Quantum Physics - Measurement/Eigenvalues(functions)

[Plutoniummatt]

Homework Statement

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

[tex]v_+ = \frac{u_+ + u_1}{\sqrt{2}}[/tex]

[tex]v_- = \frac{u_+ - u_1}{\sqrt{2}}[/tex]


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

Homework Equations

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 [tex]\pm\sqrt{2}[/tex]...I'm aware that the [tex]1/\sqrt{2}[/tex] in the eigenfunctions of B will make my answer "correct" but then they're not the eigenvalues anymore?

 

[gabbagabbahey]

Measured values are just the eigenvalues, in this case, the eigenvalues of C are just 2 and -2?

Why do you say this?

 

[Plutoniummatt]

Why do you say this?

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

 

[gabbagabbahey]

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

No, why would you think this was true?

 

[Plutoniummatt]

No, why would you think this was true?

then how do I do this question?

 

[gabbagabbahey]

then how do I do this question?

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

 

[Plutoniummatt]

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

helpful. Much thanks!...

 

[Plutoniummatt]

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

 

[gabbagabbahey]

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

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

 

[Plutoniummatt]

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

yes...

 

[gabbagabbahey]

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

 

[Plutoniummatt]

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

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?

 

[gabbagabbahey]

yes I can do put A in matrix form,

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

for B do I need to use the transformation matrix and transform B into A basis?

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

 

[Plutoniummatt]

[itex]\begin{pmatrix} 1 & 1\\1 & -1 \end{pmatrix}[/itex] 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?

 

[gabbagabbahey]

[itex]\begin{pmatrix} 1 & 1\\1 & -1 \end{pmatrix}[/itex] for A

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

 

[Plutoniummatt]

[itex]\begin{pmatrix} 1 & 0\\0 & -1 \end{pmatrix}[/itex] for A

sorry i messed up the typing

 

[gabbagabbahey]

That's better, so I see you are representing the eigenfunctions of [itex]A[/itex] as

[tex]u_{+}\to\begin{pmatrix}1 \\ 0 \end{pmatrix}[/tex] and [tex]u_{-}\to\begin{pmatrix} 0 \\ 1 \end{pmatrix}[/tex]

correct?

What does this make [itex]v_{\pm}[/itex] in this representation?

 

[Plutoniummatt]

That's better, so I see you are representing the eigenfunctions of [itex]A[/itex] as

[tex]u_{+}\to\begin{pmatrix}1 \\ 0 \end{pmatrix}[/tex] and [tex]u_{-}\to\begin{pmatrix} 0 \\ 1 \end{pmatrix}[/tex]

correct?

What does this make [itex]v_{\pm}[/itex] in this representation?

yes

my [itex]v_{\pm}[/itex] would be:

[itex]\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ \pm 1 \end{pmatrix}[/itex]



so B would be [itex]\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}[/itex]

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

 

[gabbagabbahey]

my [itex]v_{\pm}[/itex] would be:

[itex]\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ \pm 1 \end{pmatrix}[/itex]

Good.

so B would be [itex]\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}[/itex]

No, this isn't quite correct.

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?

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

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

(Where [itex]f_i^\dagger[/itex] is the adjoint of [itex]f_i[/itex])

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

 

[Plutoniummatt]

Good.



No, this isn't quite correct.



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

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

(Where [itex]f_i^\dagger[/itex] is the adjoint of [itex]f_i[/itex])

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

for B i used the transformation matrix:

[itex]\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\1 & -1 \end{pmatrix}[/itex]

which means B =

[itex]\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}[/itex]

which is [itex]\begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}[/itex]