- #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:

- the possible results of a measurement of ##A## are equiprobable
- the possible results of a measurement of ##B## are equiprobable
- the average value of the operator ##C=i[A,B]## is ##\langle C\rangle=ab##.

**Solution Outline**

The solution outline is straightforward:

- compute eigenvalues and eigenvectors of ##A## and ##B##
- 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## - 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## - 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:

- 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
- 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 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] - 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]