dispersion

Second update of aggregate notes for phy1520, Graduate Quantum Mechanics

October 20, 2015 phy1520 No comments , , , , , , , , , , , ,

I’ve posted a second update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains lecture notes up to lecture 9, my ungraded solutions for the second problem set, and some additional worked practise problems.

Most of the content was posted individually in the following locations, but those original documents will not be maintained individually any further.

Time evolution of spin half probability and dispersion

October 15, 2015 phy1520 No comments , , ,

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

Question: Time evolution of spin half probability and dispersion ([1] pr. 2.3)

A spin \( 1/2 \) system \( \BS \cdot \ncap \), with \( \ncap = \sin \beta \xcap + \cos\beta \zcap \), is in state with eigenvalue \( \Hbar/2 \), acted on by a magnetic field of strength \( B \) in the \( +z \) direction.

(a)

If \( S_x \) is measured at time \( t \), what is the probability of getting \( + \Hbar/2 \)?

(b)

Evaluate the dispersion in \( S_x \) as a function of t, that is,

\begin{equation}\label{eqn:spinTimeEvolution:20}
\expectation{\lr{ S_x – \expectation{S_x}}^2}.
\end{equation}

(c)

Check your answers for \( \beta \rightarrow 0, \pi/2 \) to see if they make sense.

Answer

(a)

The spin operator in matrix form is
\begin{equation}\label{eqn:spinTimeEvolution:40}
\begin{aligned}
S \cdot \ncap
&=
\frac{\Hbar}{2} \lr{ \sigma_z \cos\beta + \sigma_x \sin\beta } \\
&=
\frac{\Hbar}{2} \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \cos\beta + \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \sin\beta } \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\beta & \sin\beta \\
\sin\beta & -\cos\beta
\end{bmatrix}.
\end{aligned}
\end{equation}

The \( \ket{S \cdot \ncap ; + } \) eigenstate is found from

\begin{equation}\label{eqn:spinTimeEvolution:60}
\lr{ S \cdot \ncap – \Hbar/2}
\begin{bmatrix}
a \\
b
\end{bmatrix}
= 0,
\end{equation}

or

\begin{equation}\label{eqn:spinTimeEvolution:80}
\begin{aligned}
0
&=
\lr{ \cos\beta – 1 } a + \sin\beta b \\
&=
\lr{ -2 \sin^2(\beta/2) } a + 2 \sin(\beta/2) \cos(\beta/2) b \\
&=
\lr{ – \sin(\beta/2) } a + \cos(\beta/2) b,
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:spinTimeEvolution:100}
\ket{ S \cdot \ncap ; + }
=
\begin{bmatrix}
\cos(\beta/2) \\
\sin(\beta/2) \\
\end{bmatrix}.
\end{equation}

The Hamiltonian is

\begin{equation}\label{eqn:spinTimeEvolution:120}
H
= – \frac{e B}{m c} S_z
= – \frac{e B \Hbar}{2 m c} \sigma_z,
\end{equation}

so the time evolution operator is

\begin{equation}\label{eqn:spinTimeEvolution:140}
U
= e^{-i H t/\Hbar}
= e^{ \frac{i e B t }{2 m c} \sigma_z }.
\end{equation}

Let \( \omega = e B/(2 m c) \), so

\begin{equation}\label{eqn:spinTimeEvolution:160}
\begin{aligned}
U
&=
e^{i \sigma_z \omega t} \\
&=
\cos(\omega t) + i \sigma_z \sin(\omega t) \\
&=
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\cos(\omega t)
+
i \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \sin(\omega t) \\
&=
\begin{bmatrix}
e^{i \omega t} & 0 \\
0 & e^{-i \omega t}
\end{bmatrix}.
\end{aligned}
\end{equation}

The time evolution of the initial state is

\begin{equation}\label{eqn:spinTimeEvolution:180}
\begin{aligned}
\ket{S \cdot \ncap ; + }(t)
&=
U \ket{S \cdot \ncap ; + }(0) \\
&=
\begin{bmatrix}
e^{i \omega t} & 0 \\
0 & e^{-i \omega t}
\end{bmatrix}
\begin{bmatrix}
\cos(\beta/2) \\
\sin(\beta/2) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\beta/2) e^{i \omega t} \\
\sin(\beta/2) e^{-i \omega t} \\
\end{bmatrix}.
\end{aligned}
\end{equation}

The probability of finding the state in \( \ket{S \cdot \xcap ; + } \) at time \( t \) (i.e. measuring \( S_x \) and finding \( \Hbar/2 \)) is

\begin{equation}\label{eqn:spinTimeEvolution:200}
\begin{aligned}
\Abs{\braket{S \cdot \xcap ; + }{S \cdot \ncap ; + }}^2
&=
\Abs{\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
\cos(\beta/2) e^{i \omega t} \\
\sin(\beta/2) e^{-i \omega t} \\
\end{bmatrix}
}^2 \\
&=
\inv{2}
\Abs{
\cos(\beta/2) e^{i \omega t} +
\sin(\beta/2) e^{-i \omega t} }^2 \\
&=
\inv{2} \lr{ 1 + 2 \cos(\beta/2) \sin(\beta/2) \cos(2 \omega t) } \\
&=
\inv{2} \lr{ 1 + \sin(\beta) \cos( 2 \omega t) }.
\end{aligned}
\end{equation}

(b)

To calculate the dispersion first note that

\begin{equation}\label{eqn:spinTimeEvolution:300}
S_x^2
= \lr{ \frac{\Hbar}{2} }^2 \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}^2
= \lr{ \frac{\Hbar}{2} }^2,
\end{equation}

so only the first order expectation is non-trivial to calculate. That is

\begin{equation}\label{eqn:spinTimeEvolution:320}
\begin{aligned}
\expectation{S_x}
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos(\beta/2) e^{-i \omega t} &
\sin(\beta/2) e^{i \omega t}
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
\cos(\beta/2) e^{i \omega t} \\
\sin(\beta/2) e^{-i \omega t} \\
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos(\beta/2) e^{-i \omega t} &
\sin(\beta/2) e^{i \omega t}
\end{bmatrix}
\begin{bmatrix}
\sin(\beta/2) e^{-i \omega t} \\
\cos(\beta/2) e^{i \omega t} \\
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\sin(\beta/2) \cos(\beta/2) \lr{ e^{-2 i \omega t} + e^{ 2 i \omega t} } \\
&=
\frac{\Hbar}{2} \sin\beta \cos( 2 \omega t ).
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:spinTimeEvolution:340}
\boxed{
\expectation{(\Delta S_x)^2}
=
\lr{ \frac{\Hbar}{2} }^2 \lr{ 1 – \sin^2\beta \cos^2( 2 \omega t ) }.
}
\end{equation}

(c)

For \( \beta = 0 \), \( \ncap = \zcap \), and \( \beta = \pi/2 \), \( \ncap = \xcap \). For the first case, the state is in an eigenstate of \( S_z \), so must evolve as

\begin{equation}\label{eqn:spinTimeEvolution:220}
\ket{S \cdot \ncap ; + }(t) = \ket{S \cdot \ncap ; + }(0) e^{i \omega t}.
\end{equation}

The probability of finding it in state \( \ket{S \cdot \xcap ; + } \) is therefore

\begin{equation}\label{eqn:spinTimeEvolution:240}
\begin{aligned}
\Abs{
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
e^{i \omega t} \\
0
\end{bmatrix}
}^2
&=
\inv{2} \Abs{ e^{i\omega t} }^2 \\
&=
\inv{2} \\
&=
\inv{2} \lr{ 1 + \sin(0) \cos(2 \omega t) }.
\end{aligned}
\end{equation}

This matches \ref{eqn:spinTimeEvolution:200} as expected.

For \( \beta = \pi/2 \) we have

\begin{equation}\label{eqn:spinTimeEvolution:260}
\begin{aligned}
\ket{S \cdot \xcap ; + }(t)
&=
\inv{\sqrt{2}}
\begin{bmatrix}
e^{i \omega t} & 0 \\
0 & e^{-i \omega t}
\end{bmatrix}
\begin{bmatrix}
1 \\
1
\end{bmatrix} \\
&=
\inv{\sqrt{2}}
\begin{bmatrix}
e^{i \omega t} \\
e^{-i \omega t}
\end{bmatrix}.
\end{aligned}
\end{equation}

The probability for the \( \Hbar/2 \) \( S_x \) measurement at time \( t \) is
\begin{equation}\label{eqn:spinTimeEvolution:280}
\begin{aligned}
\Abs{
\inv{2}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
e^{i \omega t} \\
e^{-i \omega t}
\end{bmatrix}
}^2
&=
\inv{4} \Abs{ e^{i \omega t} + e^{-i \omega t} }^2 \\
&=
\cos^2(\omega t) \\
&=
\inv{2}\lr{ 1 + \sin(\pi/2) \cos( 2 \omega t )}.
\end{aligned}
\end{equation}

Again, this matches the expected value.

For the dispersions, at \( \beta = 0 \), the dispersion is

\begin{equation}\label{eqn:spinTimeEvolution:360}
\lr{\frac{\Hbar}{2}}^2
\end{equation}

This is the maximum dispersion, which makes sense since we are measuring \( S_x \) when the initial state is \( \ket{S \cdot \zcap ; + } \). For \( \beta = \pi/2 \) the dispersion is

\begin{equation}\label{eqn:spinTimeEvolution:380}
\lr{\frac{\Hbar}{2}}^2 \sin^2 ( 2 \omega t ).
\end{equation}

This starts off as zero dispersion (because the initial state is \( \ket{ S \cdot \xcap ; + } \), but then oscillates.

References

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