1
$\begingroup$

The task is:

We are performing measurements on hydrogen atom, that is in an unknown state $\psi$. $\psi$ is a superposition of $n=1$ and $n=2$ pure states and is orthogonal to $|n=2,l=1,m=0,s=1/2\rangle$. Determine $\psi$ if we measure $\langle E\rangle=\frac{7}{16}E_0$, $\langle l^2\rangle=\frac{\hbar^2}{2}$, $\langle l_z\rangle=\frac{\hbar}{16}$ and $\langle s_z\rangle=\frac{\hbar}{2}$. ($E_0=-13.6$ eV)

I tried to solve this without spin and got $\psi=\frac{\sqrt3}{2}\psi_{200}+\sqrt{\frac{5}{2^5}}\psi_{211}+\sqrt{\frac{3}{2^5}}\psi_{21-1}$ but I didn't use the $\langle s_z\rangle$ anywhere. My question: does $\langle s_z\rangle=\frac{\hbar}{2}$ mean that electron's spin in this state is $+\frac12$ or should I write $\psi$ as linear combination of all pure states including those with opposite spins?
($\psi=A\psi_{100\frac12}+\psi_{100-\frac12}+\cdots$)

$\endgroup$
  • $\begingroup$ By the way I think your answer is wrong, because there should be a component in the $n=1$ subspace. The solution should go like this. First off, according to my answer, there should only be states in the $s_z=\hbar /2$ subspace. Also, since $\psi$ is orthogonal to the $n=2,l=1,m=0,s=1/2$ state, there cannot be contributions from this state. Thus our state only has components of $\psi_{100}$, $\psi_{200}$, $\psi_{21-1}$, and $\psi_{211}$. Now let $A_1$ be the square amplitude in the $n=1$ subspace and $A_2$ be the square amplitude in the $n=2$ subspace... $\endgroup$ – Brian Moths Jan 31 '15 at 14:29
  • $\begingroup$ Then normalization tells us that $A_1+A_2 =1$, while $\langle E \rangle = \frac{7}{16} E_0$ tells us $A_1+A_2/4 = 7 /16$. These two equations can be solved to give us $A_1$ and $A_2$. Now if $A_{20}$ is the square magnitude in the $n=2$, $l=0$ subspace and $A_{21}$ is the square magnitude in the $n=2$, $l=1$ subspace. Then we know $A_{20} + A_{21} = A_2$. Also, the condition on $\langle l^2 \rangle$ gives us a linear equation relating $A_{20}$ and $A_{21}$, analogous to the equation we got from $\langle E \rangle$. From these two linear equations we can solve for $A_{20}$ and $A_{21}$... $\endgroup$ – Brian Moths Jan 31 '15 at 14:34
  • $\begingroup$ Lastly define $A_{21\pm1}$ to be the square amplitude of the $n=2$, $l=1$, $m=\pm1$ subspace. Then we know $A_{211}+A_{21-1}=A_{21}$, meanwhile $\langle l_z \rangle$ gives us another linear equation relating $A_{211}$ and $A_{21-1}$. Thus we can solve for these two variables as well. Thus we have the magnitudes of all the coefficients. It will be important to note though that the quantum state is not determined uniquely because the complex phase of each coefficient is undetermined by the given information. $\endgroup$ – Brian Moths Jan 31 '15 at 14:39
2
$\begingroup$

The maximum $s_z$ eigenvalue an electron can have is $\hbar/2$. Therefore, the only way that a quantum state can have $\langle s_z \rangle = \hbar/2$ is if the state lies completely within the $s_z = \hbar/2$ eigenspace. Thus your answer may only include pure states with $s_z = \hbar /2$.

$\endgroup$
  • 2
    $\begingroup$ Having said which, it is perfectly possible for any atom to be in a state that's a superposition of states of different spin. The only consequence is that spin is no longer well-defined, which is perfectly OK in quantum mechanics. $\endgroup$ – Emilio Pisanty Jan 31 '15 at 14:17
  • $\begingroup$ So if I understand correctly, if measurement of spin would be somewhere between $-\hbar/2$ and $\hbar/2$, I would have to write linear combination of states with different spin, but in my case all spins must be $+1/2$ because measurement is maximum $\hbar/2$? $\endgroup$ – Jan Jan 31 '15 at 14:28
  • $\begingroup$ That is correct. $\endgroup$ – Brian Moths Jan 31 '15 at 14:42

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.