# Will a uniform magnetic field be considered as measurment?

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Any pure state of spin-1/2 particle can be represented by a superposition of spin up and spin-down relative to an arbitrarily direction.
lw>=a l+> + b l-> where a^2+b^2=1
If there is no magnetic field, the measurement of the spin is random and we get 50-50 chance to be spin up and spin down along that direction. The system state collapses into the one with eigen value after the measurement (interaction with SG-device). The system afterword can not be a superposition of two base states.
If there is a magnetic field and the particle enters it, a time evolution operator will act on it. The time evolution unitary operator is a function of the Hamiltonian which is a function of B and S. So, logically because of S, the state should also collapse into the one with the eigen value after that interaction. However, the time dependent state is still a superposition of the two base vectors lw(t)>=a exp(iyBt/2) l+> + b exp(-iyBt/2) l->
How a state is still pure after being acted upon by S part of the unitary operator?

Some observations:

If there is a magnetic field and the particle enters it, a time evolution operator will act on it
As long as the system has energy, its state will evolve with time, even that there are no magnetic fields present.
However, the time dependent state is still a superposition of the two base vectors lw(t)>=a exp(iyBt/2) l+> + b exp(-iyBt/2) l->
The state at time ##t## will be given by ##|w(t)> \ \propto \ \exp(Ht) |w> \ = \ a \exp(Ht) |+> + \ b \exp(Ht) |->##, where I used the proportionality symbol, because I omitted the constants in the exponential.
If the states ##\{|+>, |->\}## are eigenstates of the Hamiltonian, the earlier relation can be expressed as ##\exp(Ht) |w> \ = \ a \exp(E_{+}t) |+> + \ b \exp(E_{-}t) |-> \ = \ a' |+ (t)> + \ b' |- (t) >##.
We see that the general state at time ##t## is equivalent to the earlier general state, but not equal.

Some observations:

As long as the system has energy, its state will evolve with time, even that there are no magnetic fields present.

The state at time ##t## will be given by ##|w(t)> \ \propto \ \exp(Ht) |w> \ = \ a \exp(Ht) |+> + \ b \exp(Ht) |->##, where I used the proportionality symbol, because I omitted the constants in the exponential.
If the states ##\{|+>, |->\}## are eigenstates of the Hamiltonian, the earlier relation can be expressed as ##\exp(Ht) |w> \ = \ a \exp(E_{+}t) |+> + \ b \exp(E_{-}t) |-> \ = \ a' |+ (t)> + \ b' |- (t) >##.
We see that the general state at time ##t## is equivalent to the earlier general state, but not equal.
My question is, why the general state is still general? The unitary operator is like any other operator, once acts on the system, the state should collapse into the one with a definite value not a state of superposition of base vectors!

My question is, why the general state is still general?
Ah, ok. That's because we supposedly have made no measurements. Think of it as if you had just allowed the state to evolve with time.
once acts on the system, the state should collapse into the one with a definite value
The postulate actually states that the collapse will happen and the measured value will be one of the eigenvalues of the operator after a measurement is made, not before.

So just to act with the operator on the general state does not cause its collapse.

Ah, ok. That's because we supposedly have made no measurements. Think of it as if you have just allowed the state to evolve with time.
But ##exp \frac{-iHt}{h} = exp \frac{i\gamma B_0 S_z t}{h}##. Now ##S_z## should act on l+> and l-> which is a measurement process.

which is a measurement process.
No. A measurement is a thing that you can eventually make, not your equations --it's a physical process. You accept the mentioned postulate of QM and do what it says to do, to see what you should get from the measurement.

So just to act with the operator on the general state does not cause its collapse.
What is then the difference between the collapse of the state and acting on it by an operator? The operator is a matrix and the state is a vector, so acting on a vector by a matrix given a value times the vector.

What is then the difference between the collapse of the state and acting on it by an operator?

Sorry, but it is not clear for me the difference. How can the measurement be represented mathematically in QM other than eigen value problem?

So, you must keep in mind that theories have a physical part and a mathematical part.

In our case, the physical part is the measurement. It's a thing that is up to you or others to do. Suppose you perform an experiment and obtain a numerical result.

Now, the math part of the theory says this result you got is the eigenvalue of an operator, and that this operator represents the physical observable phenomenum in your experiment, say, the electron spin.

Any pure state of spin-1/2 particle can be represented by a superposition of spin up and spin-down relative to an arbitrarily direction.
lw>=a l+> + b l-> where a^2+b^2=1
If there is no magnetic field, the measurement of the spin is random and we get 50-50 chance to be spin up and spin down along that direction.
That's not right. The probability of getting spin-up in the basis direction is ##|a|^2## while the probability of getting spin-down is ##|b|^2##; only in the case of ##|a|=|b|=\sqrt{2}/2## will you get that 50-50 chance.
The system afterword can not be a superposition of two base states.
It's not in a superposition of spin-up and spin-down along the direction of measurement (although the state still is ##a|+\rangle+b|-\rangle##; one of ##a## and ##b## will be zero and by convention we don't call it a superposition of those two states when one of the constants is zero). It is, however, still a superposition of spin-up and spin-down at any other angle.
So, logically because of S, the state should also collapse into the one with the eigen value after that interaction.
You could choose not to use a collapse interpretation, and then you wouldn't have to worry about this. But if you do choose to use a collapse interpretation, then you need to accept the problem with collapse interpretations: Some interactions cause collapse and some keep the time-dependent evolution going forward without collapsing anything, and you just have to know which is which. There's something not to like about every interpretation, and this is the thing not to like about collapse interpretations.
However, the time dependent state is still a superposition of the two base vectors lw(t)>=a exp(iyBt/2) l+> + b exp(-iyBt/2) l->
How a state is still pure after being acted upon by S part of the unitary operator?
That time-varying state is still a superposition of ##|+\rangle## and ##|-\rangle##, and all superpositions are pure states. A state is pure if it is a single vector in the appropriate Hilbert space; that we can choose to write it as sum of two other vectors is irrelevant to its purity.

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bhobba and Leo1233783
But ##exp \frac{-iHt}{h} = exp \frac{i\gamma B_0 S_z t}{h}##. Now ##S_z## should act on l+> and l->

No. You are confusing two different operators.

The operator ##S_z## is Hermitian; it describes a measurement (specifically the measurement of spin about the ##z## axis). The eigenstates of this operator are the spin "up" and spin "down" states about the chosen axis; they correspond to eigenvalues ##+ 1/2## and ##- 1/2##. Notice that these eigenvalues are real (because the operator is Hermitian); that is why we can interpret this operator as describing a measurement, since the actual results observed from any measurement must be real numbers.

The operator ##exp \frac{i \gamma B_0 aS_z t}{t}## is unitary; it describes a time evolution (specifically the time evolution of the state of a charged particle with spin in a magnetic field). You can figure out the eigenstates and eigenvalues of this operator if you want, but they won't be real, which is why this operator can't describe a measurement (it's not Hermitian). However, you should also be able to show that this operator preserves all inner products of vectors in the corresponding Hilbert space (that's one way of saying what "unitary" means); that's why it can describe a time evolution.

You could choose not to use a collapse interpretation, and then you wouldn't have to worry about this.

There is an issue here in that many beginner and even some intermediate QM texts all talk about collapse - advanced tests like Ballentine do not.

So it may be difficult for the OP to come to grips with no collapse interpretations.

Fortunately Griffiths (not the same Griffiths that wrote the famous textbook - its different guy) has made freely available online a textbook that gives the Consistent/Decoherent histories interpretation which is non-collapse:
http://quantum.phys.cmu.edu/CQT/index.html

Good warm up for reading Ballentine where there is no collapse at all - it's from two axioms only - the eigenvalue Observable rule (the only outcomes are eigenvalues of the observsble) and the Born Rule. The science advisers had a long discussion about collapse and even defining is rather difficult - the best I could see was basically its one way of looking at the Born rule which seems a rather trivial thing to worry about - but QM IMHO is full of things like that - things that IMHHO are basically trivial some get really worked up about. It one of the strange things about QM. IMHO QM is strange enough without making it stranger than it needs to be.

BTW I fell into the same trap before posting here - it really is an insidious problem.

Thanks
Bill

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vanhees71
The operator ##exp \frac{i \gamma B_0 aS_z t}{t}## is unitary; it describes a time evolution (specifically the time evolution of the state of a charged particle with spin in a magnetic field). You can figure out the eigenstates and eigenvalues of this operator if you want, but they won't be real, which is why this operator can't describe a measurement (it's not Hermitian). However, you should also be able to show that this operator preserves all inner products of vectors in the corresponding Hilbert space (that's one way of saying what "unitary" means); that's why it can describe a time evolution.
But if the particle goes inside a SG-device, it is measured and we get a real value eigenvalue. There is a gradient of the magnetic field inside the device and the time evolution operator of SG-device is ##e^\frac{-i\gamma(B_0+\nabla B.z)St}{\hbar}##. So how does this complex value operator yield a real value? There is a small typo error in your operator, the denominator should be ##\hbar## not t.

Some interactions cause collapse and some keep the time-dependent evolution going forward without collapsing anything, and you just have to know which is which.
In a previous thread about an entangled pair, I proposed a thought experiment where one particle goes into a uniform magnetic field and the other set free. I remember you told me that the one goes into the field emits photon if it is anti-parallel to the field and this can be considered a type of measurement. Now I am confused because this thread told there is no measurement when the particle goes into a uniform field.

But if the particle goes inside a SG-device, it is measured and we get a real value eigenvalue. There is a gradient of the magnetic field inside the device and the time evolution operator of SG-device is ##e^\frac{-i\gamma(B_0+\nabla B.z)St}{\hbar}##. So how does this complex value operator yield a real value? There is a small typo error in your operator, the denominator should be ##\hbar## not t.
In a Stern-Gerlach device you have not only a (strong) homogeneous magnetic field in the direction of the spin component you want to measure (or even prepare) but also an inhomogeneous magnetic field, which leads to a force on the particle as a whole and thus a split of the particle beam dependent on the value of the spin component, i.e., the value of the spin component becomes entangled with the position of the particle. So far everything is described by unitary time evolution (and this case for a simple model magnetic field can be completely calculated semi-analytically without making reference to semiclassical methods you usually find in textbooks, although the exact calculation shows that this approximation is well justified under the usual SG setup). Everything is described as a closed system consisting of a single particle in an external magnetic field.

Now to get a clean beam of particles which has a determined spin component you have to block all partial beams except one which contains the particles with the wanted value of the spin component. At this moment you introduce some "beam dump" absorbing the unwanted particles, i.e., the quantum system is no longer closed, and the time evolution when looked still further only at the particle, becomes non-unitary. For all practical purposes (FAPP) that's described in a very superficial way by the early quantum physicists around Bohr and Heisenberg as "the collapse of the state".

I'm not a proponent of collapse interpretations, but that's another topic. I strongly recommend to read a good discussion on interpretations: One is found in Weinberg, Lectures on Quantum Mechanics (coming to the conclusion that there is no final interpretation for QT yet). Another one is very comprehensive in Ballentine, Quantum Mechanics. Ballentine is a proponent of the minimal statistical interpretation, which is the one which almost all practitioners of QT use.

Now to get a clean beam of particles which has a determined spin component you have to block all partial beams except one which contains the particles with the wanted value of the spin component. At this moment you introduce some "beam dump" absorbing the unwanted particles, i.e., the quantum system is no longer closed, and the time evolution when looked still further only at the particle, becomes non-unitary.
I understand that the gradient of the field inside SG-device applies a force on the particle and separate the beam with particular spin in one direction from all other beam with different spin in the other direction. But, does it mean the time evolution operator I write above, with the gardient component, is wrong? If it is right, then the operator should have a complex value against what we observe. If it is wrong, then do you propose a different sort of operator discribing the inhomogeneous magnetic field, what will be the formalism of it? or may be the physical process of measurment can not be described by any operator at all in this scenario?

The time-evolution operator (for a time-independent Hamiltonian, i.e., for the SG apparatus with a time-independent magnetic field written in Coulomb gauge) is given by
$$\hat{U}(t)=\exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} \right).$$
A nice paper on the subject is

https://journals.aps.org/pra/abstract/10.1103/PhysRevA.83.012109

Mentz114
The time-evolution operator (for a time-independent Hamiltonian, i.e., for the SG apparatus with a time-independent magnetic field written in Coulomb gauge) is given by
$$\hat{U}(t)=\exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} \right).$$
A nice paper on the subject is

https://journals.aps.org/pra/abstract/10.1103/PhysRevA.83.012109

This paper is also avaible freely here.

I look forward to reading it.

if the particle goes inside a SG-device, it is measured and we get a real value eigenvalue

Yes. But what operator describes that measurement process? The answer is, the operator ##S_z##. Not the operator ##exp \left( i \gamma B_0 S_z / \hbar \right)##.

the time evolution operator of SG-device

Which is not the operator that describes the measurement. Time evolution and measurement are not the same thing.

does it mean the time evolution operator I write above, with the gardient component, is wrong?

No, it means that operator describes time evolution, not measurement.

@PeterDonis I'm a bit puzzled by your statement ##\hat{S}_z## "describes the measurement (of the spin-z component)". In which sense do you mean this? Is this a clever didactical trick, I'm not aware of?

In the standard formulation the operator represents an observable, but has nothing to do with measuring this observable. The measurement itself is done via an interaction of the measured system with the measurement apparatus, which is usually very difficult to describe quantum mechanically, but that is fortunately not necessary to use quantum theory to real-world phenomena. The SG experiment is a particularly simple example, where you can analytically evaluate the time evolution of the systems quantum theoretically. It explains in a quite simple way from first principles, how the spin-z component becomes entangled with the position of the particle and thus how you can prepare spin-z eigenstates by filtering out the corresponding partial beam. This is done by "dumping" all other beams, and you just need to know that you can put material in the way of particles getting them absorbed. There's no need to describe this process in full microscopic detail (which however is also possible, see "Bethe Bloch formula" in standard textbooks on experimental particle physics).

bhobba
I'm a bit puzzled by your statement ^SzS^z\hat{S}_z "describes the measurement (of the spin-z component)". In which sense do you mean this?

I mean that the observable "spin about the z axis" is represented by ##S_z##, not by the exponentiation of ##S_z##. You're right, my use of language was sloppy, your statement is more precise.

That's for sure correct. Nevertheless the point of the self-adjoint operators used to represent observables is that they have a complete set of (generalized) orthonormal eigenvectors, which admits the usual interpretation of the quantum state in terms of Born's rule, i.e., if ##\hat{\rho}## represents the state and you measure the observable ##A## with the outcome ##a## (necessarily an eigenvalue of the representing opertor ##\hat{A}##) and if ##|a,\beta \rangle## is a complete set of orthonormal eigenvectors of ##\hat{A}## with eigenvalue ##a##, then the probability is
$$P(a|\rho)=\sum_{\beta} \langle a,\beta|\hat{\rho}|a,\beta \rangle.$$
Nevertheless, sometimes it can be advantageous to choose other kinds of operators to represent observables. E.g., angle variables are better represented by the corresponding unitary operators, i.e., instead of representing the angle in polar coordinates by a self-adjoint operator you can use its exponential, i.e., ##\exp(\mathrm{i} \hat \phi)##. Again the point is that unitary operators have complete sets of (generalized) orthonormal eigenvectors, and Born's rule again makes sense. See, e.g.,

https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.40.411

if ##\hat{\rho}## represents the state and you measure the observable ##A## with the outcome ##a## (necessarily an eigenvalue of the representing opertor ##\hat{A}##) and if ##|a,\beta \rangle## is a complete set of orthonormal eigenvectors of ##\hat{A}## with eigenvalue ##a##, then the probability is
$$P(a|\rho)=\sum_{\beta} \langle a,\beta|\hat{\rho}|a,\beta \rangle.$$
Why there is a sum? is this the probability to find the system in the state ##|a,\beta \rangle##?

This is the probability to find the value ##a## when measuring the observable ##A## (as explained in #24) provided the measured system is prepared in the state ##\hat{\rho}##.

the point is that unitary operators have complete sets of (generalized) orthonormal eigenvectors

But with eigenvalues that might not be real, which is why operators representing observables are normally required to be Hermitian.

The paper you linked to does not appear to actually claim that the operator ##\exp(i \hat{\phi})## represents the phase angle observable. Rather, it appears to be saying that there is no "phase angle" observable conjugate to the number operator, but there are "cosine" and "sine" observables (i.e., Hermitian operators with real eigenvalues) that can be used instead.

This is the probability to find the value ##a## when measuring the observable ##A## (as explained in #24) provided the measured system is prepared in the state ##\hat{\rho}##.
I thought it is in the form ##\frac{\mid\langle a,\beta|\rho\rangle\mid^2}{\sum \mid\langle a_k,\beta|\rho\rangle\mid^2}##

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But with eigenvalues that might not be real, which is why operators representing observables are normally required to be Hermitian.

The paper you linked to does not appear to actually claim that the operator ##\exp(i \hat{\phi})## represents the phase angle observable. Rather, it appears to be saying that there is no "phase angle" observable conjugate to the number operator, but there are "cosine" and "sine" observables (i.e., Hermitian operators with real eigenvalues) that can be used instead.
Yes, sure, and instead of using ##\cos## and ##\sin## you can use ##\exp## to define an angle or phase operator. The eigenvalues of unitary operators are of course all of the form ##\exp(\mathrm{i} \varphi)## with ##\varphi \in \mathbb{R} \; \text{mod} \; 2 \pi##.

I thought it is in the form ##\frac{\mid\langle a,\beta|\rho\rangle\mid^2}{\sum \mid\langle a_k,\beta|\rho\rangle\mid^2}##
How did you come to this conclusion? I don't know, what this might represent. The notation doesn't make any sense.

##\hat{\rho}## is the Statistical Operator uniquely representing the quantum state the system is prepared in. For a pure state, there is a normalized vector ##|\psi \rangle## such that ##\hat{\rho}=|\psi \rangle##. Then, of course, the probability discussed is
$$P(a|\psi)=\sum_{\beta} |\langle a,\beta|\psi \rangle|^2.$$
Here I assumed, you just measure the observable ##A## and not additionally any other compatible observables. Then you have to sum (integrate) over all degeneracies.

So what's the problem then? That's standard QT, as I've written above (I considered the general case that usually applies, i.e., degenerate eigenstates of a single observable operator).

So what's the problem then? That's standard QT, as I've written above (I considered the general case that usually applies, i.e., degenerate eigenstates of a single observable operator).
No great problem, just I am not familiar with this expression of probability. The probability is, in general, written in term of a value over the sum of all values representing the system. So the sum should be in the denominator. And the value in our case is the square of the projection of the general state on the eigen state (the square of the dot product).

I still don't understand your problem.

Let's take the simplest case of a nondegenerate discrete spectrum. Then the operator ##\hat{A}## representing the observable ##A## has a complete set of orthonormal vectors ##|a \rangle##, i.e., each eigenspace is onedimensional. Now let the system be prepared in a pure state, represented by a unit ray in Hilbert space and let ##|\psi \rangle## be a representant of this ray. Then the probability to find the value ##a## when measuring the observable ##A## is [corrected in view of #35]
$$P(a|\psi)=|\langle a|\psi \rangle|^2.$$
These are non-negative real numbers. Since the vector ##|\psi \rangle## by definition is normalized and since the orthonormal set ##|a \rangle## is complete, you have
$$1=\langle \psi |\psi \rangle=\sum_a \langle \psi|a \rangle \langle a|\psi \rangle=\sum_a |\langle a|\psi \rangle|^2=\sum_a P(a|\psi),$$
i.e., ##P(a|\psi)## are indeed a correct set of probabilities for the outcome of measurements of observable ##A## provided the system is prepared in the said pure state.

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Mentz114
I still don't understand your problem.

Let's take the simplest case of a nondegenerate discrete spectrum. Then the operator ##\hat{A}## representing the observable ##A## has a complete set of orthonormal vectors ##|a \rangle##, i.e., each eigenspace is onedimensional. Now let the system be prepared in a pure state, represented by a unit ray in Hilbert space and let ##|\psi \rangle## be a representant of this ray. Then the probability to find the value ##a## when measuring the observable ##A## is
$$P(a|\psi)=|\langle a|\psi \rangle|.$$
These are non-negative real numbers. Since the vector ##|\psi \rangle## by definition is normalized and since the orthonormal set ##|a \rangle## is complete, you have
$$1=\langle \psi |\psi \rangle=\sum_a \langle \psi|a \rangle \langle a|\psi \rangle=\sum_a |\langle a|\psi \rangle|^2=\sum_a P(a|\psi),$$
i.e., ##P(a|\psi)## are indeed a correct set of probabilities for the outcome of measurements of observable ##A## provided the system is prepared in the said pure state.
Although this is excursion from the main point of the thread, I still think the first probability must be ##P(a|\psi)=|\langle a|\psi \rangle|^2## not ##|\langle a|\psi \rangle|## . Example, spin-1/2 has the state ##|\psi \rangle=\frac{1}{\sqrt2} |up \rangle+\frac{1}{\sqrt2} |down\rangle## So the probability that the system in anyone of those base states is 1/2