# 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.