QQuestionMathematics
QuestionMathematics
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_1 e^{-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##??
4 months agoReport content
Answer
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Step 1:Find the eigenvalues
\det(A - \sigma I) = 0
Set up the characteristic equation for the matrix to find its eigenvalues.
Step 2:Solve the characteristic equation
\det\left(\lambda \begin{pmatrix} -\sigma/\lambda & 1 & 0 \\ 1 & -\sigma/\lambda & 0 \\ 0 & 0 & 2 - \sigma/\lambda \end{pmatrix}\right) = 0
Rewrite the matrix and solve for the roots of the determinant.
Step 3:Find the eigenvalues explicitly
(2\lambda - \sigma)((-\sigma/\lambda)^2 - 1) = 0
Expand the determinant and solve for the eigenvalues.
Step 4:Eigenvalues are
\sigma_1 = 2\lambda, \quad \sigma_2 = \lambda, \quad \sigma_3 = -\lambda
The matrix has three eigenvalues: $2\lambda$, $\lambda$, and $-\lambda$.
Step 5:Find the eigenvectors
Substitute each eigenvalue back into $(A - \sigma I)\mathbf{v} = 0$ to find the corresponding eigenvectors.
Step 6:Eigenvectors
\mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \quad \mathbf{v}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \quad \mathbf{v}_3 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}
The normalized eigenvectors corresponding to the eigenvalues are as shown.
Final Answer
\text{Eigenvalues: } 2\lambda, \lambda, -\lambda$$
$$\text{Eigenvectors: } \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}
The eigenvalues and normalized eigenvectors of $A$ are listed above.
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