# I Applying an observable operator on the current state

#### saar321412

hey :)

assume I have an operator A with |ai> eigenstates and matching ai eigenvalues, and assume my system is in state
|Ψ> = Σci|ai>
I know that applying the measurement that corresponds to A will collapse the system into one of the |ai>'s with probability
|<Ψ|ai>|2.

with that being said, what does applying the operator on |Ψ> (A|Ψ> ) means? what is the meaning of the new state A|Ψ>?
it won't be the state of the system after the measurement because that will be one of the eigenstates.

also, what is the meaning of A eigenvalues?

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#### vanhees71

Gold Member
Well, answering this question in this forum is dangerous for you, since I predict there will be 100 more postings in no time contradicting me, because many people follow other interpretations of quantum theory than I do. I follow the minimal statistical interpretation, and in my opinion that's what should be taught to beginning quantum mechanic, because it describes in most simple terms what's related to physics and how QT is applied to real-world phenomena. Everything beyond it is subject of philosophy (if not personal belief and religion) of the individual physicist.

To formalism goes with the following postulates (kinematical part, which is all you need to answer your question)

(1) A (pure) quantum state is described as a unit ray $[|\psi \rangle]$ in an appropriate Hilbert space, i.e., a unit-vector $|\psi \rangle$ with all other unit vectors of the form $|\psi' \rangle=\exp(\mathrm{i} \varphi) |\psi \rangle$ with $\varphi \in \mathbb{R}$ considered to represent the same state.

(2) An observable $A$ is represented by a self-adjoint operator $\hat{A}$. The (generalized) eigenvalues (spectral values) of this operators are the possible values the observable can take when (accurately) measured. Let $|a,\beta \rangle$ denote a complete set of orthonormal eigenvectors of $\hat{A}$.

(3) If the system is prepared in a state $[|\psi \rangle]$, then the probability to find one of the possible eigenvalues $a$ of the operator $\hat{A}$ when then observable $A$ is measured is given by
$$P_{\psi}(a)=\sum_{\beta} |\langle a|\psi \rangle|^2.$$

Note that I don't make any statement what happens to the system when I do a measurement of the observable $A$ on it, because this I can't say without knowing precisely the measurement device you use. It's not necessary to make this collapse postulate you mention. It can happen that you have an experimental setup allowing to achieve such a "collapse-like" behavior in the sense of an effective description of the (usually quite complicated) dynamics describing the interaction between the measured system and the measurement device, but taken literally the collapse postulate leads to direct contradictions with the very foundations of relativistic quantum field theory (linked-cluster property of the S-matrix) and is thus to be abandoned from the foundations.

That said, you see that the vector $\hat{A} |\psi \rangle$ has no specific physical meaning. Also the acting with the operator $\hat{A}$ on $|\psi \rangle$ has no phsycal meaning. The only thing it provides is a calculational tool to evaluate the expectation value of the observable $A$ under the condition that the system is prepared in the state $[|\psi \rangle]$,
$$\langle A \rangle = \sum_{a} P_{\psi}(a) a =\sum_{a,\beta} \langle \psi|a,\beta \rangle \langle a,\beta|\psi \rangle a = \sum_{a,\beta} \langle \psi|\hat{A}| a,\beta \rangle \langle a,\beta|\psi \rangle = \langle \psi|\hat{A}|\psi \rangle,$$
where I have used first that
$$\hat{A}|a,\beta \rangle=a |a,\beta \rangle$$
and the completeness of the eigenstates,
$$\sum_{a,\beta} |a,\beta \rangle \langle a,\beta|=\hat{1}.$$

#### saar321412

thank you!!!!!

and about the "collapse to eigenvalue" thing, if it is not always true, how do I know the state of the system after some measurement?
to be more specific, if I think about the Stern Gerlach experiment I have learned that measuring +1 resulting in a spin up |+> state, is this statement correct or I have misunderstood?

#### vanhees71

Gold Member
As I said, it depends on the measurement apparatus, what happens to the measured system.

The Stern-Gerlach experiment (SGE) is a paradigmatic example for an experiment, which you can indeed interpret in a kind of "collapse to the eigenstate".

In the SGE Stern and Gerlach prepared a beam of silver atoms by just letting out some silver vapor from an oven with a little hole in it. In modern terms they prepared a state which discribes a silver atom that goes with some uncertainty in a direction perpendicular to the opening (i.e., its momentum has a value around some given value $\vec{p}_0$. This you can translate, using the usual rules of quantum theory to the position representation, leading to an atom in a state running with some spread along the classical trajetory $\vec{x}(t)=\vec{x}_0+\vec{p}_0/m t$.

This atom now runs through a magnetic field, tuned such that it has a large homogeneous component in some direction, which we'll call the $z$ direction and some field gradient. Now an Ag atom has a magnetic moment given by the magnetic moment of its single valence electron, i.e., it has magnetic moment of one Bohr magneton (with the gyrofactor of 2 for an elementary spin-1/2 particle).

This setup lets to a motion of the silver atoms which is governed by the force due to the interaction of the atom's magnetic dipole moment with the inhomogeneous magnetic field. Due to the strong homogeneous part of the field in the $z$ direction the spin and thus its magnetic moment precesses pretty quickly around the z-direction, and this it does much quicker than typical timescales within which the atom moves around (i.e., the time $\Delta t=|\Delta \vec{x}|/\vec{v}_0$). Thus one can, in good approximation, take into account only the magnetic field gradient in $z$ direction, and the force is thus in direction of the $z$-component of the magnetic moment and thus the $z$-component of the spin of the Ag atom. This component can take two values $\pm \hbar/2$. Thus the silver atom going initially along, say, the $x$ direction, is now deflected, depending on the value of the $z$-component of the spin. I.e., the component of the wave function along the eigenvector to the eigenvalue $\sigma_z=+\hbar/2$ is deflected in the opposite direction thatn that component along the eigenvector to the eigenvalue $\sigma_z=-\hbar/2$.

Of course, at the same time the wave function broadens in space, but you can choose the magnetic field such that the position uncertainty stays still small compared to the deflection of the parts of the wave function in different $\sigma_z$-direction. Thus taking many Ag atoms you get two pretty sharp beams, of which one carries (nearly) only Ag atoms with $\sigma_z=+\hbar/2$ and the other one atoms with $\sigma_z=-\hbar/2$. This means that the position and the spin-z component become entangled, i.e., having initially a state which has equal probability to measure $\sigma_z=\hbar/2$ or $\sigma_1=-\hbar/2$ now you measure with certainty $\sigma_z=\hbar/2$ for atoms at locations around one of the partial beams and with certainty $\sigma_z=-\hbar/2$ for atoms at locations around the other partial beam.

Thus just blocking one beam of silver atoms you have prepared a beam of atoms with a certain spin-z component. Of course you through away on average half of the original atoms, but you have prepared a state described by the ray $[|\sigma_z =\hbar/2 \rangle]$, of in the usual slang, a $\sigma_z=+\hbar/2$ state.

This is the paradigmatic example for a von Neumann filter meausurement, and it's effectively somehow something like a "collapse of the state" to an "eigenstate" of $\hat{\sigma}_z$. You should however note, that the relevant piece of the analysis is just using the quantum dynamics to prepare the two partial beams in a (nearly perfectly) $\sigma_z$-position entangled state, and in some sense "the collapse" simply consists in blocking one of the partial beams of Ag atoms out (e.g., by just letting them run on some material). Nowhere we need to envoke some esoteric mechanism outside of the laws of quantum theory to describe this "collapse".

Although the narrative above sounds pretty much like describing the motion of classical particles, all this can be completely described within quantum theory in the approximation where the force according to the $\vec{B}$ field not along the $z$ direction is neglected (and then using simple 1st-order time-dependent perturbation theory to also take into account that the inhomogenous magnetic field cannot be directed only in $z$ direction (because of $\vec{\nabla} \cdot \vec{B}=0$), which leads to (for an appropriately designed magnetic field very small) $\sigma_z=-\hbar/2$ admixture at locations of the $\sigma_z=+\hbar/2$ partial beam).

For a numerical exact evaluation of this, see

https://arxiv.org/abs/quant-ph/0409206

"Applying an observable operator on the current state"

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