If an electron is in a state that the probability of measuring spin along the +x axis is $P(+x)=\dfrac{1}{2}$ and the probability of measuring spin along the +y axis is $P(+y)=\dfrac{1}{2}$, what is the wavefunction of the electron?

I tried to solve the above problem using a general spinor of the form $(a, b)^{T}$ but I' can't figure out the answer.

I used the equations: $|a|^{2}+|b|^{2}=1,\quad P(+x)=|<\Psi_{+x}|\Psi>|^{2}=\dfrac{1}{2} \quad and \quad P(+y)=|<\Psi_{+y}|\Psi>|^{2}=\dfrac{1}{2} $, where $\Psi_{+x}=\dfrac{1}{\sqrt{2}}(1,1)^{T}, \quad \Psi_{+y}=\dfrac{1}{\sqrt{2}}(1,i)^{T} \quad and \quad \Psi=(a, b)^{T}$ but I couldn't find the values of a, b.

Here's exactly what I tried:

I wrote $\Psi$ as $\Psi=(cos\frac{\theta}{2}, sin\frac{\theta}{2}e^{i\phi})^{T}$ so that the normalization is right. Then $$ |<\Psi_{+x}|\Psi>|^{2}=\dfrac{1}{2} \Rightarrow \\ \Big|\dfrac{1}{\sqrt{2}}(1,1)(cos\frac{\theta}{2}, sin\frac{\theta}{2}e^{i\phi})^{T}\Big|^{2}=\dfrac{1}{2} \Rightarrow \\ \Big( cos\frac{\theta}{2}+ sin\frac{\theta}{2}e^{i\phi}\Big)\Big( cos\frac{\theta}{2}+sin\frac{\theta}{2}e^{-i\phi}\Big)=1 \Rightarrow \\ $$

$$ cos^{2}\frac{\theta}{2}+sin^{2}\frac{\theta}{2}+cos\frac{\theta}{2}sin\frac{\theta}{2}e^{-i\phi}+cos\frac{\theta}{2}sin\frac{\theta}{2}e^{i\phi}=1 \Rightarrow $$ $$\\ cos\frac{\theta}{2}sin\frac{\theta}{2}cos\phi=0 \Rightarrow \\$$ $$\dfrac{sin\theta}{2}cos\phi=0 \Rightarrow \\$$ \begin{equation} \boxed{sin\theta cos\phi=0} \end{equation} Following the same steps, this time starting from $|<\Psi_{+y}|\Psi>|^{2}=\frac{1}{2}$ I also found that $$\boxed{sin\theta sin\phi=0}$$

So if $\theta=0$ then how can I find the value of $\phi$? Also, what about $\theta=\pi$?


closed as off-topic by John Rennie, user36790, Rob Jeffries, knzhou, user10851 Jun 13 '16 at 17:41

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  • $\begingroup$ You mentioned what you tried, but it would be helpful to explain why it didn't work for you. Is there some particular part of the work that's confusing? $\endgroup$ – Luke Pritchett Jun 13 '16 at 16:18
  • $\begingroup$ I hope it is clear now $\endgroup$ – Bsmith123 Jun 13 '16 at 17:00
  • $\begingroup$ Two states are mutually exclusive. Hence +x state will have both $\pm y$ states hence if you choose one of them probability again reduce by $1/\sqrt2$. I hope this will help. $\endgroup$ – hsinghal Jun 13 '16 at 17:32

In such cases it is very helpful to write a and b in such a way that your first equation is automatically satisfied and the overall phase is neglected. I would take: $$ \Psi = (\cos\theta,e^{i\phi}\sin\theta)^T $$ You only have to find 2 free parameters now.

Concerning your edit. $\theta$ (your definition) is either 0 or $\pi$, which leaves the two possibilities $$ \Psi_1 = (1,0)^T\\ \Psi_2 = (0,1)^T $$

  • $\begingroup$ Can I somehow narrow it down any further into one spinor? $\endgroup$ – Bsmith123 Jun 13 '16 at 22:05
  • $\begingroup$ No because both wavefunction fulfill the requirements. $\endgroup$ – Jannick Jun 14 '16 at 9:00
  • $\begingroup$ Not even as a linear combination of the two spinors? $\endgroup$ – Bsmith123 Jun 14 '16 at 10:51
  • $\begingroup$ No. A nontrivial (both coeffecients unequal zero) combination of the two does not fulfill the requirements. You don't have enough information to decide. $\endgroup$ – Jannick Jun 14 '16 at 14:56

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