Two-level Quantum System - initial state

In summary, the initial state of a two-level quantum system is found by summing the eigenvectors of the two observables.
  • #1
Coelum
97
32
TL;DR Summary: Find the initial state of a two-level quantum system, given the probability of measurements for two observables and the expected value of an operator.

Dear PFer's,
I have been struggling with the following problem. It was assigned at an exam last year.

Problem Statement
For a two-level quantum system we have:
[tex]
H=\epsilon \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix}, ~ ~
A=a \begin{pmatrix} 1 & \sqrt 2 \\ \sqrt 2 & 0 \\ \end{pmatrix}, ~ ~
B=b \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}, ~ ~
[/tex] where ##H## represents the Hamiltonian, ##A## and ##B## represent two observables.
At ##t=0## the system is in a state such that:
  1. the possible results of a measurement of ##A## are equiprobable
  2. the possible results of a measurement of ##B## are equiprobable
  3. the average value of the operator ##C=i[A,B]## is ##\langle C\rangle=ab##.
Find the state of the system at ##t=0##.

Solution Outline
The solution outline is straightforward:
  1. compute eigenvalues and eigenvectors of ##A## and ##B##
  2. express the initial state ##\Psi_0## as the sum, with equal probability, of the eigenvectors of ##A## (say ##A_0## and ##A_1##):
    [itex] \Psi_0 = N_a (A_0+e^{i\phi_a}A_1)[/itex] where ##N_a## is a normalization factor and we arbitrarily set the phase of ##A_0## to ##0##
  3. express the initial state ##\Psi_0## as the sum, with equal probability, of the eigenvectors of ##B## (say ##B_0## and ##B_1##):
    [itex] \Psi_0 = N_b (B_0+e^{i\phi_b}B_1)[/itex] where ##N_b## is a normalization factor and we arbitrarily set the phase of ##B_0## to ##0##
  4. equate the two expressions above to get a system of two equations (one per dimension) in two unknowns (the phases ##\phi_a## and ##\phi_b##).
Solution Attempt
Following that outline, I get:
  1. the eigenvalues of ##A## are ##-a## and ##2a##, with eigenvectors ##A_0=a(-1,\sqrt 2)^T## and ##A_1=a(\sqrt 2,1)^T## respectively
  2. the eigenvalues of ##B## are ##-b## and ##b##, with eigenvectors ##B_0=b(1,-1)^T## and ##B_1=b(1,1)^T## respectively
  3. the initial state, expressed in the eigenvectors of ##A## and ##B##, gives the vector equation
    [tex]
    \Psi_0 = \frac{1}{\sqrt{3(1+e^{i2\phi_a})}} \begin{pmatrix} \sqrt 2-e^{i\phi_a} \\ 1+\sqrt 2 e^{i\phi_a} \end{pmatrix}
    = \frac{\sqrt 2}{\sqrt{3(1+e^{i2\phi_b})}} \begin{pmatrix} 1+e^{i\phi_b} \\ 1-e^{i\phi_b} \end{pmatrix}.
    [/tex]
  4. Letting ##e^{i\phi_a}=u## and ##e^{i\phi_b}=v##, the system of equations becomes
    [tex] \begin{cases}
    (1-u\sqrt 2)\sqrt{1+v^2}=(1+v)\sqrt{1+u^2} \\
    (1/\sqrt 2+u)\sqrt{1+v^2}=(1-v)\sqrt{1+u^2}. \end{cases}
    [/tex]
I cannot solve the system. I guess the third condition (on ##\langle C\rangle##) will help disambiguate the solution, but I can't see how it may help simplify the system. Maybe there is an easier way to answer the question?
 
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  • #2
You must incorporate the third condition. That's why it's there. I don't have time at the moment to look at this myself.
 
  • #3
Okay. I think I've cracked this tough nut! First, the suggested solution is misleading. I'm on my phone, so I can't type as much as I'd like. Your eigenvectors are correct, although I would have normalised them.

We can assume that$$\Psi_0 = \frac 1 {\sqrt 2} \big ( A_1 + \alpha A_2 \big )$$Where, ##\alpha## is a complex phase factor. Then, we have$$\Psi_0 = \frac 1 {\sqrt 6}[1+\alpha \sqrt 2, \alpha - \sqrt 2]$$Then, I calculated the commutator ##[A,B]##, and using the given expected value solved for ##\alpha##. Which came out rather surprisingly, I must admit.
 
  • #4
Now, we can assume that $$\Psi_0 = \frac 1 {\sqrt 2} \big (\beta B_1 + \gamma B_2 \big )$$Note that we cannot take ##\beta =1## here. Then, equating this with ##\Psi_0## above, we can solve for ##\beta## and ##\gamma##. Note, however, that the solution was obtained without considering the condition on operator ##B##.

Which, of course, means that the problem as stated appears to be over specified?

Hmm?
 
Last edited:
  • #5
@PeroK: thanks a lot for your remarks.
Let me fix my errors:
  1. You pointed out one error: I cannot take an arbitrary phase for the expression of ##\Psi_0## in terms of the eigenvectors of ##B## since I already took that liberty.
  2. I found another (minor) error: I forgot the ##a## and ##b## factors for the eigenvectors of ##A## and ##B##. That was just a typo - my notes are correct - and was not influential.
I do not understand how you computed the normalization factor for ##\Psi_0##. It should contain the phase factor, shouldn't it?

Following your advice, I computed the average value of ##C## in both representations. Here I use your notation but I start indices from 0, not from 1.
First, represent ##\Psi_0## in the two basis as
\begin{align*}
\Psi_0 & = N_a a ( A_0+\alpha A_1) & = N_a a \begin{pmatrix}-1+\sqrt 2\alpha \\ \sqrt 2+\alpha \end{pmatrix} \\
\Psi_0 & = N_b b/\sqrt 2(\beta B_0+\gamma B_1) & = N_b b\sqrt 2\begin{pmatrix}\beta+\gamma \\ -\beta+\gamma \end{pmatrix}.
\end{align*}
Then, compute the representation of ##C## as
[tex]
C = i[A,B] = iab\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.
[/tex]
Finally, compute ##\langle C\rangle## in both bases. In the base of ##A## we get
[tex]\begin{align*}
ab & = ibN_a^2 a^3\begin{pmatrix}-1+\sqrt 2\alpha^* & \sqrt 2+\alpha^* \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
\begin{pmatrix}-1+\sqrt 2\alpha \\ \sqrt 2+\alpha \end{pmatrix} \\
& = 6N_a^2 a^3 b\Re(\alpha).
\end{align*}[/tex]
and in the base of ##B## we get
[tex]\begin{align*}
ab & = ibN_b^2 ab^3/2\begin{pmatrix}\beta^*+\gamma^* & -\beta^*+\gamma^* \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
\begin{pmatrix}\beta+\gamma \\ -\beta+\gamma \end{pmatrix}\\
& = 2N_b^2 a b^3(\Re(\gamma)\Im(\beta)-\Re(\beta)\Im(\gamma)).
\end{align*}[/tex]
I still do not see how to solve.
 
  • #6
You have too many variables there. There is an arbitrary phase factor that we can apply to the state, but this should not change any of the three conditions. I think you've got that. Having a variable normalisation factor confuses things, IMO. You can just normalised the eigenvectors you found. Then, if ##\Psi_0## has equal probability of both eigenvalues, the equation for ##\Psi_0## in terms of A's eigenvectors can be significantly simplified. See my previous post. There is only a single complex factor of unit modulus, namely ##\alpha##.

Plugging that into the third equation (expectation of that commutator) gave me ##\alpha =1##.

It would be good if you could corroborate that.

But, then, the equal probability for B fails. So, I'm not totally sure that I'm correct in this.
 
  • #7
PeroK said:
But, then, the equal probability for B fails. So, I'm not totally sure that I'm correct in this.
That said, if the normalisation factors are taken to be real (in your original post), then the solution outline is clearly wrong. As you cannot assume the same simplification for both eigenvector expansions. Possibly, whoever did the question and solution outline overlooked that?
 
  • #8
PeroK said:
You have too many variables there. There is an arbitrary phase factor that we can apply to the state, but this should not change any of the three conditions. I think you've got that. Having a variable normalisation factor confuses things, IMO. You can just normalised the eigenvectors you found. Then, if ##\Psi_0## has equal probability of both eigenvalues, the equation for ##\Psi_0## in terms of A's eigenvectors can be significantly simplified. See my previous post. There is only a single complex factor of unit modulus, namely ##\alpha##.
In fact, I did not normalize the eigenvectors, just the state. I cannot see any further simplification of ##\Psi_0## beyond what I already did.
PeroK said:
Plugging that into the third equation (expectation of that commutator) gave me ##\alpha =1##.

It would be good if you could corroborate that.

But, then, the equal probability for B fails. So, I'm not totally sure that I'm correct in this.
I cannot replicate your results. I'm still stuck with the two nonlinear equations for the three phases ##\phi_a## and ##\phi_b## from the first two conditions. The third condition provides another two equations for the same three unknowns. Given the nonlinearity of all the equations, I don't see how to solve the problem yet.
PeroK said:
That said, if the normalisation factors are taken to be real (in your original post), then the solution outline is clearly wrong. As you cannot assume the same simplification for both eigenvector expansions. Possibly, whoever did the question and solution outline overlooked that?
The question comes from an exam assigned one year ago at an Italian university for the Quantum Mechanics class. I do not expect it contains any error.
 
  • #9
Coelum said:
In fact, I did not normalize the eigenvectors, just the state.
That won't work. You need to normalise the eigenvectors first.
Coelum said:
I cannot see any further simplification of ##\Psi_0## beyond what I already did.
You should be able to get what i got in post #3. You know the eigenvectors, so you ought to use what you have. Keeping the eigenvectors in a general (especually unnormalised form) is not going to work.
Coelum said:
I cannot replicate your results. I'm still stuck with the two nonlinear equations for the three phases ##\phi_a## and ##\phi_b## from the first two conditions. The third condition provides another two equations for the same three unknowns. Given the nonlinearity of all the equations, I don't see how to solve the problem yet.
You have too many variables.
Coelum said:
The question comes from an exam assigned one year ago at an Italian university for the Quantum Mechanics class. I do not expect it contains any error.
Nor do I. I can take another look later today. It would be better, of course, if you could make progress yourself.
 
  • #10
Coelum said:
Solution Attempt
Following that outline, I get:
  1. the eigenvalues of ##A## are ##-a## and ##2a##, with eigenvectors ##A_0=a(-1,\sqrt 2)^T## and ##A_1=a(\sqrt 2,1)^T## respectively
  2. the eigenvalues of ##B## are ##-b## and ##b##, with eigenvectors ##B_0=b(1,-1)^T## and ##B_1=b(1,1)^T## respectively
The normalized eigenvectors cannot depend on ##a## or ##b##.
 
  • #11
I have just solved it for myself, and the answer is simpler to get than I thought. It turns out that there is a single state (apart from a global complex phase) that satisfies conditions 1 and 3 (or 2 and 3).
 
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  • #12
DrClaude said:
I have just solved it for myself, and the answer is simpler to get than I thought. It turns out that there is a single state (apart from a global complex phase) that satisfies conditions 1 and 3 (or 2 and 3).
But not all three conditions?
 
  • #13
DrClaude said:
I have just solved it for myself, and the answer is simpler to get than I thought. It turns out that there is a single state (apart from a global complex phase) that satisfies conditions 1 and 3 (or 2 and 3).
Yes, I got it. Using PeroK's notation, it is ##\alpha=-i##. In fact, one of the two conditions 1 and 2 is redundant. I got the solution using conditions 1 and 3 only. It automatically checks condition 2. Thanks everybody for your help!
 
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  • #14
Coelum said:
Yes, I got it. Using PeroK's notation, it is ##\alpha=-i##. In fact, one of the two conditions 1 and 2 is redundant. I got the solution using conditions 1 and 3 only. It automatically checks condition 2. Thanks everybody for your help!
I had an error at the end of my calculations that gave ##\alpha =1##, by confusing real and imaginary parts!
 
  • #15
PeroK said:
But not all three conditions?
As @Coelum said, conditions 1 and 2 are redundant. Either will lead you to the solution. It makes sense since
$$
\Psi_0 = c_1 \phi_1 + c_2 \phi_2
$$
where the ##\phi##'s are the eigenstates of the Hamiltonian has only two actual unknowns, ##|c_1| / |c_2|## and the relative complex phase between ##c_1## and ##c_2##, so only two conditions are needed.
 
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