Timeline for Aren't measurements in the Stern Gerlach experiment inherently intrusive to the states of particles?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 24 at 18:48 | vote | accept | UVcatastrophe | ||
| Apr 2, 2021 at 16:39 | vote | accept | UVcatastrophe | ||
| Apr 11 at 14:13 | |||||
| Mar 18, 2021 at 2:18 | comment | added | UVcatastrophe | Okay that does clarify things a lot. Thank you! | |
| Mar 17, 2021 at 16:42 | comment | added | user1379857 | In reality, the position of the particle is described by a wave function, little "bump" of probability, and the maximum position of this bump travels more or less like a classical particle would, feeling accelerations from "forces" you could say. That is how the quantum notion of position is related to the classical one in this situation. | |
| Mar 17, 2021 at 16:40 | comment | added | user1379857 | The important this is that we do not pick and choose when to apply the classical model and when to apply the quantum model. If we are being rigorous, we always apply the quantum model and never apply the classical model. The real question is, when we want to explain it to first-time students, do we explain it using the classical or quantum explanations? Usually we use a sort of hybrid explanation, where we treat the SPIN as a quantum object, but the POSITION as a sort of classical object which feels "forces" etc. This is probably easier for students overall but does cause confusion. | |
| Mar 17, 2021 at 16:11 | comment | added | UVcatastrophe | Is it that quantum uncertainty goes along the lines of saying that 'it is not physically possible to design an experimental set-up that is better than this'? | |
| Mar 17, 2021 at 16:08 | comment | added | UVcatastrophe | I do understand and appreciate the points you made. But how can we pick and choose where to apply the classical model and where not to? Because the moment we do consider it here, we'd conclude that the experimental set up is physically affecting these states and hence, ascribing the loss of information to quantum uncertainty would be disingenuous. Only after eliminating these classical effects can we truly say that quantum uncertainty is the reason why we can't measure the components of the spin in x and z directions. | |
| Mar 17, 2021 at 15:47 | comment | added | user1379857 | ... where $\vec L = \frac{\hbar}{2}( \sigma_x, \sigma_y, \sigma_z)$ and $\sigma_x, \sigma_y, \sigma_z$ are the $2 \times 2$ Pauli matrices. The average energy of the state $| \psi \rangle$ is $\langle \psi | U | \psi \rangle$. So the energy $U$ is a $2 \times 2$ matrix, but its expectation value is indeed a single number. Schroinger's equation says that the state $| \psi \rangle$ evolves in time according to $$ i \hbar \frac{d}{dt} | \psi \rangle = U | \psi \rangle.$$ Or anyway, this is how the spin degree of freedom works. If you want to describe the position there's more to it. | |
| Mar 17, 2021 at 15:45 | comment | added | user1379857 | There is a difference between the classical story and the quantum story. The classical story can be used as a useful mental model to explain some features of the Stern Gerlach experiment, but not all of them. For instance, there isn't really a well defined notion of "force" or "torque" in quantum mechanics, everything just evolves according the the Schordinger equation. In fact, the energy $U = - \vec{\mu} \cdot \vec{B}$ is not even a number, but rather a $2 \times 2$ matrix in QM! Here, $\vec{\mu} = - g \frac{q}{2m} \vec{L}$... | |
| Mar 17, 2021 at 15:39 | comment | added | UVcatastrophe | To be able to define $\mu$, don't we need to consider spin as a three dimensional vector in space? Since $\mu=g_s\frac{q}{2m}\vec S$. We need to consider the three dimensional magnetic dipole moment to describe Larmor precession using the equation $$\tau=\mu \times \vec B$$. So why is it that we can write the equation involving $\mu$ you wrote to describe deviation but not this one that describes precession? | |
| Mar 16, 2021 at 15:29 | comment | added | user1379857 | What you describe is really the crux of quantum spin. You can't think of it as a three dimensional vector $(v_x, v_y, v_z)$ like you might classically. Please see my recent edit to the answer | |
| Mar 16, 2021 at 15:29 | history | edited | user1379857 | CC BY-SA 4.0 |
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| Mar 16, 2021 at 11:24 | comment | added | UVcatastrophe | I understand your explanation for how we observe two beams, however, my question still remains. If I'm not wrong, it should never be possible for the spins to completely align or misalign with the magnetic field (say one that points in the z direction) as the direction of spin is quantized. If spin completely points in the angle of measurement (z), we would have the information about its x and y components too, them being zero. So the case that you describe should not be possible. | |
| Mar 16, 2021 at 6:35 | history | answered | user1379857 | CC BY-SA 4.0 |