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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.