Spin-1 Particles Results of a Measurement

[chris_avfc]

Homework Statement

Spin-1 Particles prepared in the state:

$$ |\psi> = \frac{2}{\sqrt29} |1> + \frac{i 3}{\sqrt29}|0> - \frac{4}{\sqrt29} |-1> $$

Where I'm guessing the $|#>$ represents the spin state of -1,0 or 1.

I'm looking to find the results of a measurements of the $S_x$ component and the associated probabilities

Homework Equations

Also told that modulus of the spin vector always has the same value:

$$ S^2 = S.S =s(s+1)\hbar ^2 = 2\hbar ^2 $$

Although I don't see how that fits in.

The Attempt at a Solution

First normalise via:

$$ <\psi|\psi> = \frac{4}{29} + \frac{i^2 9}{29} + \frac{16}{29} = \frac{11}{29} $$

Therefore will need to multiply by $29/11$

To get the probabilities

$$ (\frac{2}{\sqrt29})^2 \times \frac{29}{11} = \frac{4}{11} $$

$$ (\frac{i3}{\sqrt29})^2 \times \frac{29}{11} = \frac{-9}{11}$$

$$ (\frac{-4}{\sqrt29})^2 \times \frac{29}{11} = \frac{16}{11} $$

From there I'm a little stuck on how to actually get the results of a measurement.

 

[tia89]

I do not remember exactly how you do this exercises (I did these things a couple of years ago and never used them again), but anyway first of all you have to understand in which direction is projected your state. Your wave function has |+1>, |0> and |-1>, meaning that they are the states of spin 1 projected along some direction (usually $z$ but you have to check).

If they are really projected along $z$, there should be a way to express the operator you want as a combination of $S^2$ and $S_z$, and you DO know the result of these operators on these states.

 

[chris_avfc]

I do not remember exactly how you do this exercises (I did these things a couple of years ago and never used them again), but anyway first of all you have to understand in which direction is projected your state. Your wave function has |+1>, |0> and |-1>, meaning that they are the states of spin 1 projected along some direction (usually $z$ but you have to check).

If they are really projected along $z$, there should be a way to express the operator you want as a combination of $S^2$ and $S_z$, and you DO know the result of these operators on these states.

I can't see anything about which axis it is projected on, I thought at first that maybe each term represents projection on a different axis.
I need to find results of the measurement for $S_z$ and $S_x$ though.

If they are projected along the $S_z$ axis, I'm guessing the expression for $S^2$ is useful, but how do I use this in combination with the operator for $S_z$?

 

[tia89]

I can't see anything about which axis it is projected on, I thought at first that maybe each term represents projection on a different axis.

That you have to check where you found the text of the exercise... anyway usually these are the eigenstates projected along z axis (it is the one we usually use)

If they are projected along the $S_z$ axis, I'm guessing the expression for $S^2$ is useful, but how do I use this in combination with the operator for $S_z$?

That's exactly what I do not remember... but try check on your book, there should be as far as I remember some relation which permits to find out $S_x$ from the two operators you know...

Sorry if I'm not able to be of more help

 

[vela]

Homework Statement

Spin-1 Particles prepared in the state:

$$ |\psi> = \frac{2}{\sqrt29} |1> + \frac{i 3}{\sqrt29}|0> - \frac{4}{\sqrt29} |-1> $$

Where I'm guessing the $|#>$ represents the spin state of -1,0 or 1.

To be more precise, $|m\rangle$ denotes an eigenstate of $\hat{S}_z$.

I'm looking to find the results of a measurements of the $S_x$ component and the associated probabilities

Homework Equations

Also told that modulus of the spin vector always has the same value:

$$ S^2 = S.S =s(s+1)\hbar ^2 = 2\hbar ^2 $$

Although I don't see how that fits in.

The Attempt at a Solution

First normalise via:

$$ <\psi|\psi> = \frac{4}{29} + \frac{i^2 9}{29} + \frac{16}{29} = \frac{11}{29} $$

Therefore will need to multiply by $29/11$

The state is already normalized. To find the adjoint of $|\psi\rangle$, you have to take the complex conjugate of the coefficients, so
$$\langle\psi|\psi\rangle = \frac{4}{29} + \frac{(-3i)(3i)}{29} + \frac{16}{29} = \frac{29}{29} = 1.$$

To get the probabilities

$$ (\frac{2}{\sqrt29})^2 \times \frac{29}{11} = \frac{4}{11} $$

$$ (\frac{i3}{\sqrt29})^2 \times \frac{29}{11} = \frac{-9}{11}$$

$$ (\frac{-4}{\sqrt29})^2 \times \frac{29}{11} = \frac{16}{21} $$

From there I'm a little stuck on how to actually get the results of a measurement.

You should recognize these answers can not possibly be correct because you can't have a negative probability and because they don't sum to 1.

To start, you need to familiarize yourself with two basic concepts in QM:

1. How are measurements related to observables?
2. If a system is in state $|\psi\rangle$, how do you find the probability of finding it in state $|\phi\rangle$?

 

[chris_avfc]

You should recognize these answers can not possibly be correct because you can't have a negative probability and because they don't sum to 1.

I thought that at first, but googling it I saw that there was a bit about Dirac saying they are possible, so I though it may be possible.

To start, you need to familiarize yourself with two basic concepts in QM:

1. How are measurements related to observables?

Am I right in thinking that if you take a measurement or an observable the related operator acts on the system, which will push the system into a different state, this changing the observable?

2. If a system is in state $|\psi\rangle$, how do you find the probability of finding it in state $|\phi\rangle$?

I think it might be this looking at my notes:

$$ P_1(\phi_1) = \frac{|<\phi_1|\psi>|^2}{<\psi|\psi>} $$

 

[vela]

Am I right in thinking that if you take a measurement or an observable the related operator acts on the system, which will push the system into a different state, this changing the observable?

No. First, an observable is the operator. Second, measurements have nothing to do with an operator acting on a state.

I think it might be this looking at my notes:

$$ P_1(\phi_1) = \frac{|<\phi_1|\psi>|^2}{<\psi|\psi>} $$

That's right if $|\phi_1\rangle$ is normalized.

 

[chris_avfc]

No. First, an observable is the operator. Second, measurements have nothing to do with an operator acting on a state.

Oh right, that makes a lot more sense thinking about it, considering you have the momentum operator and position operator.
Have I got this completely wrong then, but I thought when you make a measurement you change the wavefunction, as in would is not change from $\psi$ to $\phi$?

That's right if $|\phi_1\rangle$ is normalized.

Awesome, at least I got one part right, so once I have calculated $phi$ I can substitute this in, which would be

$$ [(\phi|1> coefficient) \times \frac{2}{\sqrt29}] [ (\phi|0> coefficient)\times \frac{i3}{\sqrt29}] + [(\phi|-1> coefficient) \times \frac{-4}{\sqrt29}] ^2 $$

All divided by 1 as $<\psi|\psi> = 1$

Edit: Wait, do I use the $S_z$ operator

$$ S_z|s,m> =\hbar m|s,m> $$

Similar to the $S^2$ operator given.

Where s = principal spin quantum number, and I'm not sure on m, maybe the mass?

 

[vela]

Oh right, that makes a lot more sense thinking about it, considering you have the momentum operator and position operator.

I'm not sure what you're getting at here.

Have I got this completely wrong then, but I thought when you make a measurement you change the wavefunction, as in would is not change from $\psi$ to $\phi$?

Yes, it's possible the state will change, which gets back to the topic of measurements in QM.

1. You have some observable that corresponds to the measurement you want to take. How do you know what the possible outcomes of the measurement are?
2. Say you make a measurement and obtain a certain result. How are the initial state and the new state related?

 

[chris_avfc]

Yes, it's possible the state will change, which gets back to the topic of measurements in QM.

1. You have some observable that corresponds to the measurement you want to take. How do you know what the possible outcomes of the measurement are?
2. Say you make a measurement and obtain a certain result. How are the initial state and the new state related?

Well I thought that the operator would act on the system, meaning I would apply $S_z$ to $\psi$, but now I'm thinking that is wrong.

 

[vela]

Yes, that's wrong. Don't you have a textbook you can consult? The concepts of observables and taking measurements are a few of the first topics typically covered in quantum mechanics.

 

[chris_avfc]

Yes, that's wrong. Don't you have a textbook you can consult? The concepts of observables and taking measurements are a few of the first topics typically covered in quantum mechanics.

Going through my lecture notes from the start of the year looking for the relevant stuff.

I think this is where I might be getting my idea of the operator acting from
"3. In classical mechanics it is always assumed that the value of an observable may
be measured without acting the state of the system. In quantum mechanics,
the act of measuring an observable usually causes an irreversible change in the
state of the system. One consequence of this is that for an arbitrary pair of
observables, it matters fundamentally in which order they are measured."

I think this statement may be more useful
"Every observable is represented by a linear hermitian operator whose eigenvalues are
the possible values of the observable and whose eigenfunctions are the wavefunctions
of the states in which the observable has a de finite value. When an observable is
measured, the state of the system changes discontinuously so that its wavefunction
becomes the eigenfunction corresponding to the value measured"

I may be interpreting this extremely wrong, but do I want the eigenvalue instead?

 

[vela]

Going through my lecture notes from the start of the year looking for the relevant stuff.

I think this is where I might be getting my idea of the operator acting from
"3. In classical mechanics it is always assumed that the value of an observable may
be measured without acting the state of the system. In quantum mechanics,
the act of measuring an observable usually causes an irreversible change in the
state of the system. One consequence of this is that for an arbitrary pair of
observables, it matters fundamentally in which order they are measured."

This just means that observing will cause the state to change. It doesn't mean applying the operator to the state.

I think this statement may be more useful
"Every observable is represented by a linear hermitian operator whose eigenvalues are
the possible values of the observable and whose eigenfunctions are the wavefunctions
of the states in which the observable has a definite value. When an observable is
measured, the state of the system changes discontinuously so that its wavefunction
becomes the eigenfunction corresponding to the value measured"

I may be interpreting this extremely wrong, but do I want the eigenvalue instead?

Yes. The eigenvalues of the operator are the possible outcomes. So if you know the eigenvalues of $S_z$, you know the possible outcomes of a measurement of the z-component of the spin.

The equation $\hat{S}_z | s, m\rangle = m\hbar | s, m\rangle$ tells you that $|s, m\rangle$ is an eigenstate of $\hat{S}_z$ with eigenvalue $m\hbar$. Usually the value of s is understood, so it's omitted, and you write simply $|m\rangle$. That's the notation being used in this problem.

The probability of measuring the spin to be $m\hbar$ is the probability of finding the system in state $|m\rangle$. You should be able to calculate this now. For $\hat{S}_z$, the calculations are straightforward because $|\psi\rangle$ was given in terms of the eigenstates of $\hat{S}_z$. For $\hat{S}_x$, you have a bit more work to do.

 

[chris_avfc]

This just means that observing will cause the state to change. It doesn't mean applying the operator to the state.


Yes. The eigenvalues of the operator are the possible outcomes. So if you know the eigenvalues of $S_z$, you know the possible outcomes of a measurement of the z-component of the spin.

The equation $\hat{S}_z | s, m\rangle = m\hbar | s, m\rangle$ tells you that $|s, m\rangle$ is an eigenstate of $\hat{S}_z$ with eigenvalue $m\hbar$. Usually the value of s is understood, so it's omitted, and you write simply $|m\rangle$. That's the notation being used in this problem.

The probability of measuring the spin to be $m\hbar$ is the probability of finding the system in state $|m\rangle$. You should be able to calculate this now. For $\hat{S}_z$, the calculations are straightforward because $|\psi\rangle$ was given in terms of the eigenstates of $\hat{S}_z$. For $\hat{S}_x$, you have a bit more work to do.

Is $m$ just an constant then?

Does that mean the actual measured value is just $m\hbar$ then, where the m changed depending on which spin state it is?

 

[vela]

Yup.

 

[chris_avfc]

Yup.

Is the relationship needed for to get the $S_x$, the formula to get the $S_x$ eigenstates in the terms of the $S_z$ basis?
I thought it would have to do with the relationship given, but I'm struggling to find anything.

Also am I wrong in thinking $m = +1,-1, 0$?

 

[vela]

You can express the eigenstates of $\hat{S}_x$ in terms of the $\hat{S}_z$ basis if you want. That's a straightforward way of doing it because you don't have to deal with a change of basis.

Why don't you start by finding the matrix that represents $\hat{S}_x$ in this basis?

 

[chris_avfc]

You can express the eigenstates of $\hat{S}_x$ in terms of the $\hat{S}_z$ basis if you want. That's a straightforward way of doing it because you don't have to deal with a change of basis.

Why don't you start by finding the matrix that represents $\hat{S}_x$ in this basis?

$$\frac{\hbar}{\sqrt2} \left(\begin{array}{cc} 0&1&0\\1&0&1\\0&1&0\end{array}\right) $$

Mod note: fixed your LaTeX for you.

 

[vela]

$$\frac{\hbar}{\sqrt2} \left(\begin{array}{cc} 0&1&0\\1&0&1\\0&1&0\end{array}\right) $$

Good. Now find the eigenvalues and eigenvectors of this matrix.

 

[chris_avfc]

Good. Now find the eigenvalues and eigenvectors of this matrix.

Haha, thanks for that was having a little trouble with that code.

Hmm, I think I might have been really stupid here.
Previously I calculated the eigenvalues and eigenvectors for the three spin operators $S_x , S_y , S_z$, which I presumed was unrelated, but thinking about it they probably are related.

So if I have the eigenvalues of $S_z$ and $S_x$ I already have the possible outcomes?

 

[vela]

Yes, the possible outcomes of a measurement are the eigenvalues of the corresponding observable.

 

[chris_avfc]

Yes, the possible outcomes of a measurement are the eigenvalues of the corresponding observable.

Are these values affected by the initial state of the system though?

 

[vela]

The probability of seeing a specific outcome depends on the state of the system. The state could result in the probability of an outcome being equal to 0.

 

[chris_avfc]

The probability of seeing a specific outcome depends on the state of the system. The state could result in the probability of an outcome being equal to 0.

Ah of course, so I calculated the eigenvalues of $S_z$ were $1, 0, -1$ all multiplied by $\hbar$, therefore the possible results of a measurement would be $\hbar, 0, -\hbar$.

But then I'm unsure how you calculate the probability because I don't get what you substitute in for $\phi$ in the equation mentioned before.

Using the example I'm looking at they have

$$ P(E_1) = |\langle\phi|\psi\rangle|^2 = |c \langle\phi|\phi\rangle x n = c^2 n$$

$c = Constant before |\phi\rangle$
$n = Normalisation Constant$

If I applied that in this situation I would end up with probabilities of $\frac{4}{29}, \frac{16}{29} and \frac{-9}{29}$, which is clearly wrong because I'm back to the negative probabilities as $i^2 = -1$ and $3^2 = 9$.

 

[vela]

Reread what I said in post 5.