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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.