Quantum Physics: observables, eigenstates and probability

1. Oct 14, 2012

sgsurrey

1. The problem statement, all variables and given/known data
Observable $\widehat{A}$ has eigenvalues $\pm$1 with corresponding eigenfunctions $u_{+}$ and $u_{-}$. Observable $\widehat{B}$ has eigenvalues $\pm$1 with corresponding eigenfunctions $v_{+}$ and $v_{-}$.

The eigenfunctions are related by:

$v_{+} = (u_{+} + u_{-})/\sqrt{2}$
$v_{-} = (u_{+} - u_{-})/\sqrt{2}$

Show that $\widehat{C}$ =$\widehat{A}$ + $\widehat{B}$ is an observable and find the possible results of a measurement of $\widehat{C}$.
Find the probability of obtaining each result when a measurement of $\widehat{C}$ is performed on an atom in the state $u_{+}$ and the corresponding state of the atom immediately after the measurement in terms of $u_{+}$ and $u_{-}$ .

2. Relevant equations
$\widehat{A}u_{\pm}=\pm u_{\pm}$
$\widehat{B}v_{\pm}=\pm v_{\pm}$
$v_{+} = (u_{+} + u_{-})/\sqrt{2}$
$v_{-} = (u_{+} - u_{-})/\sqrt{2}$
$u_{+} = (v_{+} + v_{-})/\sqrt{2}$
$u_{-} = (v_{+} - v_{-})/\sqrt{2}$

3. The attempt at a solution
Showing that C is an observable I have assumed is as simple as the fact that it is a linear combination of the A and B operators, which are observables matching the requirement to have real eigenvalues.

I'm slightly puzzled by the measurement part of the question; I have tried simply:
$\widehat{C}v_{+} = \widehat{A}v_{+} + \widehat{B}v_{+} = \widehat{A}(u_{+} + u_{-})/\sqrt{2} + v_{+} = v_{-} + v_{+} = \sqrt{2}u_{+}$

I'm not sure I fully understand this however; I expected a result of the format:
$\widehat{C}v_{+} = c_{+}v_{+}$
...where c is a 'result of the measurement', the eigenvalue of C. I'm not sure if it is correct to say that $\sqrt{2}$ is the eigenvalue if the resulting eigenfunction is not the same as the starting eigenfunction.

The answer I have been given for this part of the question is $\pm\sqrt{2}$, but I feel I have now simply found the simplest way to get this in an answer, without understanding why (had I not been lost from the outset I would have avoided looking at the answer in the first place).

For the later part of the question my attempted answer is so far from the solution that I am clueless on how to approach it at this point. I am hoping that with some insight into this first part of the question I might understand how to approach the rest of the question. I would very much appreciate any guidance.

2. Oct 14, 2012

TSny

Hi sgsurrey.

Since the question wants you to express the outcome of a measurement of $\widehat{C}$ in terms of $u_{+}$ and $u_{-}$ it might be a good idea to work in the basis $u_{+}$, $u_{-}$.
Can you find the matrix representations of $\widehat{A}$, $\widehat{B}$ , and $\widehat{C}$ with respect to the $u_{+}$, $u_{-}$ basis?

Once you have the matrix representation of $\widehat{C}$, you can use it to find the possible results of a measurement of $\widehat{C}$ and the probability of each of those results if the initial state is $u_{+}$.

Last edited: Oct 14, 2012
3. Oct 14, 2012

sgsurrey

I'm a few weeks into this QM module, I've yet to cover the matrix representations and thus my understanding is lacking in this approach. Since you've suggested this I have realised that this method is outlined a few pages after what I had previously studied in my textbook so far; I will read this now. I'll hopefully then proceed with an attempt at the solution.

4. Oct 14, 2012

TSny

It's possible to get the answer to this question without matrix representations of the operators. The matrix approach is just kind of a nice way to go about it in my opinion.

5. Oct 14, 2012

sgsurrey

The matrix method sounds useful, so I'm reading up on it.

As far as I can see so far I can represent each eigenfunction as a vector in terms of the 'u' basis:

$u_{+} = \left(\begin{array}{c} 1 \\ 0 \end{array}\right)$
$u_{-} = \left(\begin{array}{c} 0 \\ 1 \end{array}\right)$
$v_{+} = \frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ 1 \end{array}\right)$
$v_{-} = \frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ -1 \end{array}\right)$

Then it seems I can express operator A as:

$\widehat{A} = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$

Now stuck on operator B...

Are my above assumptions correct?

6. Oct 14, 2012

sgsurrey

It would appear that B is:

$\widehat{B} = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$

and thus C:

$\widehat{C} = \left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right)$

There is probably a much easier way to obtain those matrices than the way I just used (trial and error)...

7. Oct 14, 2012

TSny

Yes, that looks correct. Formally, you can get the matrix elements for C by noting that the C11 element is <u+|C|u+> = <u+|A|u+> + <u+|B|u+>. To evaluate <u+|B|u+>, express u+ in terms of the v vectors. Likewise, the C12 matrix element is <u+|C|u->, etc.

8. Oct 14, 2012

TSny

Sorry, it's easier to first find the matrix for B and then just add A and B to find C as you did! You can find the matrix elements for B using the method I described for C.

9. Oct 14, 2012

sgsurrey

I'm now struggling to understand what this means.

If I operate C on u+, or rather matrix multiply C with the vector, I get the vector:

$\widehat{C}u_{+} = \left(\begin{array}{c} 1 \\ 1 \end{array} \right)$

Which clearly has a magnitude $\pm\sqrt{2}$ (which I have in the solutions for this problem)

I'm unsure of how I find the state that the 'atom' is now in... any hints? I feel I am out of my depth with this problem, perhaps I should return to it at a later stage.

10. Oct 14, 2012

TSny

Operating on u+ with C does not correspond to measuring C when the systems is in the state u+, so that's not going to be of much help.

Can you fill in the blank here:

A fundamental postulate of QM is that the result of a measurement of an observable C is always one of the ___________ of C.

11. Oct 14, 2012

sgsurrey

Ah.. got it, sorry, matrix notation getting me confused.

A measurement of an observable will be one of the eigenvalues:

$(A-λI)\psi = 0$

$\left|\begin{array}{cc} 1-λ & 1 \\ 1 & -1-λ \end{array}\right| = 0$

λ = ±√2

12. Oct 14, 2012

TSny

Right. And what will the state of the system be after the measurement?

13. Oct 15, 2012

sgsurrey

Then get:

$ω_{+}=\frac{1}{\sqrt{4-2\sqrt{2}}}\left(\begin{array}{c} 1-\sqrt{2} \\ 1 \end{array}\right)$
$ω_{-}=\frac{1}{\sqrt{4+2\sqrt{2}}}\left(\begin{array}{c} 1+\sqrt{2} \\ 1 \end{array}\right)$

and probabilities:

$\left(\left<ω_{+}|u_{+}\right>\right)^2 = 0.146$
$\left(\left<ω_{-}|u_{+}\right>\right)^2 = 0.854$

In the solution I have:
$ω_{+}=\frac{1}{\sqrt{4-2\sqrt{2}}}\left(\begin{array}{c} 1 \\ -(1-\sqrt{2}) \end{array}\right)$
$ω_{-}=\frac{1}{\sqrt{4+2\sqrt{2}}}\left(\begin{array}{c} 1 \\ -(1+\sqrt{2}) \end{array}\right)$
and the probabilities for ω+ and ω- are opposite...

Do you know if this is equivalent; or have I made a mistake?

Last edited: Oct 15, 2012
14. Oct 15, 2012

sgsurrey

I'm now aware that I have made a mistake and will hopefully figure this out later.

15. Oct 15, 2012

TSny

Looks good! That's what I got anyway.