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

## (a)

Forming

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

the position expectation is

\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} }.

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

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

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

\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}

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

The derivative of

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

is

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

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

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

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

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

That maximized position expectation is

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

## (b)

The time evolution is given by

\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}

## (c)

The position expectation in the Schrodinger representation is

\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}

## (d)

\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}

matching the calculation using the Schrodinger picture.

## (e)

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

\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}

The uncertainty is
\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}

# References

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

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

A correlation function can be defined as

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

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

The time dependent Heisenberg picture position operator was found to be

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

so the correlation function is

\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}

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

so

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

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

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

Define the partition function as

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

Show that the ground state energy is given by

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

The propagator evaluated at the same point is

\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}

The derivative is
\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}.

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

\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.

# References

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

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

Using

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

show that

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

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.

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

\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}

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

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

with the momentum representation

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

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

\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}

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

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

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

\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}

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

\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}

This means that the potential matrix element is

\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}

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

\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’).
}

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

\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}

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

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

so
\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}

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

\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}

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

\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 },

the momentum space wave functions are

\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 }.

# References

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

## SHO translation operator expectation

September 2, 2015 phy1520 , , , ,

## 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

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

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

Recall that the Heisenberg picture position operator expands to

\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),

so the expectation of the position operator is
\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}

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

\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}

The position expectation with respect to this translated state is

\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}

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].

## Question: Hermite polynomial normalization constant ([1] pr. 2.21)

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

\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},

by deriving the orthogonality relationship using generating functions

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

Start by working out the integral

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

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

First using the exponential definition of the generating function

\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}

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

\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}

Let

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

and equate the two expansions of this integral

\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},

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

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

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

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

and

\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!.

The SHO normalization is fixed by

\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!,

or

\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}

# References

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