Introductory Quantum Mechanics exercise.

[gasar8]

Homework Statement

We describe particle's movement with the Hamiltonian:
H=- \frac{\Delta E}{2} |0\rangle \langle0| + \frac{\Delta E}{2} |1\rangle \langle1|,

where |0\rangle and |1\rangle are the ortonormal basis. Let:
|a\rangle = \frac{1}{\sqrt{2}}|0\rangle +\frac{i}{\sqrt{2}} |1\rangle.

a) Find |b\rangle state, so that it would form orthonormal basis of a Hilbert space with |a\rangle.
b) Find eigenvalues and eigenstates of a projector P_b=|b \rangle \langle b|.
c) Let a particle be in an |a \rangle state at t=0. Find the time evolution of a wave function.
d) At t>0 we do a measurement of an operatorP_b. What are possible results of a measurement and what are their chances?

The Attempt at a Solution

a) We can write:
|b \rangle = A |0 \rangle + B |1 \rangle\\ \langle a|b \rangle=0 \\ \langle b|b \rangle=1.
We get:
A=iB \\ |A|^2 + |B|^2=1,
and finally:
|b \rangle = \frac{i}{\sqrt{2}} |0 \rangle +\frac{1}{\sqrt{2}} |1 \rangle.

b) I am not sure what to do here? Do I have to use a projector on states?
P_b |b \rangle =|b \rangle \langle b|b \rangle
I get only b state (because \langle b|b \rangle=1) as an eigenfunction (?), but I'm not sure what are the eigenvalues then?
P_b |a \rangle =0c) | a,t \rangle = e^{-i \frac{H}{\hbar} t} | a,0 \rangle \\ H|0\rangle=-\frac{\Delta E}{2} |0\rangle \\ H|1\rangle= \frac{\Delta E}{2} |1\rangle \\ | a,t \rangle = \frac{1}{\sqrt{2}} e^{i \frac{\Delta E}{2 \hbar} t} |0 \rangle + \frac{i}{\sqrt{2}} e^{-i \frac{\Delta E}{2 \hbar} t} |1 \rangle
Is this OK?

d) Here, I have got some problems. I am thinking - we can get 0, if the wave function is still in |a\rangle state or (I don't know what) if the wave function gets into |b\rangle state.
But on the other hand I think here must be something with |0\rangle and |1\rangle states, so that I use only square of absolute values of coefficients from c) for their chances (which are 50:50?).

Thank you for your answers.

 

[kontejnjer]

Your part a) seems to be correct, although you've ignored a potential phase factor, making the general solution for |b\rangle=\frac{i}{\sqrt 2} e^{i\phi}+\frac{1}{\sqrt 2}e^{i\phi} (not -really- an issue, just being pedantic here).

For part b), you're on the right track: if P_b |b\rangle=|b\rangle, what does that tell you about the eigenvalues and the eigenvectors (hint: it's almost trivial)? A similar conclusion can be obtained from P_b|a\rangle=0.

Part c) looks good.

Regarding the last part, I think you can just rewrite state |a,t\rangle in terms of states |a,0\rangle and |b,0\rangle instead, and then apply the projector to it and see which coefficients you obtain.

 

[gasar8]

a) Ok, but what does this phase factor mean? I imagine those two states |a\rangle and |b\rangle in a plane, and they are perpendicular (like x and y-axis in coordinate plane). Does this only mean that they are randomly directed in a plane like on the next pic?

b) I think I have got problems with this precisely because it is trivial and don't really understand what are eigenvalues and -vectors of such kind of operators. :)
From the definition of an eigenvector and eigenvalue (A v=\lambda v, where A is operator, v is eigenvector and \lambda is eigenvalue), I would say that in our case P_b |b\rangle=|b\rangle, eigenvalue is 1 and eigenvector is |b\rangle and in P_b|a\rangle=0 eigenvalue is 0 and eigenvector still |a\rangle? Or vector \vec{0}?

d) Big thanks! I rewrited:
| a,t \rangle = A |a,0 \rangle + B |b,0 \rangle \\ =\frac{1}{\sqrt{2}}(A+iB)|0 \rangle +\frac{1}{\sqrt{2}} (iA+B)|1 \rangle,
from which I nicely get:
| a,t \rangle = \cos(\frac{\Delta E}{2 \hbar}t) |a,0 \rangle + \sin(\frac{\Delta E}{2 \hbar}t) |b,0 \rangle
If I now apply P_b to this, I get:
P_b |a,t\rangle=0 |a,0\rangle + \sin(\frac{\Delta E}{2 \hbar}t) |b,0 \rangle.
So the results of a measurement are 0 or sine?
Chances are just squares of absolute values of coefficients in |a,t\rangle and the wave function collapses to the measured state?

 

[kontejnjer]

An absolute phase has no physical significance, it's a sort of a redundancy in the description of the state, and doesn't influence any observables, that is, expectation values of hermitian operators. However, a relative phase difference between states can be detected: see for instance the Aharonov-Bohm effect (Sakurai - Modern quantum mechanics (2010), ch. 2.7 has some info).

You are right for part b), the eigenvalues of the operator are just 0 and 1, and the eigenvectors are |a\rangle and |b\rangle respectively (for the first one just note that P_b|a\rangle=0\cdot |a\rangle). Note that the null vector cannot be an eigenvector by definition, since any linear map acting on it would result in 0 trivially.

The last part is right, but it seems a bit strange that the probabilities do not add up to 1 at all times. However, the operator in question is a projector, so given a basis of a Hilbert space denoted as |c_i\rangle, the completeness relation is \sum\limits_i|c_i\rangle\langle c_i|=1, in your case being |a\rangle\langle a|+|b\rangle\langle b|=1, so nothing is violated, see Sakurai ch. 1.3 for more details.

 

[gasar8]

it seems a bit strange that the probabilities do not add up to 1 at all times.

Why? Isn't the sum of squares of sine and cosine in | a,t \rangle always 1?
Thank you!

Another question.
We've got a free particle with spin 1/2. We mesure z-component of its spin and get \frac{\hbar}{2}.
a) After this, we measure the x-component, too. What are possible results and what are their probabilities?
b) What if we measure the component of its spin at angle \theta about z axis?

a) I am not sure here. Firstly, I would say that its spin wave function collapses after first measurement in z component so then we get 0 with 100%, but in other hand if we use S_x operator, we get something:
S_x |\uparrow \rangle = \frac{S_++S_-}{2} |\uparrow \rangle = \frac{\hbar}{2} |\downarrow \rangle

b) At lectures we said:
| \psi \rangle = \cos(\frac{\theta}{2}) |\uparrow \rangle + \sin(\frac{\theta}{2}) e^{i \phi} |\downarrow \rangle

 

[blue_leaf77]

a) After this, we measure the x-component, too. What are possible results and what are their probabilities?

Are you familiar with working with Pauli matrices? If yes, you can first try to find the eigenvectors of $\sigma_x$ in the basis of eigenvectors of $\sigma_z$.

 

[gasar8]

OK.
<br /> \sigma_{x} = \left(<br /> \begin{array}{cc}<br /> 0 &amp; 1\\<br /> 1 &amp; 0<br /> \end{array}<br /> \right)\\<br /> \sigma_{3} = \left(<br /> \begin{array}{cc}<br /> 1 &amp; 0\\<br /> 0 &amp; -1<br /> \end{array}<br /> \right)<br />
So, the eigenvectors of \sigma_x are x_1=<br /> \left(\begin{array}{cc} <br /> 1\\<br /> 1<br /> \end{array}<br /> \right)<br /> and x_2=<br /> \left(\begin{array}{cc} <br /> -1\\<br /> 1<br /> \end{array}<br /> \right)<br />. The eigenvectors of\sigma_z are x_3=<br /> \left(\begin{array}{cc} <br /> 1\\<br /> 0<br /> \end{array}<br /> \right)<br /> and x_4=<br /> \left(\begin{array}{cc} <br /> 0\\<br /> 1<br /> \end{array}<br /> \right)<br />.
So if I write eigenvectors of \sigma_x in the basis of eigenvectors of \sigma_z, I get:
x_1=x_3+x_4\\x_2=-x_3+x_4.
What does these eigenvectors tell me?

 

[blue_leaf77]

OK.
<br /> \sigma_{x} = \left(<br /> \begin{array}{cc}<br /> 0 &amp; 1\\<br /> 1 &amp; 0<br /> \end{array}<br /> \right)\\<br /> \sigma_{3} = \left(<br /> \begin{array}{cc}<br /> 1 &amp; 0\\<br /> 0 &amp; -1<br /> \end{array}<br /> \right)<br />
So, the eigenvectors of \sigma_x are x_1=<br /> \left(\begin{array}{cc}<br /> 1\\<br /> 1<br /> \end{array}<br /> \right)<br /> and x_2=<br /> \left(\begin{array}{cc}<br /> -1\\<br /> 1<br /> \end{array}<br /> \right)<br />. The eigenvectors of\sigma_z are x_3=<br /> \left(\begin{array}{cc}<br /> 1\\<br /> 0<br /> \end{array}<br /> \right)<br /> and x_4=<br /> \left(\begin{array}{cc}<br /> 0\\<br /> 1<br /> \end{array}<br /> \right)<br />.
So if I write eigenvectors of \sigma_x in the basis of eigenvectors of \sigma_z, I get:
x_1=x_3+x_4\\x_2=-x_3+x_4.
What does these eigenvectors tell me?

Almost correct, you just forgot that a state vector must be normalized. After normalizing them, express $x_3$ (not $x_4$ because initially the state is known to be corresponding to the eigenvalue $\hbar/2$) in terms of $x_1$ and $x_2$. This will allow you to calculate the probability of obtaining the probabilities of the measurement as mentioned in your problem statement.

 

[gasar8]

So
x_1={1 \over \sqrt{2}}<br /> \left(\begin{array}{cc}<br /> 1\\<br /> 1<br /> \end{array}<br /> \right) and
x_2={1 \over \sqrt{2}}<br /> \left(\begin{array}{cc}<br /> -1\\<br /> 1<br /> \end{array}<br /> \right). From this, I get:
x_3={1 \over 2} x_1 - {1 \over 2} x_2.

So, if I understand this correctly, possible results of a x-component measurement are again \pm \frac{\hbar}{2} with the probabilities of square of the coefficients, so both with \frac{1}{4} chance.

express x_3 (not x_4 because initially the state is known to be corresponding to the eigenvalue ℏ/2) in terms of x1 and x_2.

Aha, so I would use x_4 if the result of a measurement would be -\hbar \over 2 which is the other eigenvalue?
But why do I have to express x_3 with x_1 and x_2? -->

If yes, you can first try to find the eigenvectors of σx in the basis of eigenvectors of σz.

Do I just understand this wrong? Because I first expressed just the opposite. :)

 

[blue_leaf77]

x_3={1 \over 2} x_1 - {1 \over 2} x_2.

The coefficients are still off although their ratio is correct, remember any state vector must be normalized - the sum of the modulus square of each coefficient must add up to unity.

Aha, so I would use x4x4x_4 if the result of a measurement would be −ℏ2−ℏ2 -\hbar \over 2 which is the other eigenvalue?

Yes.

But why do I have to express x3x3x_3 with x1x1x_1 and x2x2x_2? -->

Because you want to measure $S_x$ and the only possible outcomes are given by its eigenvalues. Therefore, you need to know how $x_3$ (the initial state) looks like when expressed in the basis of eigenvectors of $S_x$ (or $\sigma_x$).

Do I just understand this wrong? Because I first expressed just the opposite. :)

No you didn't do something wrong. It's just because the most Pauli matrices are written in the basis of eigenvectors of $\sigma_z$. Therefore, when you calculated the eigenvectors of $\sigma_x$ in this form you will automatically get these eigenvectors in basis of eigenvectors of $\sigma_z$.