What does it mean to influence a quantum measurement's outcome?

[entropy1]

This just occurred to me and I don't expect to be the first one to address it:

It is said that in a specific measurement basis, the outcome of a measurement in this basis is determined by chance.

But in how far is this the case, since if the eigenvectors are for example $\overrightarrow{A}$ and $\overrightarrow{B}$, the choice of outcomes is narrowed to two possibilities we can freely choose?

For example, we can freely choose the measurement basis and thus which outcomes are possible. Isn't that a kind of influence on the measurement outcome?

In the same way you could manipulate the correlation between Alice and Bob in an entanglement experiment.

I am wondering if choice of measurement basis is partly determining the (possible) outcome, and in how far the outcome is random? It seems to me it is partly both!

 

[PeterDonis]

It is said that in a specific measurement basis, the outcome of a measurement in this basis is determined by chance.

It would be nice if you would give a more specific source than "it is said", but this is more or less ok (as long as we adopt a suitable interpretation of "by chance").

how far is this the case, since if the eigenvectors are for example $\overrightarrow{A}$ and $\overrightarrow{B}$, the choice of outcomes is narrowed to two possibilities we can freely choose?

You can't freely choose which of the two eigenvectors the outcome will be. Freely choosing which two eigenvectors are possible outcomes (for the specific case where the measurement has only two possible outcomes) is just choosing the basis. Physically, that means choosing what measurement device you are using and what its settings are. Standard QM does assume that the experimenter can freely choose that, yes. And that is perfectly consistent with your first statement above; I don't see why you would think it wasn't.

we can freely choose the measurement basis and thus which outcomes are possible. Isn't that a kind of influence on the measurement outcome?

Not the kind that is being ruled out in your first statement above.

In the same way you could manipulate the correlation between Alice and Bob in an entanglement experiment.

Which you can't, so I don't know what you are trying to say here.

 

[vanhees71]

Well, let's take the example of a spin measurement for a particle with spin 1 (as an example take the $\omega$ meson). Say we measure the component $s_z$ of the spin. The orthonormal eigenvectors are $|\sigma_z \rangle$ with $\sigma_z \in \{-1,0,1\}$. Suppose further that the particle's spin has been prepared in the state
$$|\psi \rangle=\alpha |s_z=1 \rangle + \beta |s_z=-1 \rangle.$$
We assume the state to be normalized, i.e.,
$$\langle \psi|\psi \rangle=|\alpha|^2+|\beta|^2=1.$$
Now what does QT tell us about the outcome of the measurement of $s_z$? Nothing more than the probabilities for finding each of the three possible outcomes:
$$P(\sigma_z)=|\langle \sigma_z|\psi \rangle|^2,$$
which in our case leads to
$$P(+1)=|\alpha|^2, \quad P(0)=0, \quad P(-1)=|\beta|^2.$$
That's all you can say about the outcome of an $s_z$ measurement. In this case the only thing that's for sure is that you never find $s_z=0$. Whether you get +1 or -1 for an individual measurement you can't say. You only know the probabilities for the outcome being $|\alpha|^2$ and $|\beta|^2$ respectively.

The only influence you have here is the choice of the observable you want to measure and the preparation procedure described by the initial state $|\psi \rangle$.

 

[entropy1]

Thinking out loud:

I think my argument would then become that the outcome is part of the set $\{-1,0,1\}$. So yes, you are right that you can't predict which of the three it will be. But it will be one of the three. It is like a preparation.

Perhaps the similarity to a preparation is what I would consider a bit manipulative. But then the fact that theoretically all three outcomes are possible, is the totally random part.

It is like a single measurement outcome is rounded to a unit, one of the eigenvectors, but an ensemble of identical measurements shows an average that appoaches exactly $|\psi_0 \rangle$. To me this means the measurement is a form of quantisation ("We have rounded off the value for you"). If you look at it that way, the outcomes are randomly picked, but also linked to $|\psi_0 \rangle$.

The strange part is the preparation aspect, that $|\psi \rangle$ gets changed by the measurement along one of the eigenvectors. It is as if this new state is now an element of the approach of $|\psi_0 \rangle$.

I notice I am not getting to a point. I think the point I want to make is that measurement has a strong link to quantization. If something is quantized, it may suggest that the quantified result is the outcome (preparation), but it may also suggest that it is a part of an approach of the outcome ($|\psi_0 \rangle$). Like $|\psi_0 \rangle$ spawns a component of itself.

Well, that was my contribution for now.

 

[Demystifier]

I am wondering if choice of measurement basis is partly determining the (possible) outcome, and in how far the outcome is random? It seems to me it is partly both!

You are right that it is partly both. But the most important part to understand is that in the standard textbook presentation of QM there two very different ways how $|\psi\rangle$ can change:

1) By the Schrodinger equation. This change of $|\psi\rangle$ is determined by the Hamiltonian. This change is unitary, deterministic, predictable, interpretation-independent and non-controversial.

2) By the so-called "collapse": This change of $|\psi\rangle$ is not determined by the Hamiltonian. This change is non-unitary, non-predictable, interpretation-dependent and controversial. In some interpretations it is non-deterministic (e.g. standard textbook QM), while in other interpretations it is deterministic (e.g. Bohmian mechanics).

In particular, the basis in which the measurement is performed is determined by 1), while the actual result that one will obtain in this basis is given by 2).

 

[entropy1]

Yes, in case of 1) you get all components of the state rounded to unit size. In that case the components do not represent $|\psi_0 \rangle$. They represent different worldlines.

In case of 2) you get one component per measurement, and adding them together does represent $|\psi_0 \rangle$. Both views have their issues I think, at least concerning what I know.

 

[vanhees71]

In particular, the basis in which the measurement is performed is determined by 1), while the actual result that one will obtain in this basis is given by 2).

I think, it is misleading to talk about "the basis in which the measurement is performed" though this is usual slang among physicists.

It's clear that you can expand any state in terms of any basis (taking here "state" to mean a vector in Hilbert space representing a pure state). E.g., whether you give a non-relativistic single-particle state in terms of position-spin or momenum-spin representation doesn't change the state, it's simply a different representation. You can simply change the basis by the corresponding representation, i.e.,
$$\psi(\vec{x},\sigma)=\int_{\mathbb{R}^3} \mathrm{d}^3 p \tilde{\psi}(\vec{p},\sigma) \exp(\mathrm{i} \vec{x} \cdot \vec{p}.$$
The state is determined by a preparation procedure.

What's measured is determined by the measurement apparatus you use to measure an observable (or a set of compatible observables in the here discussed most simple (idealized) case of a complete von Neumann measurement). That uniquely determines the probabilities for the outcome of this specific measurement via Born's rule, and it tells you which basis vectors you have to choose to calculate them, namely the (common) complete set of eigenvectors of the corresponding self-adjoint operator(s) representing the (compatible) observables.

 

[entropy1]

To my understanding the chosen basis is relative to the WF measured, right? So you could rotate the WF and have the same/inverse effect of rotating the basis.

Of course you could regard the orientation of the basis part of the measurement setup.

I haven't got experience nor knowledge on much of this.

 

[Demystifier]

To my understanding the chosen basis is relative to the WF measured, right? So you could rotate the WF and have the same/inverse effect of rotating the basis.

Of course you could regard the orientation of the basis part of the measurement setup.

That's all true, but important point is that the "measurement setup" can be described theoretically by an effective Hamiltonian that describes the measuring apparatus and its interaction with the measured system. The effect of this interaction can, to a large extent, be reduced to the theory of decoherence. See e.g. the 3rd lecture in my https://www.physicsforums.com/threads/reading-materials-on-quantum-foundations.963543/post-6299524

 

[vanhees71]

I think we don't need to enter an discussion about this interpretational issues here. We first have to understand the mathematics.

A vector does not depend on any basis, i.e., $|\psi \rangle$ is just a unit vector, and it's fixed. Physically it represents the state (which is strictly speaking represented not by the vector but by the statistical operator of the corresponding pure state $\hat{\rho}=|\psi \rangle \langle \psi|$, but that's also not that important here).

The state of the system is determined by the preparation of the system before the measurement.

Now you want to measure an observable $O$. In quantum mechanics this observable is represented by a self-adjoint operator $\hat{O}$. For simplicity I assume it has a non-degenerate discrete spectrum, i.e., there's a discrete set of eigenvalues $o_n$ and for each eigenvalue the corresponding eigenspace is one-dimensional. Then there's a complete set of orthonormalized eigenvectors, i.e.,
$$\langle o_m|o_n \rangle=\delta_{mn}, \quad \sum_{n} |o_n \rangle \langle o_n|=\hat{1}.$$
These vectors are determined uniquely up to irrelevant phase factors, i.e., they are also independent of any specific basis, but they themselves define a basis, which in this physical situation is preferred, because you measure this specific observable $O$.

Then this implies that for the system being prepared in the state $\hat{\rho}=|\psi \rangle \langle \psi|$ the probability to get the result $o_n$ when measuring $O$ is given by
$$P_n=|\langle o_n|\psi \rangle|^2.$$
This value is obviously independent of the choice of any arbitrary basis. That you have to use the basis of eigenvectors of $\hat{O}$ is rather determined by your choice of which observable you measure, namely $O$. Also note that the probabilities $P_n$ do not depend on the indetermined phases of the eigenvectors $|o_n \rangle$.

 

[entropy1]

That you have to use the basis of eigenvectors of $\hat{O}$ is rather determined by your choice of which observable you measure, namely $O$.

So (if I understand you correctly), you are saying that the (orientation of the) measurement basis comes with the choice of observable measured?

I guess the measurement basis and the observable together define the probabilities of the outcomes. That way you could possibly rightly claim that the measurement basis can be expressed in terms of the probabilities and the observable.

 

[vanhees71]

I do not know, what you mean by "measurement basis". You measure observables, which are represented by self-adjoint operators in Hilbert space. The possible outcomes of a measurment are the eigenvalues of the operator representing the measured observable, and you get the probabilities by Born's rule as described in my previous posting.

The probabilities for one obserable alone do not determine the state uniquely though. How do you come to this conclusion?

 

[entropy1]

I do not know, what you mean by "measurement basis". You measure observables, which are represented by self-adjoint operators in Hilbert space. The possible outcomes of a measurment are the eigenvalues of the operator representing the measured observable, and you get the probabilities by Born's rule as described in my previous posting.

I will have to refresh my QM knowledge by re-reading Susskind. If we consider the operator representing the measurement apparatus: for instance, measuring photon polarisation with a sheet of polariser, where the orientation of the sheet can be varied (the measurement basis).

I suspect that you could also see the setup as a single system.

The probabilities for one obserable alone do not determine the state uniquely though. How do you come to this conclusion?

The projections of $|\psi \rangle$ on the eigenvectors $\langle \psi | e_i \rangle$ (magnitudes) add up to $|\psi \rangle$. So the eigenvectors must be involved too, yes.

I hope that makes sense.

 

[PeterDonis]

So (if I understand you correctly), you are saying that the (orientation of the) measurement basis comes with the choice of observable measured?

Of course. Different orientations for spin measurements are different observables. That's a basic fact about spin measurements.

 

[PeterDonis]

I guess the measurement basis and the observable together define the probabilities of the outcomes.

The measurement basis is part of the definition of the observable. They are not separate, independent things.

 

[entropy1]

Well, now I know for sure I have to re-read Susskind.

 

[entropy1]

Source

The term ‘observable’, while originally applied to the physical quantity of interest, is also applied to the associated Hermitean operator.

Why is an observable also being associated with an operator?
So since I don't talk to physisists IRL, I did not understand this until now.
Perhaps because the operator defines the outcome of the measurement as much as the WF does.
What is the thing that gets measured?

 

[PeterDonis]

Why is an observable also being associated with an operator?

Because that's how you model the observable in the math.

Perhaps because the operator defines the outcome of the measurement as much as the WF does.

Mathematically, you operate on the wave function with the operator that describes the observable in order to find out the possible measurement outcomes and their probabilities.

What is the thing that gets measured?

I'm not sure what you mean by the question. The actual outcome of the measurement will be one of the possible outcomes obtained as above. If you make the same measurement on a large number of identically prepared systems, the statistics of the outcomes will match the probabilities obtained as above.

Have you read any other treatments of basic QM besides Susskind's? If not, you might want to. QM is one of those subjects that everyone has to find their own way to understanding: you should not expect a single source to necessarily match the best way for you to understand QM.

 

[PeroK]

Why is an observable also being associated with an operator?

If you are asking that now, is it any surprise that your original post is confused about the role of eigenbases?

 

[entropy1]

If you are asking that now, is it any surprise that your original post is confused about the role of eigenbases?

Terminology can be a source of confusion, yes.

 

[Demystifier]

Have you read any other treatments of basic QM besides Susskind's? If not, you might want to. QM is one of those subjects that everyone has to find their own way to understanding: you should not expect a single source to necessarily match the best way for you to understand QM.

Perhaps the best advice (on learning QM) ever!

 

[entropy1]

Perhaps the best advice (on learning QM) ever!

Probably, at least for me!

I am doing all learning on my own. And I am handicapped by some dyslexia and fatique and meds, and I want to understand everything I read totally. But nothing brings me as much fun as QM

So, something else I've probably got dead wrong: if we have an electron, it has a single wavefunction that enables us to apply all operators we can think of, to get any thing about it we want to know. Is this correct?

 

[entropy1]

So, something else I've probably got dead wrong: if we have an electron, it has a single wavefunction that enables us to apply all operators we can think of, to get any thing about it we want to know. Is this correct?

* May I bump?? *

 

[PeroK]

* May I bump?? *

In terms of dynamic properties, the wavefunction is all there is. That represents the state of the electron. Whatever you choose to measure, the wavefunction contains the information about the possible measurement results and the probability of getting each result.