Finding <A> given the eigenvalues

[LiorSh]

[Note from mentor: This thread was originally posted in a non-homework forum, so it lacks the homework template. Even though the solution was resolved there, the thread has been moved here for future reference.]So I'm given Φ = N(φ₁+2*φ₂ + 3*φ₃)
and the operator A with eigen values λ₁ = 1, λ₂ = 3, λ₃=5.
I had to find N such that Φ is normalized which the result is 1/sqrt(14).
The I had to find a single measurement of finding A in this state which is simple the eigenvalues 1or 3 or 5.
Now have to find <A> which I'm not sure how to get. I keep on getting 3, but my professor got 20/7. What are the steps I need to take? Thank you!

 

[jambaugh]

How are you calculating your expectation value <A>?

You should calculate
$$\langle A \rangle = \Phi^\dagger A \Phi = N^2 (\varphi^\dagger_1 + 2\varphi^\dagger_2 + 3\varphi^\dagger_3)A(\varphi_1 + 2\varphi_2 + 3\varphi_3)$$
Expand and apply orthonormality of the eigen-basis to get the appropriate result.

BTW, be sure in future you submit homework in the appropriate forum section of the forum following the appropriate guidelines and template.

 

[LiorSh]

Hi, thanks for the response and sorry for posting in in the wrong place. When I act with the operator on the functions, I get the eigenvalue correct? What happens to the constant of that function?

 

[jambaugh]

Hi, thanks for the response and sorry for posting in in the wrong place. When I act with the operator on the functions, I get the eigenvalue correct? What happens to the constant of that function?

Correct, and since the operator is linear. $A c\varphi_k =c A\varphi_k = c \lambda_k \varphi_k$

 

[LiorSh]

So for that A(φ1+2φ2+3φ3) I'm getting (1*1)φ1+(3*2)φ2+(3*5)φ3= φ1+(6)φ2+(15)φ3
does that make sense?

 

[jambaugh]

So for that A(φ1+2φ2+3φ3) I'm getting (1*1)φ1+(3*2)φ2+(3*5)φ3= φ1+(6)φ2+(15)φ3
does that make sense?

Yes, good. Now apply $\Phi^\dagger$ to this. ($\varphi^\dagger_j \varphi_k = 1$ if $j=k$ else zero.) (And don’t forget the normalization factor occurs twice.)

 

[LiorSh]

That's what I did... I'm getting (1/14) * (1+12 + 45) = 29/7. The answer though is 20/7. What did I forget?

 

[jambaugh]

I'm getting 29/7 as well. At this stage I'd suggest carefully checking the original problem statement.

Looking for obvious transposition errors. Let's see... is there a permutation of {1,3,5} which gives: $\lambda_1 + 4\lambda_2 + 9\lambda_3 = 40$? ... Nope, I just went through the 6 permutations plus combined sign changes on the eigen-values and none yield 40/14 = 20/7. It is possible that the solution has a typo? The 0 and 9 key are right next to each other on most keyboards!? Recheck the original problem.

But your solution is sound. You can also get a qualitative estimation by considering the case where each eigen-state is equally likely. The expected value of A would then be the straight average of the eigen-values which is 3 = 21/7. The answer 20/7 implies the lower eigen-value states are more probable but you'll note that, as you've stated the problem it is the highest eigen-value that is more likely with probability 9/14, vs 1/14 and 4/14 probability for the lower and middle values. You should thus get an answer significantly higher than 3 = 21/7, something like... oh... say... 29/7!

From where did you get this problem?

 

[LiorSh]

Thanks for your help! it's from Berman, see original problem below:

Also, for part d he got 13/49, and I get something completely different... What am I missing?

 

[jambaugh]

Hmmm... variance. $\sigma^2_A = \langle (A-\langle A\rangle)^2\rangle = \langle A^2\rangle - \langle A\rangle^2$
$$\langle A^2 \rangle = \frac{1}{14}(1\lambda_1^2 + 4\lambda_2^2 + 9\lambda_3^2)= \frac{1+36+225}{14}=\frac{262}{14}=\frac{131}{7}$$
Given our value 29/7 for $\langle A \rangle$, I get:
$$\sigma_A^2 = \frac{917}{49} - \frac{841}{49} = 76/49$$.

Putting that in context, the root variance = std is $\sqrt{76}/7 \approx 1.237$. That might be a wee bit high but fits within the range of 1 to 5.

If you want to work this out a different way, you have three values with three probabilities:
Value a= 1 with probability p = 1/14, value a = 3 with probability p = 4/14 and value a = 5 with probability p = 9/14.
Once that's established and since we're all within a single "frame" (all referenced states are in eigen-states) we can treat it as a classical probability calculation.

Find the mean and variance of the random variable with these values and probabilities. Easy peasy lemon squeezy: mean = 29/7, var = 76/49.

Are you finding the different values from the text? or from your professor?

 

[LiorSh]

Wow, thanks for the explanation. The answers are from my professor which he might be a little off. I wanted to make sure that my answers are correct before I tell him that he is wrong. He asked me to go over his solution and confirm if he is correct or not. The funny thing is that we have been solving really complex problems, which I don't have issues with - but when it comes to this little simple problems I get lost and question myself. I'm going to do the work again and hopefully, I can show him my way of solving it tomorrow. Thank you very much again!

 

[jambaugh]

... I wanted to make sure that my answers are correct before I tell him that he is wrong...

Always a good policy, and even then be humble and polite about it.

As for simple problems... sometimes the triviality of it causes us to pay too little attention to the simple details. I know that's my experience. Good luck in the course!

 

[LiorSh]

Of course - thank you very much for your help :)