## Question: Quantum Virial Theorem ( 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.

\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

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

## A symmetric real Hamiltonian

August 31, 2015 phy1520 , ,

## Question: A symmetric real Hamiltonian ( 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}

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

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

## Operator matrix element ## Weird dreams

I woke up today having a dream still in my head from the night, but it was a strange one. I was expanding out the Dirac notation representation of an operator in matrix form, but the symbols in the kets were elaborate pictures of Disney princesses that I was drawing with forestry scenery in the background, including little bears. At the point that I woke up from the dream, I noticed that I’d gotten the proportion of the bears wrong in one of the pictures, and they looked like they were ready to eat one of the princess characters.

## Guts

As a side effect of this weird dream I actually started thinking about matrix element representation of operators.

When forming the matrix element of an operator using Dirac notation the elements are of the form $$\bra{\textrm{row}} A \ket{\textrm{column}}$$. I’ve gotten that mixed up a couple of times, so I thought it would be helpful to write this out explicitly for a $$2 \times 2$$ operator representation for clarity.

To start, consider a change of basis for a single matrix element from basis $$\setlr{\ket{q}, \ket{r} }$$, to basis $$\setlr{\ket{a}, \ket{b} }$$

\begin{equation}\label{eqn:operatorMatrixElement:20}
\begin{aligned}
\bra{q} A \ket{r}
&=
\braket{q}{a} \bra{a} A \ket{r}
+
\braket{q}{b} \bra{b} A \ket{r} \\
&=
\braket{q}{a} \bra{a} A \ket{a}\braket{a}{r}
+ \braket{q}{a} \bra{a} A \ket{b}\braket{b}{r} \\
&+ \braket{q}{b} \bra{b} A \ket{a}\braket{a}{r}
+ \braket{q}{b} \bra{b} A \ket{b}\braket{b}{r} \\
&=
\braket{q}{a}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix}
+
\braket{q}{b}
\begin{bmatrix}
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix} \\
&=
\begin{bmatrix}
\braket{q}{a} &
\braket{q}{b}
\end{bmatrix}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix}.
\end{aligned}
\end{equation}

Suppose the matrix representation of $$\ket{q}, \ket{r}$$ are respectively

\begin{equation}\label{eqn:operatorMatrixElement:40}
\begin{aligned}
\ket{q} &\sim
\begin{bmatrix}
\braket{a}{q} \\
\braket{b}{q} \\
\end{bmatrix} \\
\ket{r} &\sim
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r} \\
\end{bmatrix} \\
\end{aligned},
\end{equation}

then

\begin{equation}\label{eqn:operatorMatrixElement:60}
\bra{q} \sim
{\begin{bmatrix}
\braket{a}{q} \\
\braket{b}{q} \\
\end{bmatrix}}^\dagger
=
\begin{bmatrix}
\braket{q}{a} &
\braket{q}{b}
\end{bmatrix}.
\end{equation}

The matrix element is then

\begin{equation}\label{eqn:operatorMatrixElement:80}
\bra{q} A \ket{r}
\sim
\bra{q}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\ket{r},
\end{equation}

and the corresponding matrix representation of the operator is

\begin{equation}\label{eqn:operatorMatrixElement:100}
A \sim
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}.
\end{equation}

## Question: Hermite polynomial normalization constant ( pr. 2.21)

Derive the normalization constant $$c_n$$ for the Harmonic oscillator solution

\begin{equation}\label{eqn:hermiteOrtho:20}
u_n(x) = c_n H_n\lr{ x \sqrt{\frac{m\omega}{\Hbar}} } e^{-m \omega x^2/2 \Hbar},
\end{equation}

by deriving the orthogonality relationship using generating functions

\begin{equation}\label{eqn:hermiteOrtho:40}
g(x,t) = e^{-t^2 + 2 t x} = \sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}.
\end{equation}

Start by working out the integral

\begin{equation}\label{eqn:hermiteOrtho:60}
I = \int_{-\infty}^\infty g(x, t) g(x, s) e^{-x^2} dx,
\end{equation}

consider the integral twice with each side definition of the generating function.

First using the exponential definition of the generating function

\begin{equation}\label{eqn:hermiteOrtho:80}
\begin{aligned}
\int_{-\infty}^\infty g(x, t) g(x, s) e^{-x^2} dx
&=
\int_{-\infty}^\infty
e^{-t^2 + 2 t x}
e^{-s^2 + 2 s x} e^{-x^2} dx \\
&=
e^{-t^2 -s^2}
\int_{-\infty}^\infty
e^{-(x^2- 2 t x – 2 s x)} dx \\
&=
e^{-t^2 -s^2 + (s + t)^2}
\int_{-\infty}^\infty
e^{-(x – t – s)^2} dx \\
&=
e^{2 st}
\int_{-\infty}^\infty
e^{-u^2} du \\
&= \sqrt{\pi} e^{2 st}.
\end{aligned}
\end{equation}

With the Hermite polynomial definition of the generating function, this integral is

\begin{equation}\label{eqn:hermiteOrtho:100}
\begin{aligned}
\int_{-\infty}^\infty g(x, t) g(x, s) e^{-x^2} dx
&=
\int_{-\infty}^\infty
\sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}
\sum_{m=0}^\infty H_m(x) \frac{s^m}{m!}
e^{-x^2} dx \\
&=
\sum_{n=0}^\infty \frac{t^n}{n!}
\sum_{m=0}^\infty \frac{s^m}{m!}
\int_{-\infty}^\infty H_n(x) H_m(x) e^{-x^2} dx.
\end{aligned}
\end{equation}

Let

\begin{equation}\label{eqn:hermiteOrtho:120}
\alpha_{n m} = \int_{-\infty}^\infty H_n(x) H_m(x) e^{-x^2} dx,
\end{equation}

and equate the two expansions of this integral

\begin{equation}\label{eqn:hermiteOrtho:140}
\sqrt{\pi} \sum_{n=0}^\infty \frac{(2st)^n}{n!}
=
\sum_{n=0}^\infty \frac{t^n}{n!}
\sum_{m=0}^\infty \frac{s^m}{m!}
\alpha_{n m},
\end{equation}

or, after equating powers of $$t^n$$

\begin{equation}\label{eqn:hermiteOrtho:160}
\sqrt{\pi} (2 s)^n =
\sum_{m=0}^\infty \frac{s^m}{m!} \alpha_{n m}.
\end{equation}

This requires $$\alpha_{n m}$$ to be zero for $$n \ne m$$, so

\begin{equation}\label{eqn:hermiteOrtho:180}
\sqrt{\pi} 2^n = \frac{1}{n!} \alpha_{n n},
\end{equation}

and

\begin{equation}\label{eqn:hermiteOrtho:200}
\int_{-\infty}^\infty H_n(x) H_m(x) e^{-x^2} dx = \delta_{n m} \sqrt{\pi} 2^n n!.
\end{equation}

The SHO normalization is fixed by

\begin{equation}\label{eqn:hermiteOrtho:220}
\int_{-\infty}^\infty u_n^2(x) dx
= c_n^2
\int_{-\infty}^\infty H_n^2(x/x_0) e^{-(x/x_0)^2} dx
= c_n^2 x_0 \sqrt{\pi} 2^n n!,
\end{equation}

or

\begin{equation}\label{eqn:hermiteOrtho:240}
\begin{aligned}
c_n
&= \inv{\sqrt{ \sqrt{\pi} 2^n n! \sqrt{\frac{\Hbar}{m \omega}}}} \\
&= \lr{ \frac{m \omega}{\Hbar \pi} }^{1/4} 2^{-n/2} \inv{\sqrt{n!}}
\end{aligned}
\end{equation}

# References

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

## Relation of probability flux to momentum

August 19, 2015 phy1520 ,

In  it is mentioned that the probability flux

\begin{equation}\label{eqn:fluxAndMomentum:20}
\Bj(\Bx, t) = -\frac{i\Hbar}{2 m} \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj },
\end{equation}

is related to the momentum expectation at a given time by the integral of the flux over all space

\begin{equation}\label{eqn:fluxAndMomentum:40}
\int d^3 x \Bj(\Bx, t) = \frac{\expectation{\Bp}_t}{m}.
\end{equation}

That wasn’t obvious to me at a glance, however, this can be seen by recasting the integral in bra-ket form. Let

\begin{equation}\label{eqn:fluxAndMomentum:60}
\psi(\Bx, t) = \braket{\Bx}{\psi(t)},
\end{equation}

and note that the momentum portions of the flux can be written as

\begin{equation}\label{eqn:fluxAndMomentum:80}
-i \Hbar \spacegrad \psi(\Bx, t) = \bra{\Bx} \Bp \ket{\psi(t)}.
\end{equation}

The current is therefore

\begin{equation}\label{eqn:fluxAndMomentum:100}
\begin{aligned}
\Bj(\Bx, t)
&= \frac{1}{2 m}
\lr{
\psi^\conj \bra{\Bx} \Bp \ket{\psi(t)}
+\psi {\bra{\Bx} \Bp \ket{\psi(t)} }^\conj
} \\
&= \frac{1}{2 m}
\lr{
{\braket{\Bx}{\psi(t)}}^\conj \bra{\Bx} \Bp \ket{\psi(t)}
+ \braket{\Bx}{\psi(t)} {\bra{\Bx} \Bp \ket{\psi(t)} }^\conj
} \\
&= \frac{1}{2 m}
\lr{
\braket{\psi(t)}{\Bx} \bra{\Bx} \Bp \ket{\psi(t)}
+
\bra{\psi(t)} \Bp \ket{\Bx} \braket{\Bx}{\psi(t)}
}.
\end{aligned}
\end{equation}

Integrating and noting that the spatial identity is $$1 = \int d^3 x \ket{\Bx}\bra{\Bx}$$, we have

\begin{equation}\label{eqn:fluxAndMomentum:n}
\int d^3 x \Bj(\Bx, t)
=
\bra{\psi(t)} \Bp \ket{\psi(t)},
\end{equation}

This is just the expectation of $$\Bp$$ with respect to a specific time-instance state.

# References

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