Calculate expectation value of entangled 2 state system?

Click For Summary
The discussion revolves around calculating the expectation value of an entangled two-state system using the eigenstates of operator C, specifically |B> and |R>. The user seeks guidance after struggling with the problem for hours, eventually applying a projection operator to derive the probability of measuring the states. The calculated expectation value of the operator C is determined to be 0, indicating that the average measurement result over many trials will yield this value. The distinction between expectation values and projection operators is clarified, emphasizing that expectation values represent average outcomes while projection operators relate to state probabilities. Overall, the conversation highlights the importance of understanding these concepts in quantum mechanics.
ianmgull
Messages
19
Reaction score
0

Homework Statement



Image_1.jpg

Homework Equations


I know that there are two eigenstates of the operator C:

|B> = (1 0) as a column vector with eigenvalue 1
|R> = (0 1) also a column vector with eigenvalue -1

The Attempt at a Solution


My work is shown here:

IMG_3540.jpg


If anyone could point me in the right direction, I'd greatly appreciate it. I've been stuck for hours and just can't figure out what I'm doing wrong.

thanks

Ian
 

Attachments

  • IMG_3540.jpg
    IMG_3540.jpg
    16.1 KB · Views: 658
Physics news on Phys.org
Use a projection operator. A projector on the red basis would look like P_R = |R\rangle\langle R|. Then (dropping 1 and 2 subscripts, and assuming orthogonal states):

<br /> Prob_R =\frac{1}{2} (\langle B|\langle R| - \langle R|\langle B|)|R\rangle\langle R|(|B\rangle|R\rangle -|R\rangle|B\rangle )=\frac{1}{2}<br />

The expectation value of C is the probability that the measurement produces either 1 or -1, so the average result will be 0.
 
I just worked it out and that makes much more sense.

I'm still a little unclear on (conceptually) what meaning I should attribute to taking the expectation value of an operator (like above) vs a projection operator.

Thanks so much!
 
The eigenvalues are what can be measured in a lab; an expectation value will give you the average result that will be obtained over a large number of measurements, in this case 0. A projection operator projects the state onto a basis (red or blue) according to a probability, 1/2 here.
 
Awesome.

THANK YOU
 

Similar threads

Replies
38
Views
3K
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K