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?

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.