Two-Level Quantum System, Need help Finding State at time t

[B3NR4Y]

Homework Statement

$|1>$ and $|2>$ form an orthonormal basis for a two-level system. The Hamiltonian of this system is given by:
$$\hat{H} = \epsilon \begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix}$$

a.) Is this Hamiltonian hermitian? What is the significance of a hermitian operator?
b.) Find the eigenvalues and eigenvectors of this hamiltonian.
c.) Suppose a particle is in the state $|1>$. An energy measurement is performed on the particle. What are the possible outcomes of such a measurement, and what are the probabilities?

Homework Equations

All that should be necessary is the Schrodinger equation.
$$\hat{H} |\psi> = E|\psi>$$

The Attempt at a Solution

For part a I checked that the eigenvalues of the hamiltonian are real. This kills two birds with one stone because part b asks for the eigenvectors. They were real, and given by 2ε and 0. The significance of an operator being hermitian is that hermitian operators correspond to observables.

For part b I just went through the rigor of finding the eigenvectors.
They were
$|1> = \begin{pmatrix} i \\ 1 \end{pmatrix}$ and
$|2> = \begin{pmatrix} -i \\ 1 \end{pmatrix}$ The first corresponds to the eigenvalue 2ε and the second corresponds to the eigenvalue 0.

For part C I used Schrodinger's equation and said that since |1> is an eigenvector of the hamiltonian it has a constant definite energy given by 2ε with 100% probability. This is where I feel like I am wrong, because I think this violates an uncertainty principle.

For part D I am not sure what to do. I know I should multiple by some exponent with time, but I'm not sure how to find this.

 

[blue_leaf77]

For part b I just went through the rigor of finding the eigenvectors.
They were
|1>=(i1)|1>=(i1) |1> = \begin{pmatrix} i \\ 1 \end{pmatrix} and
|2>=(−i1)|2>=(−i1) |2> = \begin{pmatrix} -i \\ 1 \end{pmatrix} The first corresponds to the eigenvalue 2ε and the second corresponds to the eigenvalue 0.

The problem doesn't say that $|1\rangle$ and $|2\rangle$ are energy eigenvectors and you shouldn't assume this way.

For part C I used Schrodinger's equation and said that since |1> is an eigenvector of the hamiltonian it has a constant definite energy given by 2ε with 100% probability. This is where I feel like I am wrong, because I think this violates an uncertainty principle.

If in part b) you don't assume that $|1\rangle$ and $|2\rangle$ are energy eigenvectors, these vectors will be some superposition states of the energy eigenvectors and thus the energy measurement will yield a multiple possibilities.

 

[B3NR4Y]

Yeah my assumption about the basis was stupid. I did some extra reading to refresh my memory from my first quantum course and realized that.

Since |1> and |2> are any two linearly independent vectors, then an arbitrary state |ψ> can be written c₁ |1> + c₂ |2>. In the case that part c says, c₂ = 0 and c₁=1. However I am not sure how to write the |1> state in terms of the eigenvectors which I think is necessary. I was thinking that if the basis vectors are given by
$|1> = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and
$|2> = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$

Writing |1> in terms of the eigenvectors is difficult to me.

 

[blue_leaf77]

Since |1> and |2> are any two linearly independent vectors, then an arbitrary state |ψ> can be written c1 |1> + c2 |2>.

Yes that's right.

In the case that part c says, c2 = 0 and c1=1. However I am not sure how to write the |1> state in terms of the eigenvectors which I think is necessary.

You actually have a good start already since you have calculated the eigenvectors of $H$ in the basis $\{| i \rangle\}$ with $i = 1,2$. However, let's rename the two eigenvectors using different notation, let's say $|a\rangle$ and $|b\rangle$. Written in vector notation, these eigenvectors of $H$ will become
$$
|a\rangle =
\begin{pmatrix}
i \\ 1
\end{pmatrix} = i
\begin{pmatrix}
1\\ 0
\end{pmatrix}+
\begin{pmatrix}
0\\ 1
\end{pmatrix} = i|1\rangle + |2\rangle
$$
and
$$
|b\rangle =
\begin{pmatrix}
-i \\ 1
\end{pmatrix} = -i
\begin{pmatrix}
1\\ 0
\end{pmatrix}+
\begin{pmatrix}
0\\ 1
\end{pmatrix} = -i|1\rangle + |2\rangle
$$
Your task is to invert these two equations such that $|1\rangle$ and $|2\rangle$ are explicitly written in terms of $|a\rangle$ and $|b\rangle$. After that, you should normalize them.

 

[B3NR4Y]

I see, that is shockingly easy.

So once I have that I should compute $<1|\hat{H}|1>$ and the coefficient of each term is the probability of that energy?

 

[blue_leaf77]

So once I have that I should compute <1|^H|1>

You are not required to calculate the average of the measurement results, only the possible outcomes and the corresponding probability are asked.

 

[B3NR4Y]

Okay I have that |1> can be written:
$$|1\rangle = \frac{1}{\sqrt{2}i} \left(|a\rangle-|b\rangle\right)$$
With this I apply H, which I can use the schrodinger equation to find. The only thing I am uneasy on now is that I think the coefficients will be imaginary, and I am not sure how to reconcile that. I remember in the past we had problems with clebsch-gordan tables and if the coefficients were imaginary we just took the length of that vector (i.e. -i/2 would just be given 1/2).

Sorry if all of these are stupid questions, I haven't done any of this all summer and am trying to get back in the swing of it.

 

[blue_leaf77]

Okay I have that |1> can be written:

|1⟩=1√2i(|a⟩−|b⟩)|1⟩=12i(|a⟩−|b⟩)​

Yes that's right.

The only thing I am uneasy on now is that I think the coefficients will be imaginary, and I am not sure how to reconcile that.

The coefficients are generally complex numbers, it's the probabilities which equal the modulus square of those coefficients that are real.

I remember in the past we had problems with clebsch-gordan tables and if the coefficients were imaginary we just took the length of that vector (i.e. -i/2 would just be given 1/2).

I don't remember one has to deal with imaginary coefficients in calculating CG coefficients. The proportionality constant between the states $|s_1s_2;SS\rangle$ and $|s_1s_2;s_1s_2\rangle$ is agreed to be 1 and also the action of the lowering and raising operators gives only real coefficient, so the CG coefficients must be real.