Day: December 16, 2015

Totally asymmetric potential

December 16, 2015 phy1520 2 comments , , ,

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Q: [1] pr 4.11

(a) Given a time reversal invariant Hamiltonian, show that for any energy eigenket

\begin{equation}\label{eqn:totallyAsymmetricPotential:20}
\expectation{\BL} = 0.
\end{equation}

(b) If the wave function of such a state is expanded as

\begin{equation}\label{eqn:totallyAsymmetricPotential:40}
\sum_{l,m} F_{l m} Y_{l m}(\theta, \phi),
\end{equation}

what are the phase restrictions on \( F_{lm} \)?

A: part (a)

For a time reversal invariant Hamiltonian \( H \) we have

\begin{equation}\label{eqn:totallyAsymmetricPotential:60}
H \Theta = \Theta H.
\end{equation}

If \( \ket{\psi} \) is an energy eigenstate with eigenvalue \( E \), we have

\begin{equation}\label{eqn:totallyAsymmetricPotential:80}
\begin{aligned}
H \Theta \ket{\psi}
&= \Theta H \ket{\psi} \\
&= \lambda \Theta \ket{\psi},
\end{aligned}
\end{equation}

so \( \Theta \ket{\psi} \) is also an eigenvalue of \( H \), so can only differ from \( \ket{\psi} \) by a phase factor. That is

\begin{equation}\label{eqn:totallyAsymmetricPotential:100}
\begin{aligned}
\ket{\psi’}
&=
\Theta \ket{\psi} \\
&= e^{i\delta} \ket{\psi}.
\end{aligned}
\end{equation}

Now consider the expectation of \( \BL \) with respect to a time reversed state

\begin{equation}\label{eqn:totallyAsymmetricPotential:120}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\bra{ \psi} \Theta^{-1} \BL \Theta \ket{\psi} \\
&=
\bra{ \psi} (-\BL) \ket{\psi},
\end{aligned}
\end{equation}

however, we also have

\begin{equation}\label{eqn:totallyAsymmetricPotential:140}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\lr{ \bra{ \psi} e^{-i\delta} } \BL \lr{ e^{i\delta} \ket{\psi} } \\
&=
\bra{\psi} \BL \ket{\psi},
\end{aligned}
\end{equation}

so we have \( \bra{\psi} \BL \ket{\psi} = -\bra{\psi} \BL \ket{\psi} \) which is only possible if \( \expectation{\BL} = \bra{\psi} \BL \ket{\psi} = 0\).

A: part (b)

Consider the expansion of the wave function of a time reversed energy eigenstate

\begin{equation}\label{eqn:totallyAsymmetricPotential:160}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} e^{i\delta} \ket{\psi} \\
&=
e^{i\delta} \braket{\Bx}{\psi},
\end{aligned}
\end{equation}

and then consider the same state expanded in the position basis

\begin{equation}\label{eqn:totallyAsymmetricPotential:180}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \ket{\Bx’}\bra{\Bx’} } \ket{\psi} \\
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} } \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \Theta \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \ket{\Bx’} \\
&=
\int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \braket{\Bx}{\Bx’} \\
&=
\int d^3 \Bx’ \braket{\psi}{\Bx’} \delta(\Bx- \Bx’) \\
&=
\braket{\psi}{\Bx}.
\end{aligned}
\end{equation}

This demonstrates a relationship between the wave function and its complex conjugate

\begin{equation}\label{eqn:totallyAsymmetricPotential:200}
\braket{\Bx}{\psi} = e^{-i\delta} \braket{\psi}{\Bx}.
\end{equation}

Now expand the wave function in the spherical harmonic basis

\begin{equation}\label{eqn:totallyAsymmetricPotential:220}
\begin{aligned}
\braket{\Bx}{\psi}
&=
\int d\Omega \braket{\Bx}{\ncap}\braket{\ncap}{\psi} \\
&=
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) \\
&=
e^{-i\delta}
\lr{
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) }^\conj \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj Y_{lm}^\conj(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj (-1)^m Y_{l,-m}(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{l,-m}(r)}^\conj (-1)^m Y_{l,m}(\theta, \phi),
\end{aligned}
\end{equation}

so the \( F_{lm} \) functions are constrained by

\begin{equation}\label{eqn:totallyAsymmetricPotential:240}
F_{lm}(r) = e^{-i\delta} \lr{ F_{l,-m}(r)}^\conj (-1)^m.
\end{equation}

References

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

Commutators for some symmetry operators

December 16, 2015 phy1520 No comments , , ,

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Q: [1] pr 4.2

If \( \mathcal{T}_\Bd \), \( \mathcal{D}(\ncap, \phi) \), and \( \pi \) denote the translation, rotation, and parity operators respectively. Which of the following commute and why

  • (a) \( \mathcal{T}_\Bd \) and \( \mathcal{T}_{\Bd’} \), translations in different directions.
  • (b) \( \mathcal{D}(\ncap, \phi) \) and \( \mathcal{D}(\ncap’, \phi’) \), rotations in different directions.
  • (c) \( \mathcal{T}_\Bd \) and \( \pi \).
  • (d) \( \mathcal{D}(\ncap,\phi)\) and \( \pi \).

A: (a)

Consider
\begin{equation}\label{eqn:symmetryOperatorCommutators:20}
\begin{aligned}
\mathcal{T}_\Bd \mathcal{T}_{\Bd’} \ket{\Bx}
&=
\mathcal{T}_\Bd \ket{\Bx + \Bd’} \\
&=
\ket{\Bx + \Bd’ + \Bd},
\end{aligned}
\end{equation}

and the reverse application of the translation operators
\begin{equation}\label{eqn:symmetryOperatorCommutators:40}
\begin{aligned}
\mathcal{T}_{\Bd’} \mathcal{T}_{\Bd} \ket{\Bx}
&=
\mathcal{T}_{\Bd’} \ket{\Bx + \Bd} \\
&=
\ket{\Bx + \Bd + \Bd’} \\
&=
\ket{\Bx + \Bd’ + \Bd}.
\end{aligned}
\end{equation}

so we see that

\begin{equation}\label{eqn:symmetryOperatorCommutators:60}
\antisymmetric{\mathcal{T}_\Bd}{\mathcal{T}_{\Bd’}} \ket{\Bx} = 0,
\end{equation}

for any position state \( \ket{\Bx} \), and therefore in general they commute.

A: (b)

That rotations do not commute when they are in different directions (like any two orthogonal directions) need not be belaboured.

A: (c)

We have
\begin{equation}\label{eqn:symmetryOperatorCommutators:80}
\begin{aligned}
\mathcal{T}_\Bd \pi \ket{\Bx}
&=
\mathcal{T}_\Bd \ket{-\Bx} \\
&=
\ket{-\Bx + \Bd},
\end{aligned}
\end{equation}

yet
\begin{equation}\label{eqn:symmetryOperatorCommutators:100}
\begin{aligned}
\pi \mathcal{T}_\Bd \ket{\Bx}
&=
\pi \ket{\Bx + \Bd} \\
&=
\ket{-\Bx – \Bd} \\
&\ne
\ket{-\Bx + \Bd}.
\end{aligned}
\end{equation}

so, in general \( \antisymmetric{\mathcal{T}_\Bd}{\pi} \ne 0 \).

A: (d)

We have

\begin{equation}\label{eqn:symmetryOperatorCommutators:120}
\begin{aligned}
\pi \mathcal{D}(\ncap, \phi) \ket{\Bx}
&=
\pi \mathcal{D}(\ncap, \phi) \pi^\dagger \pi \ket{\Bx} \\
&=
\pi \mathcal{D}(\ncap, \phi) \pi^\dagger \pi \ket{\Bx} \\
&=
\pi \lr{ \sum_{k=0}^\infty \frac{(-i \BJ \cdot \ncap)^k}{k!} } \pi^\dagger \pi \ket{\Bx} \\
&=
\sum_{k=0}^\infty \frac{(-i (\pi \BJ \pi^\dagger) \cdot (\pi \ncap \pi^\dagger) )^k}{k!} \pi \ket{\Bx} \\
&=
\sum_{k=0}^\infty \frac{(-i \BJ \cdot \ncap)^k}{k!} \pi \ket{\Bx} \\
&=
\mathcal{D}(\ncap, \phi) \pi \ket{\Bx},
\end{aligned}
\end{equation}

so \( \antisymmetric{\mathcal{D}(\ncap, \phi)}{\pi} \ket{\Bx} = 0 \), for any position state \( \ket{\Bx} \), and therefore these operators commute in general.

References

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

Plane wave and spinor under time reversal

December 16, 2015 phy1520 No comments , , , , ,

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Q: [1] pr 4.7

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

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

A: part (a)

The Hamiltonian for a plane wave is

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

Under time reversal the momentum side transforms as

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

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

Solutions to this equation are linear combinations of

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

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

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

A: part (b)

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

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

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

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

The state orthogonal to this one is claimed to be

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

We have

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

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

so we should have

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

This looks right, but we can sanity check orthogonality

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

as expected.

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

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

which is the result to be demononstrated.

References

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