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## Question: 1D SHO linear superposition that maximizes expectation ([1] pr. 2.17)

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

## Answer

## (a)

Forming

\begin{equation}\label{eqn:shoSuperposition:20}

\ket{\psi} = \frac{\ket{0} + \sigma \ket{1}}{\sqrt{1 + \Abs{\sigma}^2}}

\end{equation}

the position expectation is

\begin{equation}\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} }.

\end{equation}

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

\begin{equation}\label{eqn:shoSuperposition:60}

\lr{ a^\dagger + a } \lr{ \ket{0} + \sigma \ket{1} }

=

\ket{1} + \sqrt{2} \sigma \ket{2} + \sigma \ket{0}.

\end{equation}

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

\begin{equation}\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}

\end{equation}

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

The derivative of

\begin{equation}\label{eqn:shoSuperposition:100}

f(\sigma) = \frac{\sigma}{1 + \sigma^2},

\end{equation}

is

\begin{equation}\label{eqn:shoSuperposition:120}

f'(\sigma) = \frac{1 – \sigma^2}{(1 + \sigma^2)^2}.

\end{equation}

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

\begin{equation}\label{eqn:shoSuperposition:140}

f”(\sigma) = \frac{-2 \sigma (3 – \sigma^2)}{(1 + \sigma^2)^3}.

\end{equation}

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

\begin{equation}\label{eqn:shoSuperposition:160}

\psi = \inv{\sqrt{2}}\lr{ \ket{0} + \ket{1} }.

\end{equation}

That maximized position expectation is

\begin{equation}\label{eqn:shoSuperposition:180}

\expectation{x}

=

\frac{x_0}{\sqrt{2}}.

\end{equation}

## (b)

The time evolution is given by

\begin{equation}\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}

\end{equation}

## (c)

The position expectation in the Schrodinger representation is

\begin{equation}\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}

\end{equation}

## (d)

\begin{equation}\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}

\end{equation}

matching the calculation using the Schrodinger picture.

## (e)

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

\begin{equation}\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}

\end{equation}

The uncertainty is

\begin{equation}\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}

\end{equation}

# References

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