Sequential measurements in quantum mechanics

[hmparticle9]

Homework Statement

An operator $\hat{A}$, representing observable $A$, has two normalised eigenstates $\psi_1$ and $\psi_2$, with eigenvalues $a_1$ and $a_2$ respectively. Operator $\hat{B}$, representing observable $B$, has two normalised eigenstates $\phi_1$ and $\phi_2$, with eigenvalues $b_1$ and $b_2$. The eigenstates are related by

$$\psi_1 = (3\phi_1 + 4\phi_2)/5 \text{ and } \psi_2 = (4\phi_1 - 3\phi_2)/5$$

(a) Observable $A$ is measured, and the value $a_1$ is obtained. What is the state of the system (immediately) after this measurement?

(b) If $B$ is now measured, what are the possible results, and what are their probabilities?

(c) Right after the measurement of $B$, $A$ is measured again. What is the probability of getting $a_1$? (Note that the answer would be quite different if I had told you the outcome of the B measurement.)

Relevant Equations

$$\psi_1 = (3\phi_1 + 4\phi_2)/5 \text{ and } \psi_2 = (4\phi_1 - 3\phi_2)/5$$
$$\phi_1 = (3\psi_1 + 4\psi_2)/5 \text{ and } \phi_2 = (4\psi_1 - 3\psi_2)/5$$

From results in my book (which I think are fairly standard across quantum mechanics) the answer to a) is $\psi_1$.

I will ask about c) later. It might come to me when I understand b). I can state with confidence that if $B$ is measured then we are either going to get $b_1$ or $b_2$. What I do not know is how to obtain the probabilities. According to:

https://www.youklab.org/teaching/mites_2010/mites2010_Solution4.pdf

it comes immediately from $\psi_1 = (3\phi_1 + 4\phi_2)/5$. I agree that $A$ is in state $\psi_1$, but I don't see how/why the probs are obtained from this equation.

 

[PeroK]

That's the Born rule.

 

[hmparticle9]

Looking for this in my book there is this result called Borns statistical interpretation:
$$\int_a^b |\Phi(x,t)|^2 dx = \{ \text{prob. of finding particle between a and b at time t} \}$$
Is that it? I would like to rely on results that I have access to within my book. (Intro. to Quantum Mechanics by Griffiths)

 

[PeroK]

I'm not at home so I can't look it up in Griffiths, he definitely covers it. The coefficients are related to the probabilities by $p_n = |c_n|^2$ etc.

 

[hmparticle9]

I think I have seen that. But one of my problems is with the fact that $\psi_i$ and $\phi_i$ are eigenfunctions of DIFFERENT operators. When you get home maybe you could take a look for me please?

 

[PeroK]

I think I have seen that. But one of my problems is with the fact that $\psi_i$ and $\phi_i$ are eigenfunctions of DIFFERENT operators.

That's the whole point. You have a state $\psi_n$, which is a linear combination of the states $\phi_i$. The coefficients in that linear expansion are related to measurement probabilities.

I won't be home until the weekend, but you need to find it for yourself in any case.

 

[vela]

I would like to rely on results that I have access to within my book. (Intro. to Quantum Mechanics by Griffiths)

Read section 3.4.

 

[hmparticle9]

I have read this.

"If the spectrum of $\hat{Q}$ is discrete, the probability of getting the particular eigenvalue $q_n$ associated with the eigenfunction $f_n$ is
$$|c_n|^2, \text{ where } c_n = \langle f_n, \Psi \rangle.$$"

But I am having trouble applying it in this case

 

[hmparticle9]

For instance, (3.45) says
$$\Phi(x,t) = \sum_{n} c_n(t) f_n(x)$$

But we are dealing with eigenfunctions of different operators. I know PeroK told me that is the whole point. But I am just struggling to put it all together.

 

[vela]

You need to consider the properties of the eigenfunctions of a hermitian operator (section 3.3).

$\Psi$ is the state vector of the system. You can express it in terms of the eigenfunctions of an observable. In this case, if you use the eigenfunctions of $\hat B$, you can write
$$\Psi = c_1 \phi_1 + c_2 \phi_2,$$ where $c_1$ and $c_2$ are complex constants. If you're interested in finding the probability of measuring $b_1$, you set $f_n = \phi_1$ and find the probability amplitude by calculating inner product $\langle \phi_1 , \Psi \rangle$.
\begin{align*}
\langle \phi_1 , \Psi \rangle &= \langle \phi_1 , c_1 \phi_1 + c_2 \phi_2 \rangle = c_1 \langle \phi_1 ,\phi_1\rangle + c_2 \langle \phi_1 ,\phi_2\rangle \\
&= c_1\cdot 1 + c_2 \cdot 0 \\
&= c_1
\end{align*} Note that the inner product $\langle \phi_1 , \Psi \rangle$ is easy to calculate because we expressed $\Psi$ in terms of $\phi_1$ and $\phi_2$. In other words, if you want probabilities associated with a measurement of $B$, express the state of the system in terms of the eigenfunctions of $\hat B$. Then you get the probability amplitudes by simply picking off the coefficients of $\phi_i$.

Could you have expressed $\Psi$ in terms of the eigenfunctions of $\hat A$ instead for this calculation? Sure. (You should try it.) You should get the same answer in the end, but you can't read off the answer by inspection like you can when you use the eigenfunctions of $\hat B$.

 

[kuruman]

I would expand @vela's explanation and write the total probability that the system be in some state is 1, $$\begin{align} 1 & =\langle \Psi | \Psi\rangle \nonumber \\
& = \langle(c_1\psi_1 +c_2\psi_2 )|(c_1\psi_1 +c_2\psi_2 ) \rangle \nonumber \\
& =|c_1|^2\langle | \psi_1| \psi_1 \rangle +|c_2|^2\langle \psi_2 | \psi_2\rangle\nonumber \\
& =|c_1|^2*1 +|c_2|^2*1 \nonumber \end{align}$$ This says that that the coefficient $|c_i|^2$ is the probability or weighting factor that the system be in state $|\psi_i \rangle.$

 

[hmparticle9]

"You need to consider the properties of the eigenfunctions of a hermitian operator (section 3.3)."

You are referring to the axiom that states that we can express any function as a linear combination of eigenfunctions. Got it.

I totally understand that we can write a function $\Psi$ as:
$$\Psi = c_1\phi_1 + c_2 \phi_2.$$
Got it.

We have set some function $\Psi = c_1\phi_1 + c_2 \phi_2.$ and we have taken the inner-product with $\phi_1$ to obtain $c_1$.

All good.

Am I correct in saying that can we set $\Psi = \psi_1$ because we measured $A$ and the state of the system is $\psi_1$. Hence $c_1 = \frac{3}{5}$?

 

[kuruman]

Am I correct in saying that can we set $\Psi = \psi_1$ because we measured $A$ and the state of the system is $\psi_1$. Hence $c_1 = \frac{3}{5}$?

You are incorrect in saying that. You know that $\Psi=c_1\psi_1+c_2\psi_2$. $\Psi=\psi_1$ would be true only if $c_2=0$ which is not the case.

What you can say is what you have already said:
After the first measurement the system is in the state $\psi_1=\frac{1}{5}\left(3\phi_1+4\phi_2\right).$

You don't know whether it is in state $\phi_1$ or in state $\phi_2$ but you know the probabilities that it will be found in each state if you make a second measurement. What are these probabilities? (Hint: See post #11.)

 

[hmparticle9]

How can we conclude from
$$\Psi = c_1\phi_1 + c_2 \phi_2.$$
that $c_1 = \frac{3}{5}$?

 

[kuruman]

How can we conclude from
$$\Psi = c_1\phi_1 + c_2 \phi_2.$$
that $c_1 = \frac{3}{5}$?

We can't. First we need to measure observable $A$. After we do this, we can forget $\Psi$ because we know that what we have is described by $\psi_1$.

 

[hmparticle9]

Okay. First, thank you (all of you) for persevering with me. QM is melting my brain a bit.

So we start with $\Psi = c_1\phi_1 + c_2 \phi_2.$ Then observable A is measured, and the value $a_1$ is obtained. So the state of the system immediately after this measurement is $\psi_1$.

We can then say $\psi_1 = c_1\phi_1 + c_2 \phi_2.$ to obtain the probabilities to answer part (b).

To answer c) we just invert the equations. I can follow the solution in the link I provided.

 

[hmparticle9]

@vela

"Could you have expressed Ψ in terms of the eigenfunctions of $\hat{A}$ instead for this calculation? Sure."

We start with

$$\Psi = d_1\psi_1 + d_2\psi_2$$

but we have the equations that relate $\psi_i$ to $\phi_i$. We use those equations and take the inner product with $\phi_i$

 

[kuruman]

To answer c) we just invert the equations. I can follow the solution in the link I provided.

If you want to be comfortable with the material, I suggest that you do not follow the solution in the link. Just ask yourself what state the system is in after the second measurement and proceed the same way as you did after the first measurement. This is reinforced learning because you follow your own solution not someone else's.

You may also want to consider Griffiths's parenthetical statement
(Note that the answer would be quite different if I had told you the outcome of the B measurement.)
How would the answer be different and why?

On edit
After you post your own solution for part (c), I will post my own from when I assigned this problem to my class. it's more compact than the one in the link.

 

[vela]

"You need to consider the properties of the eigenfunctions of a hermitian operator (section 3.3)."

You are referring to the axiom that states that we can express any function as a linear combination of eigenfunctions. Got it.

And that the eigenfunctions are orthogonal.

Am I correct in saying that can we set $\Psi = \psi_1$ because we measured $A$ and the state of the system is $\psi_1$. Hence $c_1 = \frac{3}{5}$?

I was just using $\Psi$ because that's what Griffiths used in the book, but I did mean it was the state of the system after part (a) so that, indeed, $\Psi = \psi_1$. I think @kuruman instead took it to mean the state of the system initially in the problem.

 

[vela]

but we have the equations that relate $\psi_i$ to $\phi_i$. We use those equations and take the inner product with $\phi_i$

Exactly.

 

[kuruman]

I was just using $\Psi$ because that's what Griffiths used in the book, but I did mean it was the state of the system after part (a) so that, indeed, $\Psi = \psi_1$. I think @kuruman instead took it to mean the state of the system initially in the problem.

Yes, that's what I took it to mean, the starting point before any measurement was made. There is no mention of $\Psi$ in the textbook statement of the problem to clarify.

 

[hmparticle9]

"You may also want to consider Griffiths's parenthetical statement
(Note that the answer would be quite different if I had told you the outcome of the B measurement.)
How would the answer be different and why?"

@kuruman

If we are told that the result of the measurement was $b_1$ then we know what the state of the system is and we can use
$$\phi_1 = (3\psi_1 + 4 \psi_2) /5$$

to conclude that the probability of getting $a_1$ is 9/25

 

[hmparticle9]

"Just ask yourself what state the system is in after the second measurement and proceed the same way as you did after the first measurement."

I am finding hard to "unsee" the solution in the link. Could you give me a push? By second measurement, do you mean the measurement of B or the second measurement of A? I am not sure how to answer the question without inverting the equations.

We have established that when B is measured we get result $b_1$ with probability 9/25 and we get result $b_2$ with probability 16/25.

I am not sure how to proceed without inverting the equations.

 

[kuruman]

We have established that when B is measured we get result b1 with probability 9/25 and we get result b2 with probability 16/25.

I am not sure how to proceed without inverting the equations.

Inverting the equations is necessary for proceeding. So go ahead and do it, but you need to explain in your own words why this is part of the strategy for getting the solution.

 

[hmparticle9]

We know that $\{\phi_i\}$ and $\{\psi_i\}$ are complete. We can write the state of the system as a linear combination of them.

We used this fact in part b). The state of the system immediately after measurement was $\psi_1$, and since we had $\psi_1$ in terms of $\{\phi_i\}$ we could "pick out" the probabilities by observation.

But now we are measuring observable $B$, which can either be in state $\phi_1$ or $\phi_2$. In order to pick out the probabilities like we did before it would be nice to write these states in a linear combination of functions that form a complete set.

We can do this by taking the matrix that maps $\phi$ to $\psi$ and invert it to get it.

 

[kuruman]

We know that $\{\phi_i\}$ and $\{\psi_i\}$ are complete. We can write the state of the system as a linear combination of them.

We used this fact in part b). The state of the system immediately after measurement was $\psi_1$, and since we had $\psi_1$ in terms of $\{\phi_i\}$ we could "pick out" the probabilities by observation.

But now we are measuring observable $B$, which can either be in state $\phi_1$ or $\phi_2$. In order to pick out the probabilities like we did before it would be nice to write these states in a linear combination of functions that form a complete set.

We can do this by taking the matrix that maps $\phi$ to $\psi$ and invert it to get it.

What matrix? It would be simpler to say, "I am given that $$\psi_1 = (3\phi_1 + 4\phi_2)/5 \text{ and } \psi_2 = (4\phi_1 - 3\phi_2)/5.$$ I consider this to be a system of two equations and two unknowns, $\phi_1$ and $\phi_2$. Solving this system gives me the equations I already posted as relevant in #1, $$\phi_1 = (3\psi_1 + 4\psi_2)/5 \text{ and } \phi_2 = (4\psi_1 - 3\psi_2)/5.$$

Proceed and wrap it up.

 

[hmparticle9]

I think of the system as a matrix-vector equation. Suppose that the state obtained when measuring $B$ was $\phi_1$ (with probability 9/25) then the probability of obtaining $a_1$ on measuring $A$ again is given by $c^2$ where $c = \langle \phi_1 | \psi_1 \rangle$. Hence the total probability for this particular way is (9/25) * (9/25). The only other way of obtaining $a_1$ on this second measurement of A is to get a measurement of $b_2$ on measuring $B$ (the state is $\phi_2$) and then obtain $a_1$ from the second measurement of $A$. The probability of the state being $\phi_2$ immediately after the measurement of $B$ is given by $\langle \phi_2 | \psi_1 \rangle = 16/25$. Before measuring $A$ again we are in state $\phi_2$. Hence we need to evaluate $\langle \psi_1 | \phi_2 \rangle = 16/25$

This gives us (16/25)*(16/25). In total the probability is 0.5932

"On edit
After you post your own solution for part (c), I will post my own from when I assigned this problem to my class. it's more compact than the one in the link."

Could you share this with me?

 

[kuruman]

Could you share this with me?

It's not much different from yours.