Introductory Quantum Mechanics exercise.

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Homework Statement


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

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

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

The Attempt at a Solution


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

b) I am not sure what to do here? Do I have to use a projector on states?
[tex]P_b |b \rangle =|b \rangle \langle b|b \rangle [/tex]
I get only b state (because [itex]\langle b|b \rangle=1[/itex]) as an eigenfunction (?), but I'm not sure what are the eigenvalues then?
[tex]P_b |a \rangle =0 [/tex]


c) [tex] | 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 [/tex]
Is this OK?

d) Here, I have got some problems. I am thinking - we can get 0, if the wave function is still in [itex] |a\rangle[/itex] state or (I don't know what) if the wave function gets in to [itex] |b\rangle[/itex] state.
But on the other hand I think here must be something with [itex] |0\rangle[/itex] and [itex] |1\rangle[/itex] 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.
 

Answers and Replies

  • #2
72
21
Your part a) seems to be correct, although you've ignored a potential phase factor, making the general solution for [itex]|b\rangle=\frac{i}{\sqrt 2} e^{i\phi}+\frac{1}{\sqrt 2}e^{i\phi}[/itex] (not -really- an issue, just being pedantic here).

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

Part c) looks good.

Regarding the last part, I think you can just rewrite state [itex]|a,t\rangle[/itex] in terms of states [itex]|a,0\rangle[/itex] and [itex]|b,0\rangle[/itex] instead, and then apply the projector to it and see which coefficients you obtain.
 
  • #3
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a) Ok, but what does this phase factor mean? I imagine those two states [itex]|a\rangle[/itex] and [itex]|b\rangle[/itex] 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?
rotacija.jpg


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 ([itex] A v=\lambda v[/itex], where [itex] A[/itex] is operator, [itex] v[/itex] is eigenvector and [itex] \lambda [/itex] is eigenvalue), I would say that in our case [itex]P_b |b\rangle=|b\rangle[/itex], eigenvalue is 1 and eigenvector is [itex]|b\rangle[/itex] and in [itex]P_b|a\rangle=0[/itex] eigenvalue is 0 and eigenvector still [itex]|a\rangle[/itex]? Or vector [itex]\vec{0}[/itex]?

d) Big thanks! I rewrited:
[tex]| 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, [/tex]
from which I nicely get:
[tex]| a,t \rangle = \cos(\frac{\Delta E}{2 \hbar}t) |a,0 \rangle + \sin(\frac{\Delta E}{2 \hbar}t) |b,0 \rangle[/tex]
If I now apply [itex]P_b [/itex] to this, I get:
[tex]P_b |a,t\rangle=0 |a,0\rangle + \sin(\frac{\Delta E}{2 \hbar}t) |b,0 \rangle.[/tex]
So the results of a measurement are 0 or sine?
Chances are just squares of absolute values of coefficients in [itex]|a,t\rangle[/itex] and the wave function collapses to the measured state?
 
  • #4
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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 [itex]|a\rangle[/itex] and [itex]|b\rangle[/itex] respectively (for the first one just note that [itex]P_b|a\rangle=0\cdot |a\rangle[/itex]). 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 [itex]|c_i\rangle[/itex], the completeness relation is [itex]\sum\limits_i|c_i\rangle\langle c_i|=1[/itex], in your case being [itex]|a\rangle\langle a|+|b\rangle\langle b|=1[/itex], so nothing is violated, see Sakurai ch. 1.3 for more details.
 
  • #5
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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 [itex]| a,t \rangle[/itex] 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 [itex]\frac{\hbar}{2}[/itex].
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 [itex]\theta[/itex] 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 [itex]S_x[/itex] operator, we get something:
[tex]S_x |\uparrow \rangle = \frac{S_++S_-}{2} |\uparrow \rangle = \frac{\hbar}{2} |\downarrow \rangle[/tex]

b) At lectures we said:
[tex]| \psi \rangle = \cos(\frac{\theta}{2}) |\uparrow \rangle + \sin(\frac{\theta}{2}) e^{i \phi} |\downarrow \rangle[/tex]
 
  • #6
blue_leaf77
Science Advisor
Homework Helper
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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##.
 
  • #7
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OK.
[tex]
\sigma_{x} = \left(
\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}
\right)\\
\sigma_{3} = \left(
\begin{array}{cc}
1 & 0\\
0 & -1
\end{array}
\right)
[/tex]
So, the eigenvectors of [itex]\sigma_x[/itex] are [itex] x_1=
\left(\begin{array}{cc}
1\\
1
\end{array}
\right)
[/itex] and [itex] x_2=
\left(\begin{array}{cc}
-1\\
1
\end{array}
\right)
[/itex]. The eigenvectors of[itex]\sigma_z[/itex] are [itex] x_3=
\left(\begin{array}{cc}
1\\
0
\end{array}
\right)
[/itex] and [itex] x_4=
\left(\begin{array}{cc}
0\\
1
\end{array}
\right)
[/itex].
So if I write eigenvectors of [itex]\sigma_x[/itex] in the basis of eigenvectors of [itex]\sigma_z[/itex], I get:
[tex]x_1=x_3+x_4\\x_2=-x_3+x_4[/tex].
What does these eigenvectors tell me?
 
  • #8
blue_leaf77
Science Advisor
Homework Helper
2,637
785
OK.
[tex]
\sigma_{x} = \left(
\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}
\right)\\
\sigma_{3} = \left(
\begin{array}{cc}
1 & 0\\
0 & -1
\end{array}
\right)
[/tex]
So, the eigenvectors of [itex]\sigma_x[/itex] are [itex] x_1=
\left(\begin{array}{cc}
1\\
1
\end{array}
\right)
[/itex] and [itex] x_2=
\left(\begin{array}{cc}
-1\\
1
\end{array}
\right)
[/itex]. The eigenvectors of[itex]\sigma_z[/itex] are [itex] x_3=
\left(\begin{array}{cc}
1\\
0
\end{array}
\right)
[/itex] and [itex] x_4=
\left(\begin{array}{cc}
0\\
1
\end{array}
\right)
[/itex].
So if I write eigenvectors of [itex]\sigma_x[/itex] in the basis of eigenvectors of [itex]\sigma_z[/itex], I get:
[tex]x_1=x_3+x_4\\x_2=-x_3+x_4[/tex].
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.
 
  • #9
63
0
So
[itex]x_1={1 \over \sqrt{2}}
\left(\begin{array}{cc}
1\\
1
\end{array}
\right) [/itex] and
[itex]x_2={1 \over \sqrt{2}}
\left(\begin{array}{cc}
-1\\
1
\end{array}
\right) [/itex]. From this, I get:
[tex]x_3={1 \over 2} x_1 - {1 \over 2} x_2.[/tex]

So, if I understand this correctly, possible results of a x-component measurement are again [itex]\pm \frac{\hbar}{2}[/itex] with the probabilities of square of the coefficients, so both with [itex]\frac{1}{4}[/itex] 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 [itex]x_4[/itex] if the result of a measurement would be [itex] -\hbar \over 2 [/itex] which is the other eigenvalue?
But why do I have to express [itex]x_3[/itex] with [itex]x_1[/itex] and [itex]x_2[/itex]? -->
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. :)
 
  • #10
blue_leaf77
Science Advisor
Homework Helper
2,637
785
[tex]x_3={1 \over 2} x_1 - {1 \over 2} x_2.[/tex]
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##.
 

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