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We study a quantum particle confined to a one-dimensional box with walls at the positions $\pm 1$. The Hilbert-space of this system in the Schrödinger representation is hence given by $L_2([-1; 1])$ with the scalar product $$ \left \langle f,g \right \rangle=\int_{-1}^{1}\overline{f(x)}g(x) \,\mathrm{d}x,\quad f,g\in L_2([-1,1]) $$ In this Hilbert space, we consider the two functions $f_0(x) := (1+i) \exp(i\pi x)$ and $f_1(x):=\exp(i2\pi x)$.

Now the problems are as follows

1) We now consider a general superposition $\Psi=\alpha\phi_0+\beta\phi_1$ of the two states $\phi_0:=f_0/2$ and $\phi_1:=f_1/\sqrt{2}$. Show that in order for $\Psi$ to be normalized we need $|\alpha|^2+|\beta|^2=1$.

2) Determine the probability to find the particle right of the origin if it is in the state $\Psi$, depending on the two parameters $\alpha=\left \langle \phi_0,\Psi \right \rangle$ and $\beta=\left \langle \phi_1,\Psi \right \rangle$

3) Find values for $\alpha$ and $\beta$ that maximize the probability to find the particle right of the origin. Hint: By using that $\Psi$ and $e^{i\alpha}\Psi$ represent the same physical state of the system, we can choose $0\leq \alpha\leq 1$ and $\beta=\sqrt{1-\alpha^2}e^{i\eta}$.

I have no problem with number one. I have tried with the second one, but I am not sure if I have reached the right conclusion. Because of that, I can not continue with the last problem.

2) We have to determine $\int_{0}^{1}|\Psi(x)|^2\,\mathrm{d}x$. First, we observe that $$ |\Psi|^2=|\alpha|^2|\phi_0|^2+\alpha \bar{\beta}\overline{\phi_1}\phi_0+ \bar{\alpha}\beta\overline{\phi_0}\phi_1+|\beta|^2|\phi_1|^2 $$ then $$ \int_{0}^{1} |\Psi(x)|^2\,\mathrm{d}x=|\alpha|^2\int_{0}^{1}|\phi_0(x)|^2\,\mathrm{d}x+\alpha \bar{\beta}\int_{0}^{1}\overline{\phi_1(x)}\phi_0(x)\,\mathrm{d}x\\+ \bar{\alpha}\beta\int_{0}^{1}\overline{\phi_0(x)}\phi_1(x)\,\mathrm{d}x+|\beta|^2\int_{0}^{1}|\phi_1(x)|^2\,\mathrm{d}x $$ The first and the last integrals are $1/2$. The two other are $$ \int_{0}^{1}\overline{\phi_1(x)}\phi_0(x)\,\mathrm{d}x=\frac{1+i}{2\sqrt{2}}\int_{0}^{1} e^{-i\pi x}\,\mathrm{d}x=\frac{1-i}{\sqrt{2}\pi}, $$ $$ \int_{0}^{1}\overline{\phi_0(x)}\phi_1(x)\,\mathrm{d}x=\frac{1-i}{2\sqrt{2}}\int_{0}^{1} e^{i\pi x}\,\mathrm{d}x=\frac{i+1}{\sqrt{2}\pi} $$ Altogether, we get $$ \int_{0}^{1} |\Psi(x)|^2\,\mathrm{d}x=\frac{1}{2}\left ( |\alpha|^2+|\beta|^2 \right )+\alpha \bar{\beta}\left ( \frac{1-i}{\sqrt{2}\pi} \right )+ \bar{\alpha}\beta\left ( \frac{i+1}{\sqrt{2}\pi} \right ). $$ Since $$ \alpha \bar{\beta}(1-i)+ \bar{\alpha}\beta(1+i) =\alpha \bar{\beta}(1-i)+ \overline{\alpha\bar{\beta}(1-i)} =2\Re(\alpha \bar{\beta}(1-i)) $$ and with 1), we then get $$ \int_{0}^{1} |\Psi(x)|^2\,\mathrm{d}x=\frac{1}{2}+\frac{2}{\sqrt{2}\pi}\Re(\alpha \bar{\beta}(1-i)). $$ Is this correct, or am I missing something?

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    $\begingroup$ It looks right to me! $\endgroup$ – Adrian Keister Jan 30 at 20:51
  • $\begingroup$ @AdrianKeister I'm glad to hear it... If you have time, could you please show me a way for the last problem? The hint doesn't seem helpful to me. $\endgroup$ – UnknownW Jan 30 at 21:24
  • $\begingroup$ Hmm. Well, the maximization boils down to maximizing $\operatorname{Re}(\alpha\overline{\beta}(1-i)),$ subject to $|\alpha|^2+|\beta|^2=1.$ $\endgroup$ – Adrian Keister Jan 30 at 21:35
  • $\begingroup$ @AdrianKeister That's true. I could write $|\beta|=\sqrt{1-|\alpha|^2}$, which implies $\beta=\sqrt{1-|\alpha|^2}e^{i\eta}$ for some number $\eta$. Should I then calculate the real part of $\alpha \sqrt{1-|\alpha|^2}e^{-i\eta}(1-i)$ and then finding the maximum of it with respect to $ \alpha$, as long as I assume $\alpha$ is a real number? $\endgroup$ – UnknownW Jan 30 at 21:53
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For problem 3), we need to maximize $\operatorname{Re}(\alpha\overline{\beta}(1-i))$ subject to $|\alpha|^2+|\beta|^2=1$. Let \begin{align*} \alpha&=r_{\alpha}e^{i\theta_{\alpha}} \\ \beta&=r_{\beta}e^{i\theta_{\beta}}, \\ \end{align*} which implies $|\alpha|^2+|\beta|^2=r_{\alpha}^2+r_{\beta}^2=1,$ and we're trying to maximize \begin{align*} \operatorname{Re}(\alpha\overline{\beta}(1-i))&=\operatorname{Re}(r_{\alpha}e^{i\theta_{\alpha}}r_{\beta}e^{-i\theta_{\beta}}\sqrt{2}\,e^{-i\pi/4}) \\ &=\sqrt{2}\,r_{\alpha}r_{\beta}\operatorname{Re}\left[e^{i(\theta_{\alpha}-\theta_{\beta}-\pi/4)}\right] \\ &=\sqrt{2}\,r_{\alpha}r_{\beta}\cos(\theta_{\alpha}-\theta_{\beta}-\pi/4). \end{align*} As the $r$'s show up symmetrically, choose $r_{\alpha}=r_{\beta}=1/\sqrt{2},$ and choose \begin{align*} \theta_{\alpha}-\theta_{\beta}-\frac{\pi}{4}&=0 \\ \theta_{\alpha}-\theta_{\beta}&=\frac{\pi}{4} \end{align*} to maximize the cosine.

One final point: once you do all this, you should definitely back-calculate the actual probability, and make sure it isn't bigger than $1.$ If it is, you can always dial back something (the angles would be the easiest) to make it smaller.

A quick calculation: $$\frac12+\frac{2}{\sqrt{2}\,\pi}(\sqrt{2}(1/2))=\frac12+\frac{1}{\pi}<1,$$ so you're safe.

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  • $\begingroup$ Wow, thank you. Quick question: Is it possible to choose, say $r_\alpha=1/2$, so that we have $r_\beta=1-1/4 = 3/4$? (I do not understand the symmetry part.) $\endgroup$ – UnknownW Jan 30 at 23:43
  • $\begingroup$ Well, you could, but the final result would be smaller, and hence not the maximum. The symmetry argument has to do with how the variables are showing up: you could swap them, and you couldn't tell the difference. Often in situations like that, you'll find that the variables being equal is the extremum. $\endgroup$ – Adrian Keister Jan 31 at 0:45

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