phy1520

Time evolution of spin half probability and dispersion

October 15, 2015 phy1520 , , ,

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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.

PHY1520H Graduate Quantum Mechanics. Lecture 9: Dirac equation (cont.). Taught by Prof. Arun Paramekanti

October 15, 2015 phy1520 , , , , , , , ,

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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.

Where we left off

\begin{equation}\label{eqn:qmLecture9:20}
-i \Hbar \PD{t}{}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
=
\begin{bmatrix}
-i \Hbar c \PD{x}{} & m c^2 \\
m c^2 & i \Hbar c \PD{x}{} \\
\end{bmatrix}.
\end{equation}

With a potential this would be

\begin{equation}\label{eqn:qmLecture9:40}
-i \Hbar \PD{t}{}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
=
\begin{bmatrix}
-i \Hbar c \PD{x}{} + V(x) & m c^2 \\
m c^2 & i \Hbar c \PD{x}{} + V(x) \\
\end{bmatrix}.
\end{equation}

This means that the potential is raising the energy eigenvalue of the system.

Free Particle

Assuming a form

\begin{equation}\label{eqn:qmLecture9:60}
\begin{bmatrix}
\psi_1(x,t) \\
\psi_2(x,t)
\end{bmatrix}
=
e^{i k x}
\begin{bmatrix}
f_1(t) \\
f_2(t) \\
\end{bmatrix},
\end{equation}

and plugging back into the Dirac equation we have

\begin{equation}\label{eqn:qmLecture9:80}
-i \Hbar \PD{t}{}
\begin{bmatrix}
f_1 \\
f_2
\end{bmatrix}
=
\begin{bmatrix}
k \Hbar c & m c^2 \\
m c^2 & – \Hbar k c \\
\end{bmatrix}
\begin{bmatrix}
f_1 \\
f_2
\end{bmatrix}.
\end{equation}

We can use a diagonalizing rotation

\begin{equation}\label{eqn:qmLecture9:100}
\begin{bmatrix}
f_1 \\
f_2
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta_k & -\sin\theta_k \\
\sin\theta_k & \cos\theta_k \\
\end{bmatrix}
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}.
\end{equation}

Plugging this in reduces the system to the form

\begin{equation}\label{eqn:qmLecture9:140}
-i \Hbar \PD{t}{}
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}
=
\begin{bmatrix}
E_k & 0 \\
0 & -E_k
\end{bmatrix}
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}.
\end{equation}

Where the rotation angle is found to be given by

\begin{equation}\label{eqn:qmLecture9:160}
\begin{aligned}
\sin(2 \theta_k) &= \frac{m c^2}{\sqrt{(\Hbar k c)^2 + m^2 c^4}} \\
\cos(2 \theta_k) &= \frac{\Hbar k c}{\sqrt{(\Hbar k c)^2 + m^2 c^4}} \\
E_k &= \sqrt{(\Hbar k c)^2 + m^2 c^4}
\end{aligned}
\end{equation}

See fig. 1 for a sketch of energy vs momentum. The asymptotes are the limiting cases when \( m c^2 \rightarrow 0 \). The \( + \) branch is what we usually associate with particles. What about the other energy states. For Fermions Dirac argued that the lower energy states could be thought of as “filled up”, using the Pauli principle to leave only the positive energy states available. This was called the “Dirac Sea”. This isn’t a good solution, and won’t work for example for Bosons.

fig. 1. Dirac equation solution space

fig. 1. Dirac equation solution space

Another way to rationalize this is to employ ideas from solid state theory. For example consider a semiconductor with a valence and conduction band as sketched in fig. 2.

fig. 2. Solid state valence and conduction band transition

fig. 2. Solid state valence and conduction band transition

A photon can excite an electron from the valence band to the conduction band, leaving all the valence band states filled except for one (a hole). For an electron we can use almost the same picture, as sketched in fig. 3.

fig. 3. Pair creation

fig. 3. Pair creation

A photon with energy \( E_k – (-E_k) \) can create a positron-electron pair from the vacuum, where the energy of the electron and positron pair is \( E_k \).

At high enough energies, we can see this pair creation occur.

Zitterbewegung

If a particle is created at a non-eigenstate such as one on the asymptotes, then oscillations between the positive and negative branches are possible as sketched in fig. 4.

fig. 4. Zitterbewegung oscillation

fig. 4. Zitterbewegung oscillation

Only “vertical” oscillations between the positive and negative locations on these branches is possible since those are the points that match the particle momentum. Examining this will be the aim of one of the problem set problems.

Probability and current density

If we define a probability density

\begin{equation}\label{eqn:qmLecture9:180}
\rho(x, t) = \Abs{\psi_1}^2 + \Abs{\psi_2}^2,
\end{equation}

does this satisfy a probability conservation relation

\begin{equation}\label{eqn:qmLecture9:200}
\PD{t}{\rho} + \PD{x}{j} = 0,
\end{equation}

where \( j \) is the probability current. Plugging in the density, we have

\begin{equation}\label{eqn:qmLecture9:220}
\PD{t}{\rho}
=
\PD{t}{\psi_1^\conj} \psi_1
+
\psi_1^\conj \PD{t}{\psi_1}
+
\PD{t}{\psi_2^\conj} \psi_2
+
\psi_2^\conj \PD{t}{\psi_2}.
\end{equation}

It turns out that the probability current has the form

\begin{equation}\label{eqn:qmLecture9:240}
j(x,t) = c \lr{ \psi_1^\conj \psi_1 + \psi_2^\conj \psi_2 }.
\end{equation}

Here the speed of light \( c \) is the slope of the line in the plots above. We can think of this current density as right movers minus the left movers. Any state that is given can be thought of as a combination of right moving and left moving states, neither of which are eigenstates of the free particle Hamiltonian.

Potential step

The next logical thing to think about, as in non-relativistic quantum mechanics, is to think about what occurs when the particle hits a potential step, as in fig. 5.

fig. 5. Reflection off a potential barrier

fig. 5. Reflection off a potential barrier

The approach is the same. We write down the wave functions for the \( V = 0 \) region (I), and the higher potential region (II).

The eigenstates are found on the solid lines above the asymptotes on the right hand movers side as sketched in fig. 6. The right and left moving designations are based on the phase velocity \( \PDi{k}{E} \) (approaching \( \pm c \) on the top-right and top-left quadrants respectively).

fig. 6. Right movers and left movers

fig. 6. Right movers and left movers

For \( k > 0 \), an eigenstate for the incident wave is

\begin{equation}\label{eqn:qmLecture9:261}
\Bpsi_{\textrm{inc}}(x) =
\begin{bmatrix}
\cos\theta_k \\
\sin\theta_k
\end{bmatrix}
e^{i k x},
\end{equation}

For the reflected wave function, we pick a function on the left moving side of the positive energy branch.

\begin{equation}\label{eqn:qmLecture9:260}
\Bpsi_{\textrm{ref}}(x) =
\begin{bmatrix}
? \\
?
\end{bmatrix}
e^{-i k x},
\end{equation}

We’ll go through this in more detail next time.

Question: Calculate the right going diagonalization

Prove (7).

Answer

To determine the relations for \( \theta_k \) we have to solve

\begin{equation}\label{eqn:qmLecture9:280}
\begin{bmatrix}
E_k & 0 \\
0 & -E_k
\end{bmatrix}
= R^{-1} H R.
\end{equation}

Working with \( \Hbar = c = 1 \) temporarily, and \( C = \cos\theta_k, S = \sin\theta_k \), that is

\begin{equation}\label{eqn:qmLecture9:300}
\begin{aligned}
\begin{bmatrix}
E_k & 0 \\
0 & -E_k
\end{bmatrix}
&=
\begin{bmatrix}
C & S \\
-S & C
\end{bmatrix}
\begin{bmatrix}
k & m \\
m & -k
\end{bmatrix}
\begin{bmatrix}
C & -S \\
S & C
\end{bmatrix} \\
&=
\begin{bmatrix}
C & S \\
-S & C
\end{bmatrix}
\begin{bmatrix}
k C + m S & -k S + m C \\
m C – k S & -m S – k C
\end{bmatrix} \\
&=
\begin{bmatrix}
k C^2 + m S C + m C S – k S^2 & -k S C + m C^2 -m S^2 – k C S \\
-k C S – m S^2 + m C^2 – k S C & k S^2 – m C S -m S C – k C^2
\end{bmatrix} \\
&=
\begin{bmatrix}
k \cos(2 \theta_k) + m \sin(2 \theta_k) & m \cos(2 \theta_k) – k \sin(2 \theta_k) \\
m \cos(2 \theta_k) – k \sin(2 \theta_k) & -k \cos(2 \theta_k) – m \sin(2 \theta_k) \\
\end{bmatrix}.
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:qmLecture9:320}
\begin{aligned}
E_k
\begin{bmatrix}
1 \\
0
\end{bmatrix}
&=
\begin{bmatrix}
k \cos(2 \theta_k) + m \sin(2 \theta_k) \\
m \cos(2 \theta_k) – k \sin(2 \theta_k) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
k & m \\
m & -k
\end{bmatrix}
\begin{bmatrix}
\cos(2 \theta_k) \\
\sin(2 \theta_k) \\
\end{bmatrix}.
\end{aligned}
\end{equation}

Adding back in the \(\Hbar\)’s and \(c\)’s this is

\begin{equation}\label{eqn:qmLecture9:340}
\begin{aligned}
\begin{bmatrix}
\cos(2 \theta_k) \\
\sin(2 \theta_k) \\
\end{bmatrix}
&=
\frac{E_k}{-(\Hbar k c)^2 -(m c^2)^2}
\begin{bmatrix}
– \Hbar k c & – m c^2 \\
– m c^2 & \Hbar k c
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\inv{E_k}
\begin{bmatrix}
\Hbar k c \\
m c^2
\end{bmatrix}.
\end{aligned}
\end{equation}

Question: Verify the Dirac current relationship.

Prove \ref{eqn:qmLecture9:240}.

Answer

The components of the Schrodinger equation are

\begin{equation}\label{eqn:qmLecture9:360}
\begin{aligned}
-i \Hbar \PD{t}{\psi_1} &= -i \Hbar c \PD{x}{\psi_1} + m c^2 \psi_2 \\
-i \Hbar \PD{t}{\psi_2} &= m c^2 \psi_1 + i \Hbar c \PD{x}{\psi_2},
\end{aligned}
\end{equation}

The conjugates of these are
\begin{equation}\label{eqn:qmLecture9:380}
\begin{aligned}
i \Hbar \PD{t}{\psi_1^\conj} &= i \Hbar c \PD{x}{\psi_1^\conj} + m c^2 \psi_2^\conj \\
i \Hbar \PD{t}{\psi_2^\conj} &= m c^2 \psi_1^\conj – i \Hbar c \PD{x}{\psi_2^\conj}.
\end{aligned}
\end{equation}

This gives
\begin{equation}\label{eqn:qmLecture9:400}
\begin{aligned}
i \Hbar \PD{t}{\rho}
&=
\lr{ i \Hbar c \PD{x}{\psi_1^\conj} + m c^2 \psi_2^\conj } \psi_1 \\
&+ \psi_1^\conj \lr{ i \Hbar c \PD{x}{\psi_1} – m c^2 \psi_2 } \\
&+ \lr{ m c^2 \psi_1^\conj – i \Hbar c \PD{x}{\psi_2^\conj} } \psi_2 \\
&+ \psi_2^\conj \lr{ -m c^2 \psi_1 – i \Hbar c \PD{x}{\psi_2} }.
\end{aligned}
\end{equation}

All the non-derivative terms cancel leaving

\begin{equation}\label{eqn:qmLecture9:420}
\inv{c} \PD{t}{\rho}
=
\PD{x}{\psi_1^\conj} \psi_1
+ \psi_1^\conj \PD{x}{\psi_1}
– \PD{x}{\psi_2^\conj} \psi_2
– \psi_2^\conj \PD{x}{\psi_2}
=
\PD{x}{}
\lr{
\psi_1^\conj \psi_1 –
\psi_2^\conj \psi_2
}.
\end{equation}

Expectations for SHO Hamiltonian, and virial theorem.

October 15, 2015 phy1520 , , , , ,

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Question: Expectations for SHO Hamiltonian, and virial theorem. ([1] pr. 2.3)

(a)

For a 1D SHO, compute \(
\bra{m} x \ket{n},
\bra{m} x^2 \ket{n},
\bra{m} p \ket{n},
\bra{m} p^2 \ket{n} \) and \( \bra{m} \symmetric{x}{p} \ket{n} \).

(b)

Verify the virial theorem is satisfied for energy eigenstates.

Answer

(a)

Using

\begin{equation}\label{eqn:shoExpectations:20}
\begin{aligned}
x &= \frac{x_0}{\sqrt{2}} \lr{ a + a^\dagger } \\
p &= \frac{i\Hbar}{x_0 \sqrt{2}} \lr{ a^\dagger – a} \\
a(t) &= a(0) e^{-i \omega t} \\
a(0) \ket{n} &= \sqrt{n} \ket{n-1} \\
a^\dagger(0) \ket{n} &= \sqrt{n+1} \ket{n+1} \\
x_0^2 &= \frac{\Hbar}{\omega m},
\end{aligned}
\end{equation}

we have

\begin{equation}\label{eqn:shoExpectations:40}
\begin{aligned}
\bra{m} x \ket{n}
&=
\frac{x_0}{\sqrt{2}} \bra{m} \lr{ a + a^\dagger } \ket{n} \\
&=
\frac{x_0}{\sqrt{2}} \bra{m}
\lr{
e^{-i \omega t} \sqrt{n} \ket{n-1}
+
e^{i \omega t} \sqrt{n+1} \ket{n+1}
} \\
&=
\frac{x_0}{\sqrt{2}} \lr{
\delta_{m, n-1} e^{-i \omega t} \sqrt{n}
+
\delta_{m, n+1} e^{i \omega t} \sqrt{n+1}
},
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:shoExpectations:60}
\begin{aligned}
\bra{m} x^2 \ket{n}
&=
\frac{x_0^2}{2} \bra{m} \lr{ a + a^\dagger }^2 \ket{n} \\
&=
\frac{x_0^2}{2}
\lr{
e^{i \omega t} \sqrt{m} \bra{m-1}
+
e^{-i \omega t} \sqrt{m+1} \bra{m+1}
}
\lr{
e^{-i \omega t} \sqrt{n} \ket{n-1}
+
e^{i \omega t} \sqrt{n+1} \ket{n+1}
} \\
&=
\frac{x_0^2}{2}
\lr{
\delta_{m+1,n+1} \sqrt{(m+1)(n+1)}
+\delta_{m+1,n-1} \sqrt{(m+1)n} e^{-2 i \omega t}
+\delta_{m-1,n+1} \sqrt{m(n+1)} e^{2 i \omega t}
+\delta_{m-1,n-1} \sqrt{m n}
},
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:shoExpectations:80}
\begin{aligned}
\bra{m} p \ket{n}
&=
\frac{i\Hbar}{\sqrt{2} x_0} \bra{m} \lr{ a^\dagger – a} \ket{n} \\
&=
\frac{i\Hbar}{\sqrt{2} x_0} \bra{m} \lr{
e^{i \omega t} \sqrt{n+1} \ket{n+1}
–
e^{-i \omega t} \sqrt{n} \ket{n-1}
} \\
&=
\frac{i\Hbar}{\sqrt{2} x_0} \lr{
\delta_{m,n+1} e^{i \omega t} \sqrt{n+1}
–
\delta_{m,n-1} e^{-i \omega t} \sqrt{n}
},
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:shoExpectations:100}
\begin{aligned}
\bra{m} p^2 \ket{n}
&=
\frac{\Hbar^2}{2 x_0^2} \ket{m} \lr{ a – a^\dagger } \lr{ a^\dagger – a}
\ket{n} \\
&=
\frac{\Hbar^2}{2 x_0^2}
\lr{
-e^{-i \omega t} \sqrt{m+1} \bra{m+1}
+
e^{i \omega t} \sqrt{m} \bra{m-1}
}
\lr{
e^{i \omega t} \sqrt{n+1} \ket{n+1}
–
e^{-i \omega t} \sqrt{n} \ket{n-1}
} \\
&=
\frac{\Hbar^2}{2 x_0^2}
\lr{
\delta_{m+1,n+1} \sqrt{(m+1)(n+1)}
+\delta_{m+1,n-1} \sqrt{(m+1)n} e^{-2 i \omega t}
+\delta_{m-1,n+1} \sqrt{m(n+1)} e^{2 i \omega t}
+\delta_{m-1,n-1} \sqrt{m n}
}.
\end{aligned}
\end{equation}

For the anticommutator \( \symmetric{x}{p} \), we have

\begin{equation}\label{eqn:shoExpectations:120}
\begin{aligned}
\symmetric{x}{p}
&=
\frac{i\Hbar}{2}
\lr{
\lr{ a e^{-i \omega t} + a^\dagger e^{i \omega t} } \lr{ a^\dagger e^{i \omega t} – a e^{-i \omega t} }
–
\lr{ a^\dagger e^{i \omega t} – a e^{-i \omega t} }
\lr{ a e^{-i \omega t} + a^\dagger e^{i \omega t} }
} \\
&=
\frac{i\Hbar}{2}
\lr{
– a^2 e^{- 2 i \omega t}
+ (a^\dagger)^2 e^{ 2 i \omega t}
+ a a^\dagger
– a^\dagger a
+ a^2 e^{- 2 i \omega t}
– (a^\dagger)^2 e^{ 2 i \omega t}
– a^\dagger a
+ a a^\dagger
} \\
&=
i\Hbar
\lr{
a a^\dagger – a^\dagger a
},
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:shoExpectations:140}
\begin{aligned}
\bra{m} \symmetric{x}{p} \ket{n}
&=
i\Hbar
\bra{m}
\lr{
a a^\dagger – a^\dagger a
}
\ket{n} \\
&=
i\Hbar
\bra{m}
\lr{
\sqrt{(n+1)^2}\ket{n}
-\sqrt{n^2}\ket{n}
} \\
&=
i\Hbar
\bra{m}
\lr{
2 n + 1
}
\ket{n}.
\end{aligned}
\end{equation}

(b)

For the SHO, the virial theorem requires \( \expectation{p^2/m} = \expectation{m \omega x^2} \). That momentum expectation with respect to the eigenstate \( \ket{n} \) is

\begin{equation}\label{eqn:shoExpectations:160}
\begin{aligned}
\expectation{p^2/m}
&=
\frac{\Hbar^2}{2 x_0^2 m}
\lr{
\sqrt{(n+1)(n+1)}
+
\sqrt{n n}
} \\
&=
\frac{\Hbar^2 m \omega}{2 \Hbar m} \lr{ 2 n + 1 } \\
&=
\Hbar \omega \lr{ n + \inv{2} }.
\end{aligned}
\end{equation}

For the position expectation we’ve got

\begin{equation}\label{eqn:shoExpectations:180}
\begin{aligned}
\expectation{m \omega x^2}
&=
\frac{m \omega^2 x_0^2}{2}
\lr{
\sqrt{(n+1)(n+1)}
+ \sqrt{n n}
} \\
&=
\frac{m \omega^2 \Hbar}{2 m \omega}
\lr{
\sqrt{(n+1)(n+1)}
+ \sqrt{n n}
} \\
&=
\frac{\omega \Hbar}{2 }
\lr{ 2 n + 1 } \\
&=
\omega \Hbar
\lr{ n + \inv{2} }.
\end{aligned}
\end{equation}

This shows that the virial theorem holds for the SHO Hamiltonian for eigenstates.

References

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

PHY1520H Graduate Quantum Mechanics. Lecture 8: Dirac equation in 1D. Taught by Prof. Arun Paramekanti

October 13, 2015 phy1520 , , , ,

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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.

Schrodinger Derivation

Recall that a “derivation” of the Schrodinger equation can be associated with the following equivalences

\begin{equation}\label{eqn:qmLecture8:300}
E \leftrightarrow \Hbar \omega \leftrightarrow i \Hbar \PD{t}{}
\end{equation}
\begin{equation}\label{eqn:qmLecture8:320}
p \leftrightarrow \Hbar k \leftrightarrow -i \Hbar \PD{t}{}
\end{equation}

so that the classical energy relationship

\begin{equation}\label{eqn:qmLecture8:20}
E = \frac{p^2}{2m}
\end{equation}

takes the form

\begin{equation}\label{eqn:qmLecture8:40}
i \Hbar \PD{t}{} = -\frac{\Hbar^2}{2m}.
\end{equation}

How do we do this in a relativistic context where the energy momentum relationship is

\begin{equation}\label{eqn:qmLecture8:60}
E = \sqrt{ p^2 c^2 + m^2 c^4 } \approx m c^2 + \frac{p^2}{2m} + \cdots
\end{equation}

where \( m \) is the rest mass and \( c \) is the speed of light.

Attempt I

\begin{equation}\label{eqn:qmLecture8:80}
E = m c^2 + \frac{p^2}{2m} + (…) p^4 + (…) p^6 + \cdots
\end{equation}

First order in time, but infinite order in space \( \partial/\partial x \). Useless.

Attempt II

\begin{equation}\label{eqn:qmLecture8:100}
E^2 = p^2 c^2 + m^2 c^4.
\end{equation}

This gives

\begin{equation}\label{eqn:qmLecture8:120}
-\Hbar^2 \PDSq{t}{\psi} = – \Hbar^2 c^2 \PDSq{x}{\psi} + m^2 c^4 \psi.
\end{equation}

This is the Klein-Gordon equation, which is second order in time.

Attempt III

Suppose that we have the matrix

\begin{equation}\label{eqn:qmLecture8:140}
\begin{bmatrix}
p c & m c^2 \\
m c^2 & – p c
\end{bmatrix},
\end{equation}

or

\begin{equation}\label{eqn:qmLecture8:160}
\begin{bmatrix}
m c^2 & i p c \\
-i p c & – m c^2
\end{bmatrix},
\end{equation}

These both happen to have eigenvalues \( \lambda_{\pm} = \pm \sqrt{p^2 c^2} \). For those familiar with the Dirac matrices, this amounts to a choice for different representations of the gamma matrices.

Working with \ref{eqn:qmLecture8:140}, which has some nicer features than other possible representations, we seek a state

\begin{equation}\label{eqn:qmLecture8:180}
\Bpsi =
\begin{bmatrix}
\psi_1(x, t) \\
\psi_2(x, t) \\
\end{bmatrix},
\end{equation}

where we aim to write down an equation for this composite state.

\begin{equation}\label{eqn:qmLecture8:200}
i \Hbar \PD{t}{\Bpsi} = \BH \Bpsi
\end{equation}

Assuming the matrix is the Hamiltonian, multiplying that with the composite state gives

\begin{equation}\label{eqn:qmLecture8:220}
\begin{aligned}
\begin{bmatrix}
i \Hbar \PD{t}{\psi_1} \\
i \Hbar \PD{t}{\psi_1}
\end{bmatrix}
&=
\begin{bmatrix}
\hat{p} c & m c^2 \\
m c^2 & – \hat{p} c
\end{bmatrix}
\begin{bmatrix}
\psi_1(x, t) \\
\psi_2(x, t) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\hat{p} c \psi_1 + m c^2 \psi_2 \\
m c^2 \psi_1 – \hat{p} c \psi_2
\end{bmatrix}.
\end{aligned}
\end{equation}

What happens when we square this?

\begin{equation}\label{eqn:qmLecture8:240}
\begin{aligned}
\lr{ i \Hbar \PD{t}{} }^2 \Bpsi
&= \BH \BH \Bpsi \\
&=
\begin{bmatrix}
\hat{p} c & m c^2 \\
m c^2 & – \hat{p} c
\end{bmatrix}
\begin{bmatrix}
\hat{p} c & m c^2 \\
m c^2 & – \hat{p} c
\end{bmatrix}
\Bpsi \\
&=
\begin{bmatrix}
\hat{p}^2 c^2 + m^2 c^4 & 0 \\
0 & \hat{p}^2 c^2 + m^2 c^4 \\
\end{bmatrix}.
\end{aligned}
\end{equation}

That is
\begin{equation}\label{eqn:qmLecture8:260}
– \Hbar^2 \PDSq{t}{} \Bpsi
=
\lr{ \hat{p}^2 c^2 + m^2 c^4 } \mathbf{1} \Bpsi,
\end{equation}

or more exactly

\begin{equation}\label{eqn:qmLecture8:280}
– \Hbar^2 \PDSq{t}{} \psi_{1,2}
=
\lr{ \hat{p}^2 c^2 + m^2 c^4 } \psi_{1,2}.
\end{equation}

This recovers the Klein Gordon equation for each of the wave functions \( \psi_1, \psi_2 \).

Instead of squaring the operators, lets try to solve the first order equation. To do so we’ll want to diagonalize \( \BH \).

Before doing that, let’s write out the Hamiltonian in an alternate but useful form

\begin{equation}\label{eqn:qmLecture8:340}
\BH
=
\hat{p} c
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
m c^2
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
= \hat{p} c \hat{\sigma}_z + m c^2 \hat{\sigma}_x.
\end{equation}

We have two types of operators in the mix here. We have matrix operators that act on the wave function matrices, as well as derivative operators that act on the components of those matrices.

We have

\begin{equation}\label{eqn:qmLecture8:360}
\hat{\sigma}_z
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}
=
\begin{bmatrix}
\psi_1 \\
-\psi_2 \\
\end{bmatrix},
\end{equation}

and

\begin{equation}\label{eqn:qmLecture8:380}
\hat{\sigma}_x
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}
=
\begin{bmatrix}
\psi_2 \\
\psi_1 \\
\end{bmatrix}.
\end{equation}

Because the derivative actions of \( \hat{p} \) and the matrix operators are independent, we see that these operators commute. For example

\begin{equation}\label{eqn:qmLecture8:400}
\hat{\sigma}_z \hat{p}
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}
=
\hat{\sigma}_z
\begin{bmatrix}
-i \Hbar \PD{x}{\psi_1} \\
-i \Hbar \PD{x}{\psi_2} \\
\end{bmatrix}
=
\begin{bmatrix}
-i \Hbar \PD{x}{\psi_1} \\
i \Hbar \PD{x}{\psi_2} \\
\end{bmatrix}
=
\hat{p}
\hat{\sigma}_z
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}.
\end{equation}

Diagonalizing it

Suppose the wave function matrix has the structure

\begin{equation}\label{eqn:qmLecture8:420}
\Bpsi =
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}
e^{i k x}.
\end{equation}

We’ll plug this into the Schrodinger equation and see what we get.

1D SHO linear superposition that maximizes expectation

October 7, 2015 phy1520 , , , , , ,

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Question: 1D SHO linear superposition that maximizes expectation ([1] pr. 2.17)

For a 1D SHO

(a)

Construct a linear combination of \( \ket{0}, \ket{1} \) that maximizes \( \expectation{x} \) without using wave functions.

(b)

How does this state evolve with time?

(c)

Evaluate \( \expectation{x} \) using the Schrodinger picture.

(d)

Evaluate \( \expectation{x} \) using the Heisenberg picture.

(e)

Evaluate \( \expectation{(\Delta x)^2} \).

Answer

(a)

Forming

\begin{equation}\label{eqn:shoSuperposition:20}
\ket{\psi} = \frac{\ket{0} + \sigma \ket{1}}{\sqrt{1 + \Abs{\sigma}^2}}
\end{equation}

the position expectation is

\begin{equation}\label{eqn:shoSuperposition:40}
\bra{\psi} x \ket{\psi}
=
\inv{1 + \Abs{\sigma}^2} \lr{ \bra{0} + \sigma^\conj \bra{1} } \frac{x_0}{\sqrt{2}} \lr{ a^\dagger + a } \lr{ \ket{0} + \sigma \ket{1} }.
\end{equation}

Evaluating the action of the operators on the kets, we’ve got

\begin{equation}\label{eqn:shoSuperposition:60}
\lr{ a^\dagger + a } \lr{ \ket{0} + \sigma \ket{1} }
=
\ket{1} + \sqrt{2} \sigma \ket{2} + \sigma \ket{0}.
\end{equation}

The \( \ket{2} \) term is killed by the bras, leaving

\begin{equation}\label{eqn:shoSuperposition:80}
\begin{aligned}
\expectation{x}
&=
\inv{1 + \Abs{\sigma}^2} \frac{x_0}{\sqrt{2}} \lr{ \sigma + \sigma^\conj} \\
&=
\frac{\sqrt{2} x_0 \textrm{Re} \sigma}{1 + \Abs{\sigma}^2}.
\end{aligned}
\end{equation}

Any imaginary component in \( \sigma \) will reduce the expectation, so we are constrained to picking a real value.

The derivative of

\begin{equation}\label{eqn:shoSuperposition:100}
f(\sigma) = \frac{\sigma}{1 + \sigma^2},
\end{equation}

is

\begin{equation}\label{eqn:shoSuperposition:120}
f'(\sigma) = \frac{1 – \sigma^2}{(1 + \sigma^2)^2}.
\end{equation}

That has zeros at \( \sigma = \pm 1 \). The second derivative is

\begin{equation}\label{eqn:shoSuperposition:140}
f”(\sigma) = \frac{-2 \sigma (3 – \sigma^2)}{(1 + \sigma^2)^3}.
\end{equation}

That will be negative (maximum for the extreme value) at \( \sigma = 1 \), so the linear superposition of these first two energy eigenkets that maximizes the position expectation is

\begin{equation}\label{eqn:shoSuperposition:160}
\psi = \inv{\sqrt{2}}\lr{ \ket{0} + \ket{1} }.
\end{equation}

That maximized position expectation is

\begin{equation}\label{eqn:shoSuperposition:180}
\expectation{x}
=
\frac{x_0}{\sqrt{2}}.
\end{equation}

(b)

The time evolution is given by

\begin{equation}\label{eqn:shoSuperposition:200}
\begin{aligned}
\ket{\Psi(t)}
&= e^{-iH t/\Hbar} \inv{\sqrt{2}}\lr{ \ket{0} + \ket{1} } \\
&= \inv{\sqrt{2}}\lr{ e^{-i(0+ \ifrac{1}{2})\Hbar \omega t/\Hbar} \ket{0} +
e^{-i(1+ \ifrac{1}{2})\Hbar \omega t/\Hbar} \ket{1} } \\
&= \inv{\sqrt{2}}\lr{ e^{-i \omega t/2} \ket{0} + e^{-3 i \omega t/2} \ket{1} }.
\end{aligned}
\end{equation}

(c)

The position expectation in the Schrodinger representation is

\begin{equation}\label{eqn:shoSuperposition:220}
\begin{aligned}
\expectation{x(t)}
&=
\inv{2}
\lr{ e^{i \omega t/2} \bra{0} + e^{3 i \omega t/2} \bra{1} } \frac{x_0}{\sqrt{2}} \lr{ a^\dagger + a }
\lr{ e^{-i \omega t/2} \ket{0} + e^{-3 i \omega t/2} \ket{1} } \\
&=
\frac{x_0}{2\sqrt{2}}
\lr{ e^{i \omega t/2} \bra{0} + e^{3 i \omega t/2} \bra{1} }
\lr{ e^{-i \omega t/2} \ket{1} + e^{-3 i \omega t/2} \sqrt{2} \ket{2} + e^{-3 i \omega t/2} \ket{0} } \\
&=
\frac{x_0}{\sqrt{2}} \cos(\omega t).
\end{aligned}
\end{equation}

(d)

\begin{equation}\label{eqn:shoSuperposition:240}
\begin{aligned}
\expectation{x(t)}
&=
\inv{2}
\lr{ \bra{0} + \bra{1} } \frac{x_0}{\sqrt{2}}
\lr{ a^\dagger e^{i\omega t} + a e^{-i \omega t} }
\lr{ \ket{0} + \ket{1} } \\
&=
\frac{x_0}{2 \sqrt{2}}
\lr{ \bra{0} + \bra{1} }
\lr{ e^{i\omega t} \ket{1} + \sqrt{2} e^{i\omega t} \ket{2} + e^{-i \omega t} \ket{0} } \\
&=
\frac{x_0}{\sqrt{2}} \cos(\omega t),
\end{aligned}
\end{equation}

matching the calculation using the Schrodinger picture.

(e)

Let’s use the Heisenberg picture for the uncertainty calculation. Using the calculation above we have

\begin{equation}\label{eqn:shoSuperposition:260}
\begin{aligned}
\expectation{x^2}
&=
\inv{2} \frac{x_0^2}{2}
\lr{ e^{-i\omega t} \bra{1} + \sqrt{2} e^{-i\omega t} \bra{2} + e^{i \omega t} \bra{0} }
\lr{ e^{i\omega t} \ket{1} + \sqrt{2} e^{i\omega t} \ket{2} + e^{-i \omega t} \ket{0} } \\
&=
\frac{x_0^2}{4} \lr{ 1 + 2 + 1} \\
&=
x_0^2.
\end{aligned}
\end{equation}

The uncertainty is
\begin{equation}\label{eqn:shoSuperposition:280}
\begin{aligned}
\expectation{(\Delta x)^2}
&=
\expectation{x^2} – \expectation{x}^2 \\
&=
x_0^2 – \frac{x_0^2}{2} \cos^2(\omega t) \\
&=
\frac{x_0^2}{2} \lr{ 2 – \cos^2(\omega t) } \\
&=
\frac{x_0^2}{2} \lr{ 1 + \sin^2(\omega t) }
\end{aligned}
\end{equation}

References

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