New state function after measurement

[IHateMayonnaise]

Homework Statement

This is an extension to problem 4.55 in Griffiths. The problem is:

The electron in a hydrogen atom occupies the combined spin and position state:
$$\lvert \psi \rangle = R_{21} \left(\sqrt{1/3} Y_1^0 \chi_+ + \sqrt{2/3} Y_1^1 \chi_-\right)$$

a) If $S_z$ is measured, what values might you obtain, and with what probabilities?
b) If you happen to measure $\hbar/2$, what would the new state be?
c) Do the same for $S_x$.

Homework Equations

$$\vec{S} = \hbar/2 \vec{\sigma}, \qquad \sigma_x = \left(\begin{matrix} 0 & 1\\ 1 & 0\end{matrix}\right), \quad \sigma_y = \left(\begin{matrix}0 & -i \\ i & 0 \end{matrix}\right) , \quad \sigma_z = \left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right)$$

$$\chi = \alpha \chi_+ + \beta \chi_- = \left(\begin{matrix}\alpha \\ \beta \end{matrix}\right)$$

$$\chi_+^{(z)} = \left(\begin{matrix} 1\\0 \end{matrix}\right), \qquad \chi_-^{(z)} = \left(\begin{matrix} 0\\1 \end{matrix}\right)$$

$$\chi_+^{(x)} = \frac{1}{\sqrt{2}}\left(\begin{matrix} 1 \\ 1 \end{matrix} \right), \qquad \chi_-^{(x)} = \frac{1}{\sqrt{2}} \left(\begin{matrix} 1 \\ -1 \end{matrix} \right)$$

The Attempt at a Solution

a) It seems to me that the the probabilities are

$$P(\uparrow)_{z} = \left(\sqrt{\frac{1}{3}}\right)^2 = \frac{1}{3}$$

$$P(\downarrow)_{z} = \left(\sqrt{\frac{2}{3}}\right)^2 = \frac{2}{3}$$

b) No idea

c) For the spin along the x direction,

$$P(\uparrow)_{x} = \left\lvert\left(\chi\right)^\dagger \chi_+^{(x)}\right\rvert^2 = \left\lvert\left(\alpha \;\; \beta\right) \frac{1}{\sqrt{2}} \left(\begin{matrix} 1\\1\end{matrix}\right)\right\rvert^2 = \left\lvert\frac{\alpha + \beta}{\sqrt{2}}\right\rvert^2 = \left\lvert \frac{\sqrt{1/3} + \sqrt{2/3}}{\sqrt{2}}\right\rvert^2 \approx .97$$

And similarly

$$P(\downarrow)_{x} = \left\lvert\frac{\alpha - \beta}{\sqrt{2}}\right\rvert^2 = \left\lvert \frac{\sqrt{1/3}-\sqrt{2/3}}{\sqrt{2}}\right\rvert^2 \approx .03$$I am fairly certain that a) and c) are correct, however I am not sure about c). I mean.. I know that the state of the particle does change subsequent to a measurement, but how can I know what the new function is? Clearly it will still have the form

$$\lvert \psi \rangle = \alpha \chi_+^{(z)} + \beta \chi_-^{(z)}$$

Or in some other basis, but how does it change exactly? More specifically, what are the new $\alpha$ and $\beta$?

Thanks yall

IHateMayonnaise

 

[graphene]

Hello IHateMayonnaise

(b) In case you did not understand what the question meant - You do not know the linear coefficients as in part (a) and you'll have to calculate them, given the expectation of Sz in this state.

(c) Its correct

 

[IHateMayonnaise]

Hello IHateMayonnaise

(b) In case you did not understand what the question meant - You do not know the linear coefficients as in part (a) and you'll have to calculate them, given the expectation of Sz in this state.

(c) Its correct

Thanks for the reply graphene.

I understand what b) is asking; I just don't know how to find the new coefficients.

 

[graphene]

so, $$\lvert \psi \rangle = R_{21} \left(c_1 Y_1^0 \chi_+ + c_2 Y_1^1 \chi_-\right)$$

After operating $$S_z$$ on $$\lvert \psi \rangle$$, the sum of the squares of the coefficients is $$\hbar /2$$. This is one equation for two variables, can you think of another equation that relates the coefficients.

 

[graphene]

so, $$\lvert \psi \rangle = R_{21} \left(c_1 Y_1^0 \chi_+ + c_2 Y_1^1 \chi_-\right)$$

After operating $$S_z$$ on $$\lvert \psi \rangle$$, the sum of the squares of the coefficients is $$\hbar /2$$. This is one equation for two variables, can you think of another equation that relates the coefficients.

 

[vela]

Thanks for the reply graphene.

I understand what b) is asking; I just don't know how to find the new coefficients.

You make a measurement and get a result. What does performing a measurement do to the wave function? This is one of the postulates of quantum mechanics.

 

[IHateMayonnaise]

You make a measurement and get a result. What does performing a measurement do to the wave function? This is one of the postulates of quantum mechanics.

It changes it; it collapses in according with the outcome of the measurement. In Schrodinger's cat speak, if our initial wave function is

$$\lvert \psi \rangle = \alpha \lvert \psi \rangle_{\text{dead}} + \beta \lvert \psi \rangle_{\text{alive}}$$

Where we know that the cat has to be either dead or alive, so

$$\lvert \alpha \rvert^2 + \lvert \beta \rvert^2 = 1$$

And when you open the box (kill the cat) the wave function "collapses" to whatever it is that you did to the cat. Let's say you killed the cat, then the collapsed wave function is

$$\lvert \psi \rangle = (1) \lvert \psi \rangle_{\text{dead}}$$

But, and this is my question, is the wave function after this? Eg if I were going to make another observation of the cat that we just saw to be dead (actually, you killed it , how does this change the subsequent measurements? What are the new cat eigenfunctions?

 

[vela]

As you said, the new state is $|\psi\rangle_\mathrm{dead}$, so if you were to look into the box again, you'd always find the cat is dead. There's no chance of finding the cat alive because the state has no component in the $|\psi\rangle_\mathrm{alive}$ direction.

 

[IHateMayonnaise]

As you said, the new state is $|\psi\rangle_\mathrm{dead}$, so if you were to look into the box again, you'd always find the cat is dead. There's no chance of finding the cat alive because the state has no component in the $|\psi\rangle_\mathrm{alive}$ direction.

Perhaps my perception of QM is askew but my understanding was that this was not the case. If the cat were a particle and I happened to measure spin up, to me what you're saying is that it will always be spin up with certainty for subsequent measurements. This cannot be; if for no other reason Heisenberg's principle would be violated.

 

[vela]

Yes, you will always find the spin to be up. (For simplicity, I'm assuming that spin-up is an energy eigenstate, so as the system evolves in time, it remains in that state.)

The uncertainty principle tells you about how pairs of non-commuting observables are related. You can make the uncertainty in one variable as small as you want, but that will mean the uncertainty in the other variable must grow larger.

If you think about it, you have the same situation with energy and time. You have the uncertainty principle for that pair that says $\Delta E \Delta t \ge \hbar/2$, yet according to the Schrodinger equation, the system will remain in an energy eigenstate as it evolves in time.

 

[IHateMayonnaise]

Yes, you will always find the spin to be up. (For simplicity, I'm assuming that spin-up is an energy eigenstate, so as the system evolves in time, it remains in that state.)

The uncertainty principle tells you about how pairs of non-commuting observables are related. You can make the uncertainty in one variable as small as you want, but that will mean the uncertainty in the other variable must grow larger.

If you think about it, you have the same situation with energy and time. You have the uncertainty principle for that pair that says $\Delta E \Delta t \ge \hbar/2$, yet according to the Schrodinger equation, the system will remain in an energy eigenstate as it evolves in time.

Ah; silly silly me. Sort of like successive Stern-Gerlach apparatus. If the same observable is measured, it should return the same result. In this case the new state would be


$$\lvert \psi \rangle = R_{21} Y_1^0 \chi_+$$

Is this correct? Anyway, later on in the problem (part c) it asks to do the same thing as in

However, the problem asks to do the same thing for Sx. For this, being after the original measurement of Sz with a result of $\hbar/2$, would my part c) still be correct?

Thanks for all your help!

 

[vela]

Ah; silly silly me. Sort of like successive Stern-Gerlach apparatus. If the same observable is measured, it should return the same result. In this case the new state would be$$\lvert \psi \rangle = R_{21} Y_1^0 \chi_+$$

Is this correct?

Yes.

Anyway, later on in the problem (part c) it asks to do the same thing as in

However, the problem asks to do the same thing for Sx. For this, being after the original measurement of Sz with a result of $\hbar/2$, would my part c) still be correct?

Thanks for all your help!

I think the question is asking you to do the same thing using S_x starting from the same initial state, not the state after the measurement of S_z.

If my interpretation is the correct one, the work you have up there is fine. If you want to do it the other way, just do the same thing using the state after wave function collapse in your calculations.