How Does Superposition Affect Measurements in a 1-D Harmonic Oscillator?

[docnet]

Homework Statement

psb

Relevant Equations

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Consider a one-dimensional harmonic oscillator. $\psi_0(x)$ and $\psi_1(x)$ are the normalized ground state and the first excited states.
\begin{equation}
\psi_0(x)=\Big(\frac{m\omega}{\pi\hbar}\Big)^{\frac{1}{4}}e^{\frac{-m\omega}{2\hbar}x^2}
\end{equation}
\begin{equation}
\psi_1(x)=\Big(\frac{m\omega}{\pi\hbar}\Big)^{\frac{1}{4}}\sqrt{\frac{2m\omega}{\hbar}}xe^{\frac{-m\omega}{2\hbar}x^2}
\end{equation}
(a) Construct a state for the particle that is a linear combination
$$\psi(x)=b_0\psi_0(x)+b_1\psi_1(x)$$
$$\psi(x)=b_0\Big(\frac{m\omega}{\pi\hbar}\Big)^{\frac{1}{4}}e^{\frac{-m\omega}{2\hbar}x^2}+b_1\Big(\frac{m\omega}{\pi\hbar}\Big)^{\frac{1}{4}}\sqrt{\frac{2m\omega}{\hbar}}xe^{\frac{-m\omega}{2\hbar}x^2}$$
Find $b_1$ in terms of $b_0$.
$$\int_0^a<b_0\psi_0+b_1\psi_1|b_0\psi_0+b_1\psi_1>dx=1$$
$$b_0^2+b_1^2=1$$
$$b_1=\sqrt{1-b_0^2}$$
(b) Which particular linear combination will maximize $<\psi|\hat{x}|\psi>$?
$$<\psi|\hat{x}|\psi>=\Big<b_0\psi_0(x)+b_1\psi_1(x)\Big|\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}+\hat{a}^{\dagger})\Big|b_0\psi_0(x)+b_1\psi_1(x)\Big>$$
$$=\sqrt{\frac{\hbar}{2m\omega}}\Big<b_0\psi_0(x)+b_1\psi_1(x)\Big|b_1\psi_{0}(x)+b_0\psi_1(x)+b_1\sqrt{2}\psi_2(x)\Big>$$
$$=b_0b_1\sqrt{\frac{\hbar}{2m\omega}}\int^a_0\Big(\psi_0(x)^2+\psi_1(x)^2\Big)dx$$
maximize $b_0=b_1$ $\rightarrow$ $<\psi|\hat{x}|\psi>$.
$$\frac{d}{db_0}b_0\sqrt{1-b_0^2}=\sqrt{1-b_0^2}-\frac{b_0^2}{\sqrt{1-b_0^2}}=0\Rightarrow b_0,b_1=\sqrt{\frac{1}{2}}$$
$$max(b_0,b_1)\Rightarrow (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$$

 

[TSny]

Looks good to me.