Understanding how to "tack on" the time wiggle factor

[hmparticle9]

Homework Statement

An observable $\hat{A}$ is represented by the matrix
$$A = \lambda \begin{pmatrix}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 2
\end{pmatrix}$$

(a) Find the eigenvalues and eigenvectors of $A$

(b) Suppose that the system starts out in the generic state
$$| S(0) \rangle =
\begin{pmatrix}
c_1 \\
c_2 \\
c_3
\end{pmatrix}
$$
with $|c_1|^2 + |c_2|^2 + |c_3|^2 = 1$. Find the expectation values (at $t==0$) of $\hat{A}$.

(c) What is $| S(t) \rangle$? If you measured $\hat{A}$ at time $t$, what values might you get , and what is the probability of each?

Relevant Equations

$\langle \hat{A} \rangle = \langle \Psi | \hat{A} | \Psi \rangle$

The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself.

Part (a) is quite easy. We get

$$\sigma_1 = 2\lambda, \mathbf{v}_1 =
\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}
\sigma_2 = \lambda, \mathbf{v}_2 =
\begin{pmatrix}
1/\sqrt{2} \\
1/\sqrt{2} \\
0
\end{pmatrix}
\sigma_3 = -\lambda, \mathbf{v}_3 =
\begin{pmatrix}
1/\sqrt{2} \\
-1/\sqrt{2} \\
0
\end{pmatrix}
$$

There are two ways to solve (b). The easiest is

$$\langle \hat{A} \rangle = \langle S(0) | \hat{A} | S(0) \rangle
= \lambda
\begin{pmatrix}
c_1^* & c_2^* & c_3^*\\
\end{pmatrix}
\begin{pmatrix}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 2
\end{pmatrix}
\begin{pmatrix}
c_1 \\
c_2 \\
c_3
\end{pmatrix}
= 2 \lambda |c_3|^2 + \lambda c_1^*c_2 + \lambda c_2^*c_1
$$

But I have done a problem similar to this earlier in the textbook so I "anticipated" what I thought I might need. So I wrote:
$$| S(0) \rangle =
c_3
\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}
+
\frac{c_1 + c_2}{\sqrt{2}}
\begin{pmatrix}
1/ \sqrt{2} \\
1/ \sqrt{2} \\
0
\end{pmatrix}
+
\frac{c_1 - c_2}{\sqrt{2}}
\begin{pmatrix}
1/ \sqrt{2} \\
-1/ \sqrt{2} \\
0
\end{pmatrix}
$$

We have written the state as a sum of eigenvectors, which means that we can write

$$\langle \hat{A} \rangle =
2 \lambda |c_3|^2
+
\lambda \bigg|\frac{c_1 + c_2}{\sqrt{2}}\bigg|^2
-
\lambda \bigg|\frac{c_1 - c_2}{\sqrt{2}}\bigg|^2
= 2 \lambda |c_3|^2 + \lambda c_1^*c_2 + \lambda c_2^*c_1$$

Part (c) is where I am having problems. In a previous example the author just "tacks on the time wiggle factor"

$$| S(t) \rangle =
c_3
\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix} e^{-2i\lambda t / h}
+
\frac{c_1 + c_2}{\sqrt{2}}
\begin{pmatrix}
1/ \sqrt{2} \\
1/ \sqrt{2} \\
0
\end{pmatrix} e^{-i\lambda t / h}
+
\frac{c_1 - c_2}{\sqrt{2}}
\begin{pmatrix}
1/ \sqrt{2} \\
-1/ \sqrt{2} \\
0
\end{pmatrix} e^{i\lambda t / h}
$$

The possible values we can get are the eigenvalues of $A$. The probability of obtaining each is the squared magnitude of the coefficient of the corresponding eigenvector.

This is not correct. From looking at the solution it appears that we make something like the following map:

$$
\begin{pmatrix}
c_1 \\
c_2 \\
c_3
\end{pmatrix}
\rightarrow
\begin{pmatrix}
c_1e^{-i \lambda t /h} \\
c_2 e^{i \lambda t /h}\\
c_3 e^{-2i \lambda t /h}
\end{pmatrix}
$$

I say "something like" because I don't understand how you choose which "time-wiggle-factor" to apply to which $c_i$. Making this substitution we get closer to the solution

$$| S(t) \rangle =
c_3 e^{-2i\lambda t / h}
\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}
+
\frac{c_1 e^{-i\lambda t / h} + c_2 e^{i\lambda t / h}}{\sqrt{2}}
\begin{pmatrix}
1/ \sqrt{2} \\
1/ \sqrt{2} \\
0
\end{pmatrix}
+
\frac{c_1 e^{-i\lambda t / h} - c_2 e^{i\lambda t / h}}{\sqrt{2}}
\begin{pmatrix}
1/ \sqrt{2} \\
-1/ \sqrt{2} \\
0
\end{pmatrix}
$$

That is, the square magnitude of the coefficients of eigenvectors look a lot more like the corresponding probabilities in the solution.

So if it is the case that by "tacking on" the time-wiggle factors, we tack the time-wiggle factor onto each component of the initial state vector, that makes perfect sense to me. But how do decide which wiggle factor to tack onto which coordinate of $| S(0) \rangle$??

The example I refereed to earlier in the problem is here:

the author is clearly tacking on eigenvectors rather than that of the coefficients of the initial state. Indeed if I apply my method above by tacking on the wiggle factor to the components of the initial state I get an incorrect answer...

I am quite confused.

 

[kuruman]

This is not correct.

What is not correct?
You found the correct eigenvalues $\lambda_k$ and the corresponding eigenvectors $|\psi_k\rangle.$
You tacked on the "wiggle factor" to get $$| S(t) \rangle =\sum_k c_k | \psi_k\rangle e^{-\lambda_k t/\hbar}.$$ It follows that the expectation value takes the form $$ \langle S(t)|A| S(t) \rangle =\sum_l \sum_k {c_l}^* c_k \langle \psi_l|A | \psi_k\rangle e^{+\lambda_l t/\hbar} e^{-\lambda_k t/\hbar}$$ $$ = \sum_l \sum_k {c_l}^* c_k\lambda_k \langle \psi_l | \psi_k\rangle e^{+\lambda_l t/\hbar} e^{-\lambda_k t/\hbar} $$ $$ = \sum_l \sum_k c^*_lc_k\lambda_k \delta_{kl} e^{+\lambda_l t/\hbar} e^{-\lambda_k t/\hbar} = \sum_k c^*_k c_k\lambda_k e^{+\lambda_k t/\hbar} e^{-\lambda_k t/\hbar}$$ $$=\sum_k |c_k|^2\lambda_k$$ as expected.

 

[kuruman]

I am quite confused.

Once you find the eigenstates, you tack on the "wiggle factor" to them and forget all about the initial state because the original matrix is not diagonal and its basis states are not eigenvectors of $A$. See post #2 about how it's done. Maybe that's the source of your confusion.

 

[hmparticle9]

From this expression
$$| S(t) \rangle =
c_3
\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix} e^{-2i\lambda t / h}
+
\frac{c_1 + c_2}{\sqrt{2}}
\begin{pmatrix}
1/ \sqrt{2} \\
1/ \sqrt{2} \\
0
\end{pmatrix} e^{-i\lambda t / h}
+
\frac{c_1 - c_2}{\sqrt{2}}
\begin{pmatrix}
1/ \sqrt{2} \\
-1/ \sqrt{2} \\
0
\end{pmatrix} e^{i\lambda t / h}
$$

the probability of obtaining (say) $\lambda$ is

$$\bigg|\frac{c_1 + c_2}{\sqrt{2}}\bigg|^2$$

which is not what the solutions say. I have been through the derivation you made in post 2. It is in my book. That is how I came up with the alternative solution to part (b). I am not confused with the derivation. Only the solution to (c) was not what I expected.

We have chatted about Introduction to Quantum Mechanics by Griffiths before. This is a problem from there.

 

[vela]

What is the Hamiltonian of the system? The Hamiltonian determines how a state evolves in time. I don’t think $\hat A$ is the same as the Hamiltonian. Is the matrix representing $\hat A$ written with respect to the energy eigenstates?

 

[hmparticle9]

Okay I have made a major error here. I was treating the matrices as separate problems. There are three matrices $H, A$ and $B$. How do you suggest I proceed?

Really sorry guys, I feel like a complete tool.

"The Hamiltonian determines how a state evolves in time". The penny dropped.

 

[kuruman]

What is the Hamiltonian of the system? The Hamiltonian determines how a state evolves in time. I don’t think $\hat A$ is the same as the Hamiltonian. Is the matrix representing $\hat A$ written with respect to the energy eigenstates?

Good point. I missed that.

 

[vela]

Did you figure it out, @hmparticle9?

 

[hmparticle9]

Yes I did :)

After your post it clicked immediately: The Hamiltonian determines how a state evolves in time. I was treating the observable $A$ as the Hamiltonian.