math and physics play

Correlation function. Partition function and ground state energy.

September 5, 2015 phy1520 , , , , ,

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Question: Correlation function ([1] pr. 2.16)

A correlation function can be defined as

\begin{equation}\label{eqn:correlationSHO:20}
C(t) = \expectation{ x(t) x(0) }.
\end{equation}

Using a Heisenberg picture \( x(t) \) calculate this correlation for the one dimensional SHO ground state.

Answer

The time dependent Heisenberg picture position operator was found to be

\begin{equation}\label{eqn:correlationSHO:40}
x(t) = x(0) \cos(\omega t) + \frac{p(0)}{m \omega} \sin(\omega t),
\end{equation}

so the correlation function is

\begin{equation}\label{eqn:correlationSHO:60}
\begin{aligned}
C(t)
&=
\bra{0} \lr{ x(0) \cos(\omega t) + \frac{p(0)}{m \omega} \sin(\omega t)} x(0) \ket{0} \\
&=
\cos(\omega t) \bra{0} x(0)^2 \ket{0} + \frac{\sin(\omega t)}{m \omega} \bra{0} p(0) x(0) \ket{0} \\
&=
\frac{\Hbar \cos(\omega t) }{2 m \omega} \bra{0} \lr{ a + a^\dagger}^2 \ket{0} – \frac{i \Hbar}{m \omega} \sin(\omega t),
\end{aligned}
\end{equation}

But
\begin{equation}\label{eqn:correlationSHO:80}
\begin{aligned}
\lr{ a + a^\dagger} \ket{0}
&=
a^\dagger \ket{0} \\
&=
\sqrt{1} \ket{1} \\
&=
\ket{1},
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:correlationSHO:100}
C(t) = x_0^2 \lr{ \inv{2} \cos(\omega t) – i \sin(\omega t) },
\end{equation}

where \( x_0^2 = \Hbar/(m \omega) \), not to be confused with \( x(0)^2 \).

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Question: Partition function and ground state energy ([1] pr. 2.32)

Define the partition function as

\begin{equation}\label{eqn:partitionFunction:20}
Z = \int d^3 x’ \evalbar{ K( \Bx’, t ; \Bx’, 0 ) }{\beta = i t/\Hbar},
\end{equation}

Show that the ground state energy is given by

\begin{equation}\label{eqn:partitionFunction:40}
-\inv{Z} \PD{\beta}{Z}, \qquad \beta \rightarrow \infty.
\end{equation}

Answer

The propagator evaluated at the same point is

\begin{equation}\label{eqn:partitionFunction:60}
\begin{aligned}
K( \Bx’, t ; \Bx’, 0 )
&=
\sum_{a’} \braket{\Bx’}{a’} \ket{a’}{\Bx’} \exp\lr{ -\frac{i E_{a’} t}{\Hbar}} \\
&=
\sum_{a’} \Abs{\braket{\Bx’}{a’}}^2 \exp\lr{ -\frac{i E_{a’} t}{\Hbar}} \\
&=
\sum_{a’} \Abs{\braket{\Bx’}{a’}}^2 \exp\lr{ -E_{a’} \beta}.
\end{aligned}
\end{equation}

The derivative is
\begin{equation}\label{eqn:partitionFunction:80}
\PD{\beta}{Z}
=
-\int d^3 x’ \sum_{a’} E_{a’} \Abs{\braket{\Bx’}{a’}}^2 \exp\lr{ -E_{a’} \beta}.
\end{equation}

In the \( \beta \rightarrow \infty \) this sum will be dominated by the term with the lowest value of \( E_{a’} \). Suppose that state is \( a’ = 0 \), then

\begin{equation}\label{eqn:partitionFunction:100}
\lim_{ \beta \rightarrow \infty }
-\inv{Z} \PD{\beta}{Z}
= \frac{
\int d^3 x’ E_{0} \Abs{\braket{\Bx’}{0}}^2 \exp\lr{ -E_{0} \beta}
}
{
\int d^3 x’ \Abs{\braket{\Bx’}{0}}^2 \exp\lr{ -E_{0} \beta}
}
= E_0.
\end{equation}

References

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

Momentum space representation of Schrodinger equation

September 2, 2015 phy1520 , , , , ,

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Question: momentum space representation of Schrodinger equation ([1] pr. 2.15)

Using

\begin{equation}\label{eqn:shoMomentumSpace:20}
\braket{x’}{p’} = \inv{\sqrt{2 \pi \Hbar}} e^{i p’ x’/\Hbar},
\end{equation}

show that

\begin{equation}\label{eqn:shoMomentumSpace:40}
\bra{p’} x \ket{\alpha} = i \Hbar \PD{p’}{} \braket{p’}{\alpha}.
\end{equation}

Use this to find the momentum space representation of the Schrodinger equation for the one dimensional SHO and the energy eigenfunctions in their momentum representation.

Answer

To expand the matrix element, introduce both momentum and position space identity operators

\begin{equation}\label{eqn:shoMomentumSpace:60}
\begin{aligned}
\bra{p’} x \ket{\alpha}
&=
\int dx’ dp” \braket{p’}{x’}\bra{x’}x \ket{p”}\braket{p”}{\alpha} \\
&=
\int dx’ dp” \braket{p’}{x’}x’\braket{x’}{p”}\braket{p”}{\alpha} \\
&=
\inv{2 \pi \Hbar}
\int dx’ dp” e^{-i p’ x’/\Hbar} x’ e^{i p” x’/\Hbar} \braket{p”}{\alpha} \\
&=
\inv{2 \pi \Hbar}
\int dx’ dp” x’ e^{i (p” – p’) x’/\Hbar} \braket{p”}{\alpha} \\
&=
\inv{2 \pi \Hbar}
\int dx’ dp” \frac{d}{dp”}\lr{ \frac{-i \Hbar} e^{i (p” – p’) x’/\Hbar} }
\braket{p”}{\alpha} \\
&=
i \Hbar
\int dp”
\lr{ \inv{2 \pi \Hbar}
\int dx’ e^{i (p” – p’) x’/\Hbar} } \frac{d}{dp”} \braket{p”}{\alpha} \\
&=
i \Hbar
\int dp” \delta(p”- p’)
\frac{d}{dp”} \braket{p”}{\alpha} \\
&=
i \Hbar
\frac{d}{dp’} \braket{p’}{\alpha}.
\end{aligned}
\end{equation}

Schrodinger’s equation for a time dependent state \( \ket{\alpha} = U(t) \ket{\alpha,0} \) is

\begin{equation}\label{eqn:shoMomentumSpace:80}
i \Hbar \PD{t}{} \ket{\alpha} = H \ket{\alpha},
\end{equation}

with the momentum representation

\begin{equation}\label{eqn:shoMomentumSpace:100}
i \Hbar \PD{t}{} \braket{p’}{\alpha} = \bra{p’} H \ket{\alpha}.
\end{equation}

Expansion of the Hamiltonian matrix element for a strictly spatial dependent potential \( V(x) \) gives

\begin{equation}\label{eqn:shoMomentumSpace:120}
\begin{aligned}
\bra{p’} H \ket{\alpha}
&=
\bra{p’} \lr{\frac{p^2}{2m} + V(x) } \ket{\alpha} \\
&=
\frac{(p’)^2}{2m}
+ \bra{p’} V(x) \ket{\alpha}.
\end{aligned}
\end{equation}

Assuming a Taylor representation of the potential \( V(x) = \sum c_k x^k \), we want to calculate

\begin{equation}\label{eqn:shoMomentumSpace:140}
\bra{p’} V(x) \ket{\alpha}
= \sum c_k \bra{p’} x^k \ket{\alpha}.
\end{equation}

With \( \ket{\alpha} = \ket{p”} \) \ref{eqn:shoMomentumSpace:40} provides the \( k = 1 \) term

\begin{equation}\label{eqn:shoMomentumSpace:160}
\begin{aligned}
\bra{p’} x \ket{p”}
&= i \Hbar \frac{d}{dp’} \braket{p’}{p”} \\
&= i \Hbar \frac{d}{dp’} \delta(p’ – p”),
\end{aligned}
\end{equation}

where it is implied here that the derivative is operating on not just the delta function, but on all else that follows.

Using this the higher powers of \( \bra{p’} x^k \ket{\alpha} \) can be found easily. For example for \( k = 2 \) we have

\begin{equation}\label{eqn:shoMomentumSpace:180}
\begin{aligned}
\bra{p’} x^2 \ket{\alpha}
&=
\int dp”
\bra{p’} x \ket{p”}\bra{p”} x \ket{\alpha} \\
&=
\int dp”
i \Hbar
\frac{d}{dp’} \delta(p’ – p”) i \Hbar \frac{d}{dp”} \braket{p”}{\alpha} \\
&=
\lr{ i \Hbar }^2 \frac{d^2}{d(p’)^2} \braket{p’}{\alpha}.
\end{aligned}
\end{equation}

This means that the potential matrix element is

\begin{equation}\label{eqn:shoMomentumSpace:200}
\begin{aligned}
\bra{p’} V(x) \ket{\alpha}
&=
\sum c_k \lr{ i \Hbar \frac{d}{dp’} }^k \braket{p’}{\alpha} \\
&= V\lr{ i \Hbar \frac{d}{dp’} }.
\end{aligned}
\end{equation}

Writing \( \Psi_\alpha(p’) = \braket{p’}{\alpha} \), the momentum space representation of Schrodinger’s equation for a position dependent potential is

\begin{equation}\label{eqn:shoMomentumSpace:220}
\boxed{
i \Hbar \PD{t}{} \Psi_\alpha(p’)
=
\lr{ \frac{(p’)^2}{2m} + V\lr{ i \Hbar \PDi{p’}{} } } \Psi_\alpha(p’).
}
\end{equation}

For the SHO Hamiltonian the potential is \( V(x) = (1/2) m \omega^2 x^2 \), so the Schrodinger equation is

\begin{equation}\label{eqn:shoMomentumSpace:240}
\begin{aligned}
i \Hbar \PD{t}{} \Psi_\alpha(p’)
&=
\lr{ \frac{(p’)^2}{2m} – \inv{2} m \omega^2 \Hbar^2
\frac{\partial^2}{\partial(p’)^2} } \Psi_\alpha(p’) \\
&=
\inv{2 m} \lr{ (p’)^2 – m^2 \omega^2 \Hbar^2 \frac{\partial^2}{\partial(p’)^2} } \Psi_\alpha(p’).
\end{aligned}
\end{equation}

To determine the wave functions, let’s non-dimensionalize this and compare to the position space Schrodinger equation. Let

\begin{equation}\label{eqn:shoMomentumSpace:260}
p_0^2 = m \omega \hbar,
\end{equation}

so
\begin{equation}\label{eqn:shoMomentumSpace:280}
\begin{aligned}
i \Hbar \PD{t}{} \Psi_\alpha(p’)
&=
\frac{p_0^2}{2 m} \lr{ \lr{\frac{p’}{p_0}}^2 –
\frac{\partial^2}{\partial(p’/p_0)^2} } \Psi_\alpha(p’) \\
&=
\frac{\omega \Hbar}{2}\lr{
– \frac{\partial^2}{\partial(p’/p_0)^2} +
\lr{\frac{p’}{p_0}}^2
} \Psi_\alpha(p’).
\end{aligned}
\end{equation}

Compare this to the position space equation with \( x_0^2 = m \omega/\Hbar \),

\begin{equation}\label{eqn:shoMomentumSpace:300}
\begin{aligned}
i \Hbar \PD{t}{} \Psi_\alpha(x’)
&=
\lr{ -\frac{\Hbar^2}{2m} \frac{\partial^2}{\partial(x’)^2}
+
\inv{2} m \omega^2 (x’)^2 }
\Psi_\alpha(x’) \\
&=
\frac{\Hbar^2}{2m}
\lr{ -\frac{\partial^2}{\partial(x’)^2}
+
\frac{m^2 \omega^2}{\Hbar^2} (x’)^2 }
\Psi_\alpha(x’) \\
&=
\frac{\Hbar^2 x_0^2}{2m}
\lr{
-\frac{\partial^2}{\partial(x’/x_0)^2}
+
\lr{\frac{x’}{x_0}}^2
}
\Psi_\alpha(x’) \\
&=
\frac{\Hbar \omega}{2}
\lr{
-\frac{\partial^2}{\partial(x’/x_0)^2}
+
\lr{\frac{x’}{x_0}}^2
}
\Psi_\alpha(x’).
\end{aligned}
\end{equation}

It’s clear that there is a straightforward duality relationship between the respective wave functions. Since

\begin{equation}\label{eqn:shoMomentumSpace:320}
\braket{x’}{n} =
\inv{\pi^{1/4} \sqrt{2^n n!} x_0^{n + 1/2}} \lr{ x’ – x_0^2 \frac{d}{dx’} }^n \exp\lr{ -\inv{2} \lr{\frac{x’}{x_0}}^2 },
\end{equation}

the momentum space wave functions are

\begin{equation}\label{eqn:shoMomentumSpace:340}
\braket{p’}{n} =
\inv{\pi^{1/4} \sqrt{2^n n!} p_0^{n + 1/2}} \lr{ p’ – p_0^2 \frac{d}{dp’} }^n \exp\lr{ -\inv{2} \lr{\frac{p’}{p_0}}^2 }.
\end{equation}

References

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

SHO translation operator expectation

September 2, 2015 phy1520 , , , ,

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Question: SHO translation operator expectation ([1] pr. 2.12)

Using the Heisenberg picture evaluate the expectation of the position operator \( \expectation{x} \) with respect to the initial time state

\begin{equation}\label{eqn:translationExpectation:20}
\ket{\alpha, 0} = e^{-i p_0 a/\Hbar} \ket{0},
\end{equation}

where \( p_0 \) is the initial time position operator, and \( a \) is a constant with dimensions of position.

Answer

Recall that the Heisenberg picture position operator expands to

\begin{equation}\label{eqn:translationExpectation:40}
x^{\textrm{H}}(t)
= U^\dagger x U
= x_0 \cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t),
\end{equation}

so the expectation of the position operator is
\begin{equation}\label{eqn:translationExpectation:60}
\begin{aligned}
\expectation{x}
&=
\bra{0} e^{i p_0 a/\Hbar} \lr{ x_0 \cos(\omega t) + \frac{p_0}{m \omega}
\sin(\omega t) } e^{-i p_0 a/\Hbar} \ket{0} \\
&=
\bra{0} \lr{ e^{i p_0 a/\Hbar} x_0 \cos(\omega t) e^{-i p_0 a/\Hbar} \cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t) } \ket{0}.
\end{aligned}
\end{equation}

The exponential sandwich above can be expanded using the Baker-Campbell-Hausdorff [2] formula

\begin{equation}\label{eqn:translationExpectation:80}
\begin{aligned}
e^{i p_0 a/\Hbar} x_0 e^{-i p_0 a/\Hbar}
&=
x_0
+ \frac{i a}{\Hbar} \antisymmetric{p_0}{x_0}
+ \inv{2!} \lr{\frac{i a}{\Hbar}}^2 \antisymmetric{p_0}{\antisymmetric{p_0}{x_0}}
+ \cdots \\
&=
x_0
+ \frac{i a}{\Hbar} \lr{ -i \Hbar }
+ \inv{2!} \lr{\frac{i a}{\Hbar}}^2 \antisymmetric{p_0}{-i \Hbar}
+ \cdots \\
&=
x_0 + a.
\end{aligned}
\end{equation}

The position expectation with respect to this translated state is

\begin{equation}\label{eqn:translationExpectation:100}
\begin{aligned}
\expectation{x}
&= \bra{0} \lr{ (x_0 + a)\cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t)
}\ket{0} \\
&= a \cos(\omega t).
\end{aligned}
\end{equation}

The final simplification above follows from \( \bra{n} x \ket{n} = \bra{n} p \ket{n} = 0 \).

References

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

[2] Wikipedia. Baker-campbell-hausdorff formula — wikipedia, the free encyclopedia, 2015. URL https://en.wikipedia.org/w/index.php?title=Baker\%E2\%80\%93Campbell\%E2\%80\%93Hausdorff_formula&oldid=665123858. [Online; accessed 16-August-2015].

Quantum Virial Theorem

August 31, 2015 phy1520 , , , , ,

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Question: Quantum Virial Theorem ([1] pr. 2.7)

Consider a particle with Hamiltonian

\begin{equation}\label{eqn:qmVirialTheorem:20}
H = \frac{\Bp^2}{2 m} + V(\Bx),
\end{equation}

By calculating the time evolution of \( \antisymmetric{\Bx \cdot \Bp}{H} \), identify the quantum virial theorem and show the conditions where it is satisfied.

Answer

\begin{equation}\label{eqn:qmVirialTheorem:40}
\begin{aligned}
\antisymmetric{\Bx \cdot \Bp}{H}
&=
\inv{2 m} \antisymmetric{\Bx \cdot \Bp}{\Bp^2} + \antisymmetric{\Bx \cdot \Bp}{V(\Bx)} \\
&=
\inv{2 m} \lr{ x_r p_r \Bp^2 – \Bp^2 x_r p_r}
+
\lr{ x_r p_r V(\Bx) – V(\Bx) x_r p_r } \\
&=
\inv{2 m} \antisymmetric{ x_r }{\Bp^2} p_r
+
x_r \antisymmetric{ p_r}{ V(\Bx)},
\end{aligned}
\end{equation}

Evaluating those commutators separately, gives

\begin{equation}\label{eqn:qmVirialTheorem:60}
\begin{aligned}
\antisymmetric{ x_r }{\Bp^2}
&=
\antisymmetric{ x_r }{p_r^2}\qquad \text{no sum} \\
&=
2 i \Hbar p_r,
\end{aligned}
\end{equation}

and

\begin{equation}\label{eqn:qmVirialTheorem:80}
\antisymmetric{ p_r}{ V(\Bx)}
= -i \Hbar \PD{x_r}{V(\Bx)},
\end{equation}

so
\begin{equation}\label{eqn:qmVirialTheorem:100}
\begin{aligned}
\ddt{}\lr{\Bx \cdot \Bp}
&=
\inv{i \Hbar}
\antisymmetric{\Bx \cdot \Bp}{H} \\
&=
\inv{2 m} 2 p_r p_r – x_r \PD{x_r}{V(\Bx)} \\
&=
\frac{\Bp^2}{m} – \Bx \cdot \spacegrad V(\Bx).
\end{aligned}
\end{equation}

Taking expectation values, assuming that the states are independent of time, we have

\begin{equation}\label{eqn:qmVirialTheorem:120}
\begin{aligned}
0
&= \ddt{} \expectation{ \Bx \cdot \Bp } \\
&= \expectation{\frac{\Bp^2}{m}} – \expectation{\Bx \cdot \spacegrad V(\Bx)}.
\end{aligned}
\end{equation}

Note that taking the expectation with respect to stationary states was required to reverse the order of the time derivative with the expectation operation.

The right hand side is the quantum equivalent of the virial theorem, relating the average kinetic energy to the potential

\begin{equation}\label{eqn:qmVirialTheorem:140}
2 \expectation{T} = \expectation{\Bx \cdot \spacegrad V(\Bx)}
\end{equation}

References

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

A symmetric real Hamiltonian

August 31, 2015 phy1520 , ,

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Question: A symmetric real Hamiltonian ([1] pr. 2.9)

Find the time evolution for the state \( \ket{a’} \) for a Hamiltian of the form

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:20}
H = \delta \lr{ \ket{a’}\bra{a’} + \ket{a”}\bra{a”} }
\end{equation}

Answer

This Hamiltonian has the matrix representation

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:40}
H =
\begin{bmatrix}
0 & \delta \\
\delta & 0
\end{bmatrix},
\end{equation}

which has a characteristic equation of

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:60}
\lambda^2 -\delta^2 = 0,
\end{equation}

so the energy eigenvalues are \( \pm \delta \).

The diagonal basis states are respectively

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:80}
\ket{\pm\delta} =
\inv{\sqrt{2}}
\begin{bmatrix}
\pm 1 \\
1
\end{bmatrix}.
\end{equation}

The time evolution operator is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:100}
\begin{aligned}
U
&= e^{-i H t/\Hbar} \\
&=
e^{-i \delta t/\Hbar} \ket{+\delta}\bra{+\delta}
+ e^{i \delta t/\Hbar} \ket{-\delta}\bra{-\delta} \\
&=
\frac{e^{-i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
+ \frac{e^{i \delta t/\Hbar} }{2}
\begin{bmatrix}
-1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 \\
1
\end{bmatrix} \\
&=
\frac{e^{-i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & 1 \\
1 & 1
\end{bmatrix}
+\frac{e^{i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & -1 \\
-1 & 1
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) & -i\sin(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar) & \cos(\delta t/\Hbar) \\
\end{bmatrix}.
\end{aligned}
\end{equation}

The desired time evolution in the original basis is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:140}
\begin{aligned}
\ket{a’, t}
&=
e^{-i H t/\Hbar}
\ket{a’, 0} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) & -i\sin(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar) & \cos(\delta t/\Hbar) \\
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar)
\end{bmatrix} \\
&=
\cos(\delta t/\Hbar) \ket{a’,0} -i \sin(\delta t/\Hbar) \ket{a”,0}.
\end{aligned}
\end{equation}

This evolution has the same structure as left circularly polarized light.

The probability of finding the system in state \( \ket{a”} \) given an initial state of \( \ket{a’,0} \) is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:160}
P
=
\Abs{\braket{a”}{a’,t}}^2
=
\sin^2 \lr{ \delta t/\Hbar }.
\end{equation}

References

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