# Quantum Physics - Measurement/Eigenvalues(functions)

1. Feb 21, 2010

### Plutoniummatt

1. The problem statement, all variables and given/known data

For a certain system, an observable A has eigenvalues 1 and -1, with corresponding eigenfunctions $$u_+$$ and $$u_-$$. Another observable B also has eigenvalues 1 and -1, but with corresponding eigenfunctions:

$$v_+ = \frac{u_+ + u_1}{\sqrt{2}}$$

$$v_- = \frac{u_+ - u_1}{\sqrt{2}}$$

Find the possible results of a measurement of C = A+B

2. Relevant equations

3. The attempt at a solution

Measured values are just the eigenvalues, in this case, the eigenvalues of C are just 2 and -2? but the answer is $$\pm\sqrt{2}$$...I'm aware that the $$1/\sqrt{2}$$ in the eigenfunctions of B will make my answer "correct" but then they're not the eigenvalues anymore?

2. Feb 21, 2010

### gabbagabbahey

Why do you say this?

3. Feb 21, 2010

### Plutoniummatt

if C = A + B, and A and B both have eigenvalues of 1 and -1, then the eigenvalues of C are 2 and -2?

4. Feb 21, 2010

### gabbagabbahey

No, why would you think this was true?

5. Feb 21, 2010

### Plutoniummatt

then how do I do this question?

6. Feb 21, 2010

### gabbagabbahey

The same way one usually finds the eigenvalues of an operator...

7. Feb 21, 2010

### Plutoniummatt

8. Feb 21, 2010

### Plutoniummatt

does anyone have the patience to tell me how to do this problem?

9. Feb 21, 2010

### gabbagabbahey

If I gave you an operator in matrix form and asked you to calculate its eigenvalues, could you do it?

10. Feb 21, 2010

### Plutoniummatt

yes...

11. Feb 21, 2010

### gabbagabbahey

Okay, so if you can put $C$ into matrix form, you can find its eigenvalues....do you see how to put $C$ into matrix form? How about putting $A$ into matrix form (you are given both its eigenvalues and eigenfunctions, so this should be trivial)?

12. Feb 21, 2010

### Plutoniummatt

yes I can do put A in matrix form, for B do I need to use the transformation matrix and transform B into A basis?

13. Feb 21, 2010

### gabbagabbahey

Good, and what do you get when you do that?

You can do it without a transformation matrix since you are given $B$'s eigenfunctions in terms of $A's$ eigenfunctions.

14. Feb 21, 2010

### Plutoniummatt

$\begin{pmatrix} 1 & 1\\1 & -1 \end{pmatrix}$ for A

is there a systematic way of getting B or do I just write down the eigenvectors of B in terms of the eigenvectors of A and see which numbers i should put in?

15. Feb 21, 2010

### gabbagabbahey

That doesn't look right, how did you end up with this?

16. Feb 21, 2010

### Plutoniummatt

$\begin{pmatrix} 1 & 0\\0 & -1 \end{pmatrix}$ for A

sorry i messed up the typing

17. Feb 21, 2010

### gabbagabbahey

That's better, so I see you are representing the eigenfunctions of $A$ as

$$u_{+}\to\begin{pmatrix}1 \\ 0 \end{pmatrix}$$ and $$u_{-}\to\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

correct?

What does this make $v_{\pm}$ in this representation?

18. Feb 21, 2010

### Plutoniummatt

yes

my $v_{\pm}$ would be:

$\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ \pm 1 \end{pmatrix}$

so B would be $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

but I had to just look at it and see what numbers i should assign for B, is there a better way of doing it?

oh and i got the correct eigenvalues! thanks so much...i was really confused

Last edited: Feb 21, 2010
19. Feb 21, 2010

### gabbagabbahey

Good.

No, this isn't quite correct.

Any operator $F$ can be decomposed in terms of its eigenvalues, $\lambda_{i}$ and corresponding eigenfunctions $f_i$ (provided they are orthogonal) according to the equation

[tex]F=\sum_{i}\lambda_i f_i^\dagger f_i[/itex]

(Where $f_i^\dagger$ is the adjoint of $f_i$)

Does this look familiar? If so, you can use it to construct $A$ and $B$ from their eigenvalues/eigenvectors (this is what I had thought you had done to find $A$, but apparently you used some other method)

20. Feb 21, 2010

### Plutoniummatt

for B i used the transformation matrix:

$\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\1 & -1 \end{pmatrix}$

which means B =

$\frac{1}{2}\begin{pmatrix} 1 & 1 \\1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\1 & -1 \end{pmatrix}$

which is $\begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}$