Quantum - Two State Problem in different bases

[leo_africanus]

Homework Statement

[/B]
(Working through a problem from a practice set for which I have a solution available, but still don't understand. I get the same answer as they do for part a, but get lost in part b, I think. Relevant portions below)

Consider a two-state quantum system. In the orthonormal and complete set of basis vectors $\langle 1 |$ and $\langle 2 |$, the Hamiltonian operator for the system is represented by (ω > 0):
$$H = 10 \hbar \omega |1⟩⟨1| - 3 \hbar \omega |1⟩⟨2| - 3 \hbar \omega |2⟩⟨1| + 2 \hbar \omega |2⟩⟨2|$$
Consider another complete and orthonormal basis |α⟩, |β⟩, such that $H|α⟩ = E_1|α⟩$, and $H|β⟩ = E_2|β⟩$ (with $E_1 < E_2$). Let the action of operator A on the |α⟩, |β⟩ basis vectors be given as:
$$ A|α⟩ = 2ia_0|β⟩ $$
$$ A|β⟩ = -2ia_0|α⟩ - 3a_0|β⟩$$
where $a_0 > 0$ is real.

a) Find the eigenvalues and eigenvectors of H in the |1⟩, |2⟩ basis.

b) Find the eigenvalues and eigenvectors of A in the |α⟩, |β⟩ basis.

c) Suppose a measurement of A is carried out at t=0 on an arbitrary state and the largest possible value is obtained. Calculate the probability P(t) that another measurement made at time t will yield the value as the one measured at t=0.

Homework Equations

Eigenvalue equation: $|X-\lambda I | = 0$

Orthonormality: $⟨x|y⟩=\delta(x-y)$

Probability: $P(t) = |⟨\psi_0|\psi(t)⟩|^2$

The Attempt at a Solution

[/B]
My disagreement with the available solution comes when writing the prepared state for part c in terms of the eigenvectors for the energy.

a) From the equation for H we can write out the matrix for H in the |1⟩, |2⟩ basis as:
$$ H = \begin{bmatrix}
10ℏω & -3ℏω\\
3ℏω & 2ℏω
\end{bmatrix} $$

Then $|H-\lambda I | = 0$ gives eigenvalues of $\lambda_1 = ℏω = E_1$ and $\lambda_2 = 11ℏω = E_2$,
with corresponding normalized eigenvectors:
$$|\lambda_1⟩ = |α⟩ = \frac{1}{\sqrt{10}} \begin{bmatrix} 1\\3 \end{bmatrix}$$
$$|\lambda_2⟩ = |β⟩ = \frac{1}{\sqrt{10}} \begin{bmatrix} 3\\-1 \end{bmatrix}$$

b) To get the matrix for A in the |α⟩, |β⟩ basis I use the given action of A on the basis vectors and their orthogonality:

$$ A = \begin{bmatrix} ⟨α|A|α⟩ & ⟨α|A|β⟩\\ ⟨β|A|α⟩ & ⟨β|A|β⟩ \end{bmatrix} =
\begin{bmatrix} 2ia_0⟨α|β⟩ & -2ia_0⟨α|α⟩-3a_0⟨α|β⟩\\ 2ia_0⟨β|β⟩ & -2ia_0⟨β|α⟩-3a_0⟨β|β⟩ \end{bmatrix} =
\begin{bmatrix} 0 & -2ia_0\\ 2ia_0 & -3a_0 \end{bmatrix}$$

Then $|A-\lambda I | = 0$ gives eigenvalues of $\lambda_1 = a_0$ and $\lambda_2 = -4a_0$,
with corresponding normalized eigenvectors:
$$|\lambda_1⟩ = \frac{-1}{\sqrt{3}} \begin{bmatrix} 2\\i \end{bmatrix}$$
$$|\lambda_2⟩ = \frac{1}{\sqrt{3}} \begin{bmatrix} i\\2 \end{bmatrix}$$

c) The largest possible value of A that could be measured is $a_0$, which prepares the system in state $|\psi_0⟩ = |a_0⟩$

To get the time dependence for the wave function, I need to express |a_0⟩ in terms of the eigenstates of the Hamiltonian. The way I thought this would be done is simply by writing |a_0⟩ as a linear combination of |α⟩, |β⟩ basis vectors as follows (using the vectors found in part a):
$$x|α⟩ + y|β⟩ = |a_0⟩$$
Giving a system of two linear equations,
$$\frac{x}{\sqrt{10}} + \frac{3y}{\sqrt{10}} = \frac{-2}{\sqrt{3}} $$
$$\frac{3x}{\sqrt{10}} - \frac{y}{\sqrt{10}} = \frac{-i}{\sqrt{3}} $$

Solving for x and y gives me: $x=-\frac{3i+2}{\sqrt{30}}$ and $y=\frac{i-6}{\sqrt{30}}$ for:
$$|a_0⟩ = -\frac{3i+2}{\sqrt{30}}|α⟩ + \frac{i-6}{\sqrt{30}}|β⟩$$

The solution for the problem that I have is consistent until this point, where it gives (without showing any work):
$$|a_0⟩ = \frac{1}{\sqrt{5}}|α⟩ - \frac{2i}{\sqrt{5}}|β⟩$$

Any help with where I've gone wrong here would be much appreciated!

Thanks

 

[TSny]

Then $|A-\lambda I | = 0$ gives eigenvalues of $\lambda_1 = a_0$ and $\lambda_2 = -4a_0$,
with corresponding normalized eigenvectors:
$$|\lambda_1⟩ = \frac{-1}{\sqrt{3}} \begin{bmatrix} 2\\i \end{bmatrix}$$
$$|\lambda_2⟩ = \frac{1}{\sqrt{3}} \begin{bmatrix} i\\2 \end{bmatrix}$$

Your vectors are not normalized. The $\sqrt 3$ is not correct.

Note that what you are getting here is $|\lambda_1⟩$ and $|\lambda_2⟩$ expressed in the energy-eigenstate basis $\{ |α⟩, |β⟩ \}$. So, you immediately have the expansion of $|\lambda_1⟩$, say, in terms of $|α⟩$ and $|β⟩$ without further work.

Their answer $\frac{1}{\sqrt{5}}|α⟩ - \frac{2i}{\sqrt{5}}|β⟩$ appears to correspond to the eigenstate of $A$ with eigenvalue $-4a_0$ rather than corresponding to the state with eigenvalue $a_0$.

 

[leo_africanus]

Your vectors are not normalized. The $\sqrt 3$ is not correct.

Note that what you are getting here is $|\lambda_1⟩$ and $|\lambda_2⟩$ expressed in the energy-eigenstate basis $\{ |α⟩, |β⟩ \}$. So, you immediately have the expansion of $|\lambda_1⟩$, say, in terms of $|α⟩$ and $|β⟩$ without further work.

Their answer $\frac{1}{\sqrt{5}}|α⟩ - \frac{2i}{\sqrt{5}}|β⟩$ appears to correspond to the eigenstate of $A$ with eigenvalue $-4a_0$ rather than corresponding to the state with eigenvalue $a_0$.

Got it, thanks so much for the clarification!

Normalization should be (where $λ_{1,i}$ is the $i^{th}$ component of $|\lambda_1⟩$):
$$ || λ_1 || = \sqrt{\sum_{i=1}^N λ_{1,i} λ_{1,i}^*} = \sqrt{5}$$ (and likewise for $|\lambda_2⟩$)

Thanks again!

 

[TSny]

OK. Good work!