Quantum spin superposition interpretation

[phoenix3]

Homework Statement

How to interpret superposition

Relevant Equations

$cos^2(\frac{\theta}{2})$

Here is my workings out:
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If a particle's spin of magnitude $\frac {\hbar}{2}$ is prepared along direction $\vec r_1$ and subsequently its spin is measured along direction $\vec r_2$ at an angle $\vec \theta$ to $\vec r_1$, the probability of its being found "spin up" along is $\vec r_2$ is $P(up)=cos^2(\frac{\theta}{2})$ again with full magnitude $\frac {\hbar}{2}$.
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Classically, the spin might easily have been incorrectly predicted to be of magnitude $\frac {\hbar}{2} . cos(\theta)$ along $\vec r_2$

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in units of $\frac {\hbar}{2}$, $\frac {\hbar}{2} = 1$.
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Starting with the seemingly 'incorrect' assumption that the component along $\vec r_2$ is $\frac {\hbar}{2} . cos(\theta)$
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$\frac {\hbar}{2} . cos(\theta) = 1.cos(\theta)$ in units of $\frac {\hbar}{2}$
\begin{align}
1.cos(\theta) & = cos(\frac{\theta}{2}+\frac{\theta}{2}) \nonumber \\
& = cos^2(\frac{\theta}{2})-sin^2(\frac{\theta}{2}) \nonumber \\
& = (+1).cos^2(\frac{\theta}{2}) + (-1).sin^2(\frac{\theta}{2}) \nonumber \\
\end{align}

Multiplying accross by a time interval, $\delta t$,

\begin{align}
1.cos(\theta).\delta t = (+1).cos^2(\frac{\theta}{2}).\delta t + (-1).sin^2(\frac{\theta}{2}).\delta t \nonumber \\
\end{align}

it seems that this can be interpreted to mean either:
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(A) a combination of two vectors of magnitudes $cos^2(\frac{\theta}{2})$ and $-sin^2(\frac{\theta}{2})$ respectively, acting for duration $\delta t$ or

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(B) a combination of two vectors of magnitudes 1 and -1 respectively, the first acting for a duration $cos^2(\frac{\theta}{2}).\delta t$ and the second acting for the remaining duration $sin^2(\frac{\theta}{2}).\delta t$

$$$$
with $( cos^2(\frac{\theta}{2}).\delta t + sin^2(\frac{\theta}{2}).\delta t = \delta t )$
$$$$
This latter interpretation, (B), would seem to justify quite nicely why the spin is always measured to have the same absolute unit magnitude and also why the probability of getting "spin up" along $\vec r_2$ is equal to $cos^2(\frac{\theta}{2})$ since according to this interpretation it is in fact "spin up" along $\vec r_2$ , $cos^2(\frac{\theta}{2})$ of the time.
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Obviously once measured along $\vec r_2$ it is then in an eigenstate along $\vec r_2$ and the probability of a measurement along $\vec r_1$ being "spin up" along $\vec r_1$ is $cos^2(\frac{\theta}{2})$ for the same reason.
$$$$
Is there an error in the above and if so where?

 

[DrClaude]

Is there an error in the above and if so where?

The error seems to be that you are trying to think classically about quantum mechanics. My advice is to stop doing this and fully embrace QM without trying to make too many classical pictures of what is going on (although some classical pictures can help some of the time).

 

[phoenix3]

Respectfully, DrClaude, that still doesn't explain where any error is, either in my math or in my interpretation of it. Besides, as long as the math correctly predicts the probability, where's the harm in having an interpretation of it that is non-mysterious?

 

[DrClaude]

Respectfully, DrClaude, that still doesn't explain where any error is, either in my math or in my interpretation of it.

I'm tempted to answer that it's "not even wrong," but I'll humor you.

If a particle's spin of magnitude $\frac {\hbar}{2}$ is prepared along direction $\vec r_1$ and subsequently its spin is measured along direction $\vec r_2$ at an angle $\vec \theta$ to $\vec r_1$, the probability of its being found "spin up" along is $\vec r_2$ is $P(up)=cos^2(\frac{\theta}{2})$ again with full magnitude $\frac {\hbar}{2}$.

This is technically incorrect. A spin-1/2 system has a spin magnitude of $\frac{\sqrt{3}}{2} \hbar$, with its projection on any axis being $\pm \frac{1}{2} \hbar$.

Classically, the spin might easily have been incorrectly predicted to be of magnitude $\frac {\hbar}{2} . cos(\theta)$ along $\vec r_2$ in units of $\frac {\hbar}{2}$, $\frac {\hbar}{2} = 1$.

Strange choice of units (one would usually choose $\hbar = 1$), but not incorrect.

Multiplying accross by a time interval, $\delta t$,

What is the justification in doing this?

\begin{align}
1.cos(\theta).\delta t = (+1).cos^2(\frac{\theta}{2}).\delta t + (-1).sin^2(\frac{\theta}{2}).\delta t \nonumber \\
\end{align}

it seems that this can be interpreted to mean either:

(A) a combination of two vectors of magnitudes $cos^2(\frac{\theta}{2})$ and $-sin^2(\frac{\theta}{2})$ respectively, acting for duration $\delta t$ or
$$$$
(B) a combination of two vectors of magnitudes 1 and -1 respectively, the first acting for a duration $cos^2(\frac{\theta}{2}).\delta t$ and the second acting for the remaining duration $sin^2(\frac{\theta}{2}).\delta t$

No. Just because it "looks like" the original $cos^2(\frac{\theta}{2})$ doesn't mean it is the same thing. The entire expression should be attached to measuring spin-up, even though a minus sign appears.

Try the same exercise with the probability of measuring spin-down and you will see that it doesn't work. What you did is close to what we call "numerology," where people manipulate equations and numbers until they look like something we already know, but this doesn't give any deeper understanding because this "game" can be played with any set of numbers.

Besides, as long as the math correctly predicts the probability, where's the harm in having an interpretation of it that is non-mysterious?

The harm is in having an understanding that is actually incorrect.

 

[phoenix3]

This is technically incorrect. A spin-1/2 system has a spin magnitude of , with its projection on any axis being .

This is technically incorrect. A spin-1/2 system has a spin magnitude of , with its projection on any axis being .

This is technically incorrect. A spin-1/2 system has a spin magnitude of , with its projection on any axis being .

yes i know, but it only ever 'measured' to be $\frac {\hbar}{2}$

 

[phoenix3]

What is the justification in doing this?

to include 'time' into consideration. One can't pretend time is irrelevant.

 

[phoenix3]

Try the same exercise with the probability of measuring spin-down and you will see that it doesn't work.

It does wrk: it gives it spin-down probability as $sin^2(\frac{\theta}{2})$

 

[phoenix3]

The entire expression should be attached to measuring spin-up, even though a minus sign appears.

The entire expression evaluates to $cos(\theta)$ and indicates that $cos(\theta)$ is in fact a net result of spin-up and spin-down during any non-zero interval of time $\delta t$, whereas at any single 'point' in time it is either spin-up xor spin-down.

 

[phoenix3]

The harm is in having an understanding that is actually incorrect.

To say that, while not measured along $\vec r_2$ the spin is in a superposition state of being spin-up and spin-down at each individual point in time is not only incorrrect, its meaningless.

 

[vanhees71]

The spin component in direction $\vec{n}=(\sin \vartheta \cos \varphi,\sin \vartheta \sin \varphi,\cos \vartheta)$ is represented in the usual basis of $\hat{s}_3$-eigenstates by
$$\hat{s}_n=\vec{n} \cdot \vec{\sigma}/2=\frac{1}{2} \begin{pmatrix} \cos \vartheta & \exp(-\mathrm{i} \varphi) \sin \vartheta \\ \exp(\mathrm{i} \varphi) \sin \vartheta & -\cos \vartheta \end{pmatrix}.$$
The eigenvectors for $\hat{s}_n$ are
$$|\vec{n},1/2 \rangle=\begin{pmatrix}\exp(-\mathrm{i} \varphi/2) \cos \vartheta) \\ \exp(\mathrm{i} \varphi/2) \sin \vartheta) \end{pmatrix},$$
$$|\vec{n},1/2 \rangle=\begin{pmatrix} -\exp(-\mathrm{i} \varphi/2) \sin \vartheta \\ \exp(\mathrm{i} \varphi /2) \cos \vartheta \end{pmatrix}.$$
For a particle being prepared in $|\vec{e}_z,1/2 \rangle$ the probabilities to measure $s_n=\pm 1/2$ obviously is
$$P_{1/2}=|\langle \vec{n},1/2|\vec{e}_z,1/2 \rangle|^2=\cos^2(\vartheta/2), \quad P_{-1/2} = \sin^2(\vartheta/2).$$