-1
$\begingroup$

In an exercise in a Quantum Mechanics text (Sakurai Modern Quantum Mechanics) I completed, I showed that the eigenstates $ | \mathbf{S} \cdot \hat{n}; + \rangle$ of $$\mathbf{S} \cdot \hat{n} | \mathbf{S} \cdot \hat{n}, + \rangle= (\frac{\hbar}{2})| \mathbf{S} \cdot \hat{n}; + \rangle$$ are as follows $$| \mathbf{S} \cdot \hat{n}; + \rangle = \cos(\frac{\beta}{2})| + \rangle + \sin(\frac{\beta}{2})e^{i \alpha}|-\rangle.$$

Further it states that given that $\alpha= 0$ we have normalized eigenstates $$(\frac{1 + \cos \beta}{2})^{1/2}\dbinom{1}{\frac{\sin \beta}{(1+\cos \beta)}}.$$

Can anyone see where this comes from? As I see it if $\alpha = 0$ then given that $|+ \rangle = \dbinom{1}{0}$ and $|- \rangle = \dbinom{0}{1}$, we have $|\mathbf{S} \cdot \hat{n} \rangle = \dbinom{\cos(\frac{\beta}{2})}{\sin(\frac{\beta}{2})}$ which is already normalized. What am I missing?

Thanks.

$\endgroup$
  • 1
    $\begingroup$ $\cos(\theta/2)=\sqrt{\frac{1+\cos\theta}{2}}$ etc. $\endgroup$ – ZeroTheHero Apr 24 '17 at 12:48
  • $\begingroup$ @ZeroTheHero The first term I got in that way from the trig identity you mentioned, the second component of the vector I can't seem to get with any identities and I don't see why you would represent an already normalized vector in a simple form in a more complicated form using identities? $\endgroup$ – user101311 Apr 24 '17 at 12:53
  • $\begingroup$ I can't find the form of the equation you're confused about in Sakurai; can you tell me where it is? If it's not in Sakurai, I suspect that this just a quirk of whatever solution set you're looking at. (I also suspect that the person who wrote it may have just plugged it into Mathematica without really knowing what they were doing.) $\endgroup$ – Michael Seifert Apr 27 '17 at 16:12
  • $\begingroup$ @MichaelSeifert Is you have a chance please see my post. $\endgroup$ – user101311 May 2 '17 at 19:32
0
$\begingroup$

To get the eigenvector in the form as given, use the following two identities :

$cos(\beta/2) = ((1+cos\beta)/2)^{1/2}$ and $sin(\beta/2) = ((1-cos\beta)/2)^{1/2}$

To get unity as the first component, you need to factor out the $cos$ term, and after some simplification you should get the desired vector form.

About why should one go for a more complicated expression when both have the property of normalisation, it seems that the advantage with the second form is in the $\beta$ factor, you just need to insert the given angle(whatever the context of the angle) and not divide by 2. (This help with mental calculations for typical angles like $\pi,\pi/2$ but yeah, I see that doesn't hold much ground.

There is also the 1 element which might make calculations easier when dealing with matrices(action of operators)

$\endgroup$
  • $\begingroup$ Having looked over my notes from when I actually did this problem in a QM course [mumble-mumble] years ago, it's important to note that Sakurai asks you to "treat the problem as a straightforward eigenvalue problem" for the operator $\vec{S} \cdot \hat{n}$. The more complicated expression is the one that arises naturally when you do this, since the entries for the matrix representation of $\vec{S} \cdot \hat{n}$ are all in terms of $\beta$. The simplified form in terms of $\beta/2$ is much nicer, though. $\endgroup$ – Michael Seifert Apr 27 '17 at 16:23
  • $\begingroup$ @MichaelSeifert Thanks for you response and for checking. I did the question without using any identities and the answer seems to agree with what is expected with using the identity. $\endgroup$ – user101311 Apr 29 '17 at 21:50
  • 1
    $\begingroup$ @Moses the answers are expected to match without or without application of identities because the two eigenvectors differ by only a phase factor, so to say. $\endgroup$ – user2578520 May 1 '17 at 10:39

Your Answer

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