 + \Abs{w}^2
&=
\Abs{a_0 + i a_z}^2 + \Abs{a_y + i a_x}^2 \\
&= a_0^2 + \Ba^2.
\end{aligned}
\end{equation}

The inverse of the latter is
\begin{equation}\label{eqn:unimodularAndRotation:200}
A_{-}^{-1}
=
\inv{ a_0^2 + \Ba^2 }
\begin{bmatrix}
z & w \\
-w^\conj & z^\conj
\end{bmatrix}
\end{equation}

Noting that the numerator and denominator commute the inverse can be applied in either order. Picking one, the transformation of interest, after writing \( A = a_0^2 + \Ba^2 \), is

\begin{equation}\label{eqn:unimodularAndRotation:100}
\begin{aligned}
U
&=
\inv{A}
\begin{bmatrix}
z & w \\
-w^\conj & z^\conj
\end{bmatrix}
\begin{bmatrix}
z & w \\
-w^\conj & z^\conj
\end{bmatrix} \\
&=
\inv{A}
\begin{bmatrix}
z^2 – \Abs{w}^2 & w( z + z^\conj) \\
-w^\conj (z^\conj + z ) & (z^\conj)^2 – \Abs{w}^2
\end{bmatrix}.
\end{aligned}
\end{equation}

Recall that a unimodular transformation is one that has the form

\begin{equation}\label{eqn:unimodularAndRotation:140}
\begin{bmatrix}
z & w \\
-w^\conj & z^\conj
\end{bmatrix},
\end{equation}

provided \( \Abs{z}^2 + \Abs{w}^2 = 1 \), so \ref{eqn:unimodularAndRotation:100} is unimodular if the following sum is unity, which is the case

\begin{equation}\label{eqn:unimodularAndRotation:160}
\begin{aligned}
\frac{\Abs{z^2 – \Abs{w}^2}^2}{\lr{ \Abs{z}^2 + \Abs{w}^2}^2 } + \Abs{w}^2 \frac{\Abs{z + z^\conj}^2 }{\lr{ \Abs{z}^2 + \Abs{w}^2}^2 }
&=
\frac{
\lr{ z^2 – \Abs{w}^2 } \lr{ (z^\conj)^2 – \Abs{w}^2 }
+ \Abs{w}^2 \lr{ z + z^\conj }^2
}{
\lr{ \Abs{z}^2 + \Abs{w}^2}^2
} \\
&=
\frac{
\Abs{z}^4 + \Abs{w}^4 – \Abs{w}^2 \lr{ {z^2 + (z^\conj)^2} }
+ \Abs{w}^2 \lr{ {z^2 + (z^\conj)^2} + 2 \Abs{z}^2 }
}{
\lr{ \Abs{z}^2 + \Abs{w}^2}^2
} \\
&= 1.
\end{aligned}
\end{equation}

A: rotation

The most general rotation of a vector \( \Ba \), described by Pauli matrices is

\begin{equation}\label{eqn:unimodularAndRotation:220}
e^{i \Bsigma \cdot \ncap \theta/2}
\Bsigma \cdot \Ba
e^{-i \Bsigma \cdot \ncap \theta/2}
=
\Bsigma \cdot \ncap + \lr{ \Bsigma \cdot \Ba – (\Ba \cdot \ncap) \Bsigma \cdot \ncap } \cos \theta + \Bsigma \cdot (\Ba \cross \ncap) \sin\theta.
\end{equation}

If the unimodular matrix above, applied as \( \Bsigma \cdot \Ba’ = U^\dagger \Bsigma \cdot \Ba U \) is to also describe this rotation, we want the equivalence

\begin{equation}\label{eqn:unimodularAndRotation:240}
U = e^{-i \Bsigma \cdot \ncap \theta/2},
\end{equation}

or

\begin{equation}\label{eqn:unimodularAndRotation:260}
\inv{a_0^2 + \Ba^2}
\begin{bmatrix}
a_0^2 – \Ba^2 + 2 i a_0 a_z & 2 a_0 ( a_y + i a_x ) \\
-2 a_0( a_y – i a_x ) & a_0^2 – \Ba^2 – 2 i a_0 a_z
\end{bmatrix}
=
\begin{bmatrix}
\cos(\theta/2) – i n_z \sin(\theta/2) & (-n_y -i n_x) \sin(\theta/2) \\
-( – n_y + i n_x ) \sin(\theta/2) & \cos(\theta/2) + i n_z \sin(\theta/2)
\end{bmatrix}.
\end{equation}

Equating components, that is
\begin{equation}\label{eqn:unimodularAndRotation:280}
\begin{aligned}
\cos(\theta/2) &= \frac{a_0^2 – \Ba^2}{a_0^2 + \Ba^2} \\
-n_x \sin(\theta/2) &= \frac{2 a_0 a_x}{a_0^2 + \Ba^2} \\
-n_y \sin(\theta/2) &= \frac{2 a_0 a_y}{a_0^2 + \Ba^2} \\
-n_z \sin(\theta/2) &= \frac{2 a_0 a_y}{a_0^2 + \Ba^2} \\
\end{aligned}
\end{equation}

Noting that

\begin{equation}\label{eqn:unimodularAndRotation:300}
\begin{aligned}
\sin(\theta/2)
&=
\sqrt{
1 – \frac{(a_0^2 – \Ba^2)^2}{(a_0^2 + \Ba^2)^2}
} \\
&=
\frac{
\sqrt{ (a_0^2 + \Ba^2)^2 – (a_0^2 – \Ba^2)^2 }
}
{
a_0^2 + \Ba^2
} \\
&=
\frac{\sqrt{ 4 a_0^2 \Ba^2 }}{a_0^2 + \Ba^2} \\
&=
\frac{2 a_0 \Abs{\Ba} }{a_0^2 + \Ba^2}
\end{aligned}
\end{equation}

The vector normal direction can be written

\begin{equation}\label{eqn:unimodularAndRotation:320}
\Bn
= – \frac{2 a_0}{(a_0^2 + \Ba^2) \sin(\theta/2)} \Ba,
\end{equation}

or

\begin{equation}\label{eqn:unimodularAndRotation:340}
\boxed{
\Bn = – \frac{\Ba}{\Abs{\Ba}}.
}
\end{equation}

The angle of rotation is

\begin{equation}\label{eqn:unimodularAndRotation:380}
\boxed{
\theta = 2 \tan^{-1} \frac{2 a_0 \Abs{\Ba}}{ a_0^2 – \Ba^2}.
}
\end{equation}

References

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

Some spin problems

October 30, 2015 phy1520 , ,

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

Problems from angular momentum chapter of [1].

Q: \( S_y \) eigenvectors

Find the eigenvectors of \( \sigma_y \), and then find the probability that a measurement of \( S_y \) will be \( \Hbar/2 \) when the state is initially

\begin{equation}\label{eqn:someSpinProblems:20}
\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}
\end{equation}

A:

The eigenvalues should be \( \pm 1 \), which is easily checked

\begin{equation}\label{eqn:someSpinProblems:40}
\begin{aligned}
0
&=
\Abs{ \sigma_y – \lambda } \\
&=
\begin{vmatrix}
-\lambda & -i \\
i & -\lambda
\end{vmatrix} \\
&=
\lambda^2 – 1.
\end{aligned}
\end{equation}

For \( \ket{+} = (a,b)^\T \) we must have

\begin{equation}\label{eqn:someSpinProblems:60}
-1 a – i b = 0,
\end{equation}

so

\begin{equation}\label{eqn:someSpinProblems:80}
\ket{+} \propto
\begin{bmatrix}
-i \\
1
\end{bmatrix},
\end{equation}

or
\begin{equation}\label{eqn:someSpinProblems:100}
\ket{+} =
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
i
\end{bmatrix}.
\end{equation}

For \( \ket{-} \) we must have

\begin{equation}\label{eqn:someSpinProblems:120}
a – i b = 0,
\end{equation}

so

\begin{equation}\label{eqn:someSpinProblems:140}
\ket{+} \propto
\begin{bmatrix}
i \\
1
\end{bmatrix},
\end{equation}

or
\begin{equation}\label{eqn:someSpinProblems:160}
\ket{+} =
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
-i
\end{bmatrix}.
\end{equation}

The normalized eigenvectors are

\begin{equation}\label{eqn:someSpinProblems:180}
\boxed{
\ket{\pm} =
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
\pm i
\end{bmatrix}.
}
\end{equation}

For the probability question we are interested in

\begin{equation}\label{eqn:someSpinProblems:200}
\begin{aligned}
\Abs{\bra{S_y; +}
\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}
}^2
&=
\inv{2} \Abs{
\begin{bmatrix}
1 & -i
\end{bmatrix}
\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}
}^2 \\
&=
\inv{2} \lr{ \Abs{\alpha}^2 + \Abs{\beta}^2 } \\
&=
\inv{2}.
\end{aligned}
\end{equation}

There is a 50 % chance of finding the particle in the \( \ket{S_x;+} \) state, independent of the initial state.

Q: Magnetic Hamiltonian eigenvectors

Using Pauli matrices, find the eigenvectors for the magnetic spin interaction Hamiltonian

\begin{equation}\label{eqn:someSpinProblems:220}
H = – \inv{\Hbar} 2 \mu \BS \cdot \BB.
\end{equation}

A:

\begin{equation}\label{eqn:someSpinProblems:240}
\begin{aligned}
H
&= – \mu \Bsigma \cdot \BB \\
&= – \mu \lr{ B_x \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} + B_y
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} + B_z \begin{bmatrix} 1 & 0
\\ 0 & -1 \\ \end{bmatrix} } \\
&= – \mu
\begin{bmatrix}
B_z & B_x – i B_y \\
B_x + i B_y & -B_z
\end{bmatrix}.
\end{aligned}
\end{equation}

The characteristic equation is
\begin{equation}\label{eqn:someSpinProblems:260}
\begin{aligned}
0
&=
\begin{vmatrix}
-\mu B_z -\lambda & -\mu(B_x – i B_y) \\
-\mu(B_x + i B_y) & \mu B_z – \lambda
\end{vmatrix} \\
&=
-\lr{ (\mu B_z)^2 – \lambda^2 }
– \mu^2\lr{ B_x^2 – (iB_y)^2 } \\
&=
\lambda^2 – \mu^2 \BB^2.
\end{aligned}
\end{equation}

That is
\begin{equation}\label{eqn:someSpinProblems:360}
\boxed{
\lambda = \pm \mu B.
}
\end{equation}

Now for the eigenvectors. We are looking for \( \ket{\pm} = (a,b)^\T \) such that

\begin{equation}\label{eqn:someSpinProblems:300}
0
= (-\mu B_z \mp \mu B) a -\mu(B_x – i B_y) b
\end{equation}

or

\begin{equation}\label{eqn:someSpinProblems:320}
\ket{\pm} \propto
\begin{bmatrix}
B_x – i B_y \\
B_z \pm B
\end{bmatrix}.
\end{equation}

This squares to

\begin{equation}\label{eqn:someSpinProblems:340}
B_x^2 + B_y^2 + B_z^2 + B^2 \pm 2 B B_z
= 2 B( B \pm B_z ),
\end{equation}

so the normalized eigenkets are
\begin{equation}\label{eqn:someSpinProblems:380}
\boxed{
\ket{\pm}
=
\inv{\sqrt{2 B( B \pm B_z )}}
\begin{bmatrix}
B_x – i B_y \\
B_z \pm B
\end{bmatrix}.
}
\end{equation}

References

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

PHY1520H Graduate Quantum Mechanics. Lecture 11: Symmetries in QM. Taught by Prof. Arun Paramekanti

October 29, 2015 phy1520 , ,

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

Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering \textchapref{{4}} [1] content.

Symmetry in classical mechanics

In a classical context considering a Hamiltonian

\begin{equation}\label{eqn:qmLecture11:20}
H(q_i, p_i),
\end{equation}

a symmetry means that certain \( q_i \) don’t appear. In that case the rate of change of one of the generalized momenta is zero

\begin{equation}\label{eqn:qmLecture11:40}
\ddt{p_k} = – \PD{q_k}{H} = 0,
\end{equation}

so \( p_k \) is a constant of motion. This simplifies the problem by reducing the number of degrees of freedom. Another aspect of such a symmetry is that it \underline{relates trajectories}. For example, assuming a rotational symmetry as in fig. 1.

fig. 1. Trajectory under rotational symmetry

fig. 1. Trajectory under rotational symmetry

the trajectory of a particle after rotation is related by rotation to the trajectory of the unrotated particle.

Symmetry in quantum mechanics

Suppose that we have a symmetry operation that takes states from

\begin{equation}\label{eqn:qmLecture11:60}
\ket{\psi} \rightarrow \ket{U \psi}
\end{equation}
\begin{equation}\label{eqn:qmLecture11:80}
\ket{\phi} \rightarrow \ket{U \phi},
\end{equation}

we expect that

\begin{equation}\label{eqn:qmLecture11:100}
\Abs{\braket{ \psi}{\phi} }^2 = \Abs{\braket{ U\psi}{ U\phi} }^2.
\end{equation}

This won’t hold true for a general operator. Two cases where this does hold true is when

  • \( \braket{\psi}{\phi} = \braket{ U\psi}{ U\phi} \). Here \( U \) is unitary, and the equivalence follows from

    \begin{equation}\label{eqn:qmLecture11:120}
    \braket{ U\psi}{ U\phi} = \bra{ \psi} U^\dagger U { \phi} = \bra{ \psi} 1 { \phi} = \braket{\psi}{\phi}.
    \end{equation}

  • \( \braket{\psi}{\phi} = \braket{ U\psi}{ U\phi}^\conj \). Here \( U \) is anti-unitary.

Unitary case

If an “observable” is not changed by a unitary operation representing a symmetry we must have

\begin{equation}\label{eqn:qmLecture11:140}
\bra{\psi} \hat{A} \ket{\psi}
\rightarrow
\bra{U \psi} \hat{A} \ket{U \psi}
=
\bra{\psi} U^\dagger \hat{A} U \ket{\psi},
\end{equation}

so
\begin{equation}\label{eqn:qmLecture11:160}
U^\dagger \hat{A} U = \hat{A},
\end{equation}

or
\begin{equation}\label{eqn:qmLecture11:180}
\boxed{
\hat{A} U = U \hat{A}.
}
\end{equation}

An observable that is unchanged by a unitary symmetry commutes \( \antisymmetric{\hat{A}}{U} \) with the operator \( U \) for that transformation.

Symmetries of the Hamiltonian

Given
\begin{equation}\label{eqn:qmLecture11:200}
\antisymmetric{H}{U} = 0,
\end{equation}

\( H \) is invariant.

Given

\begin{equation}\label{eqn:qmLecture11:220}
H \ket{\phi_n} = \epsilon_n \ket{\phi_n} .
\end{equation}

\begin{equation}\label{eqn:qmLecture11:240}
\begin{aligned}
U H \ket{\phi_n}
&= H U \ket{\phi_n} \\
&= \epsilon_n U \ket{\phi_n}
\end{aligned}
\end{equation}

Such a state

\begin{equation}\label{eqn:qmLecture11:260}
\ket{\psi_n} = U \ket{\phi_n}
\end{equation}

is also an eigenstate with the \underline{same} energy.

Suppose this process is repeated, finding other states

\begin{equation}\label{eqn:qmLecture11:280}
U \ket{\psi_n} = \ket{\chi_n}
\end{equation}
\begin{equation}\label{eqn:qmLecture11:300}
U \ket{\chi_n} = \ket{\alpha_n}
\end{equation}

Because such a transformation only generates states with the initial energy, this process cannot continue forever. At some point this process will enumerate a fixed size set of states. These states can be orthonormalized.

We can say that symmetry operations are generators of a \underlineAndIndex{group}. For a set of symmetry operations we can

  • Form products that lie in a closed set

    \begin{equation}\label{eqn:qmLecture11:320}
    U_1 U_2 = U_3
    \end{equation}

  • can define an inverse
    \begin{equation}\label{eqn:qmLecture11:340}
    U \leftrightarrow U^{-1}.
    \end{equation}

  • obeys associative rules for multiplication
    \begin{equation}\label{eqn:qmLecture11:360}
    U_1 ( U_2 U_3 ) = (U_1 U_2) U_3.
    \end{equation}

  • has an identity operation.

When \( H \) has a symmetry, then degenerate eigenstates form \underlineAndIndex{irreducible} representations (which cannot be further block diagonalized).

Some simple examples

Example: Inversion.

{example:qmLecture11:1}

Given a state and a parity operation \( \hat{\Pi} \), with the transformation

\begin{equation}\label{eqn:qmLecture11:380}
\ket{\psi} \rightarrow \hat{\Pi} \ket{\psi}
\end{equation}

In one dimension, the parity operation is just inversion. In two dimensions, this is a set of flipping operations on two axes fig. 2.

fig. 2. 2D parity operation

fig. 2. 2D parity operation

The operational effects of this operator are

\begin{equation}\label{eqn:qmLecture11:400}
\begin{aligned}
\hat{x} &\rightarrow – \hat{x} \\
\hat{p} &\rightarrow – \hat{p}.
\end{aligned}
\end{equation}

Acting again with the parity operator produces the original value, so it is its own inverse, and \( \hat{\Pi}^\dagger = \hat{\Pi} = \hat{\Pi}^{-1} \). In an expectation value

\begin{equation}\label{eqn:qmLecture11:420}
\bra{ \hat{\Pi} \psi } \hat{x} \ket{ \hat{\Pi} \psi } = – \bra{\psi} \hat{x} \ket{\psi}.
\end{equation}

This means that

\begin{equation}\label{eqn:qmLecture11:440}
\hat{\Pi}^\dagger \hat{x} \hat{\Pi} = – \hat{x},
\end{equation}

or
\begin{equation}\label{eqn:qmLecture11:460}
\hat{x} \hat{\Pi} = – \hat{\Pi} \hat{x},
\end{equation}

\begin{equation}\label{eqn:qmLecture11:480}
\begin{aligned}
\hat{x} \hat{\Pi} \ket{x_0}
&= – \hat{\Pi} \hat{x} \ket{x_0} \\
&= – \hat{\Pi} x_0 \ket{x_0} \\
&= – x_0 \hat{\Pi} \ket{x_0}
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:qmLecture11:500}
\hat{\Pi} \ket{x_0} = \ket{-x_0}.
\end{equation}

Acting on a wave function

\begin{equation}\label{eqn:qmLecture11:520}
\begin{aligned}
\bra{x} \hat{\Pi} \ket{\psi}
&=
\braket{-x}{\psi} \\
&= \psi(-x).
\end{aligned}
\end{equation}

What does this mean for eigenfunctions. Eigenfunctions are supposed to form irreducible representations of the group. The group has just two elements

\begin{equation}\label{eqn:qmLecture11:540}
\setlr{ 1, \hat{\Pi} },
\end{equation}

where \( \hat{\Pi}^2 = 1 \).

Suppose we have a Hamiltonian

\begin{equation}\label{eqn:qmLecture11:560}
H = \frac{\hat{p}^2}{2m} + V(\hat{x}),
\end{equation}

where \( V(\hat{x}) \) is even ( \( \antisymmetric{V(\hat{x})}{\hat{\Pi} } = 0 \) ). The squared momentum commutes with the parity operator

\begin{equation}\label{eqn:qmLecture11:580}
\begin{aligned}
\antisymmetric{\hat{p}^2}{\hat{\Pi}}
&=
\hat{p}^2 \hat{\Pi}
– \hat{\Pi} \hat{p}^2 \\
&=
\hat{p}^2 \hat{\Pi}
– (\hat{\Pi} \hat{p}) \hat{p} \\
&=
\hat{p}^2 \hat{\Pi}
-(- \hat{p} \hat{\Pi}) \hat{p} \\
&=
\hat{p}^2 \hat{\Pi}
+ \hat{p} (-\hat{p} \hat{\Pi}) \\
&=
0.
\end{aligned}
\end{equation}

Only two functions are possible in the symmetry set \( \setlr{ \Psi(x), \hat{\Pi} \Psi(x) } \), since

\begin{equation}\label{eqn:qmLecture11:600}
\begin{aligned}
\hat{\Pi}^2 \Psi(x)
&= \hat{\Pi} \Psi(-x) \\
&= \Psi(x).
\end{aligned}
\end{equation}

This symmetry severely restricts the possible solutions, making it so there can be only one dimensional forms of this problem with solutions that are either even or odd respectively

\begin{equation}\label{eqn:qmLecture11:620}
\begin{aligned}
\phi_e(x) &= \psi(x ) + \psi(-x) \\
\phi_o(x) &= \psi(x ) – \psi(-x).
\end{aligned}
\end{equation}

References

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