# Confusions I have regarding Measurements in Quantum Mechanics

• I
• ARoyC
In summary: Well, if you apply the Operator Mm to |ψ〉, then you simply get Mm |ψ〉. But if you perform the measurement represented by the collection {Mm}, then you get some m, probabilistically. But after m is given, you will always get Mm |ψ〉as new state (or rather Mm |ψ〉/ √(〈ψ|Mm† Mm|ψ〉, because normalized states are more convenient).
ARoyC
TL;DR Summary
What is a measurement in Quantum Mechanics? Is the outcome definite or probabilistic if I apply a particular measurement operator? How is it different from Classical measurements?
Hi.

This is Annwoy Roy Choudhury. I have just completed my first-year undergraduate studies in Physics. I am new to Quantum Mechanics. There are certain confusions I have regarding Quantum Measurements. It would be really kind of you to help me out.

Postulate 3 states,
Quantum measurements are described by a collection {Mm} of measurement operators. These are operators acting on the state space of the system being measured. The index m refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is |ψ〉immediately before the measurement then the probability that result m occurs is given by,
p(m) = 〈ψ|Mm† Mm|ψ〉
and the state of the system after the measurement is
Mm |ψ〉/ √(〈ψ|Mm† Mm|ψ〉 )
An example is,
A measurement on a single qubit with two outcomes defined by the two measurement operators M0 = |0〉〈0| and M1 = |1〉〈1|
Then the probability of obtaining measurement outcome 0 is
p(0) = 〈ψ|M0†M0|ψ〉 = 〈ψ|M0|ψ〉 = |a|^2
Similarly, the probability of obtaining the measurement outcome 1 is p(1) = |b|^2. The state after measurement in the two cases is therefore
M0|ψ〉/ |a| = (a/|a|) |0〉
M1|ψ〉/ |b| = (b/|b|) |1〉
Let's come to my questions.

1. For known |ψ〉, say if I apply a Measurement Operator Mm, then will I always get the projection of |ψ〉on |m〉? Or will it be probabilistic? Can the measurement outcome be a projection on some other basis state? What will I actually get from the measurement? An example would be beneficial.

2. What happens for unknown |ψ〉?

3. What is essentially the difference between Classical Measurements and Quantum Measurements?

Sincere Regards

Annwoy

ARoyC said:
1. For known |ψ〉, say if I apply a Measurement Operator Mm, then will I always get the projection of |ψ〉on |m〉? Or will it be probabilistic?
Well, if you apply the Operator Mm to |ψ〉, then you simply get Mm |ψ〉. But if you perform the measurement represented by the collection {Mm}, then you get some m, probabilistically. But after m is given, you will always get Mm |ψ〉as new state (or rather Mm |ψ〉/ √(〈ψ|Mm† Mm|ψ〉, because normalized states are more convenient).
ARoyC said:
Can the measurement outcome be a projection on some other basis state? What will I actually get from the measurement? An example would be beneficial.
The measurement outcome itself is just m. Even so the state after the measurement often depends on m, it is not the measurement outcome. A measurement's primary purpose is to give you information about your system just before the measurement. The other case (information about your system just after the procedure) would rather be called a preparation.

ARoyC said:
2. What happens for unknown |ψ〉?
There is a density matrix formalism that allows you to describe how much or how little you know about |ψ〉. It also just gives you probabilities for the different outcomes m, and how the density matrix after you got the measurement result m will have changed.

ARoyC said:
3. What is essentially the difference between Classical Measurements and Quantum Measurements?
An idealized classical measurement can be arbitrarily accurate, and have negligible impact on the measured system. An idealized quantum measurement is described by a set of postulates like the ones you gave, which is limited both in its ability to exactly determine |ψ〉, as well as in its ability to leave |ψ〉unchanged.

topsquark
gentzen said:
Well, if you apply the Operator Mm to |ψ〉, then you simply get Mm |ψ〉. But if you perform the measurement represented by the collection {Mm}, then you get some m, probabilistically. But after m is given, you will always get Mm |ψ〉as new state (or rather Mm |ψ〉/ √(〈ψ|Mm† Mm|ψ〉, because normalized states are more convenient).

The measurement outcome itself is just m. Even so the state after the measurement often depends on m, it is not the measurement outcome. A measurement's primary purpose is to give you information about your system just before the measurement. The other case (information about your system just after the procedure) would rather be called a preparation.
Thank you. Will it be possible for you to give an example? Like what an experimentalist would do during a measurement step-by-step (mathematically, not experimental nitty-gritty)?

There is a density matrix formalism that allows you to describe how much or how little you know about |ψ〉. It also just gives you probabilities for the different outcomes m, and how the density matrix after you got the measurement result m will have changed.
Then is the postulate only for known |ψ〉?

An idealized classical measurement can be arbitrarily accurate, and have negligible impact on the measured system. An idealized quantum measurement is described by a set of postulates like the ones you gave, which is limited both in its ability to exactly determine |ψ〉, as well as in its ability to leave |ψ〉unchanged.
Could you please elaborate a bit more?

ARoyC said:
1. For known |ψ〉, say if I apply a Measurement Operator Mm, then will I always get the projection of |ψ〉on |m〉?
No.

ARoyC said:
Or will it be probabilistic?
Yes.

ARoyC said:
Can the measurement outcome be a projection on some other basis state?
Yes

ARoyC said:
What will I actually get from the measurement?
One of the states from this other basis.

ARoyC said:
An example would be beneficial.
You can measure spin in direction ##x##, or you can measure spin in direction ##y##.

ARoyC said:
2. What happens for unknown |ψ〉?
You still know the basis in which you measure, so the result will be one of the states from this basis. But you cannot compute the probability of each possible state in the basis.

ARoyC said:
3. What is essentially the difference between Classical Measurements and Quantum Measurements?
Contextuality, namely the fact that quantum measurement changes the properties of the measured system. For example, suppose that you measure the spin of the same particle twice. You can first measure spin in ##x##-direction, and then in ##y##-direction. Or you can measure in ##y##-direction twice. The results of the second measurement in ##y##-direction will be different in the two cases.

ARoyC said:
Thank you. Will it be possible for you to give an example? Like what an experimentalist would do during a measurement step-by-step (mathematically, not experimental nitty-gritty)?
The double-slit experiment and the Stern-Gerlach experiment are two standard textbook examples for quantum measurements. Not sure what you mean by "mathematically step-by-step" vs "experimental nitty-gritty".
You could use a CCD detector with a given spatial and temporal resolution, and a well controlled laser light source for the double-slit experiment. Then the different pixels of the detector would correspond to m, and the time resolution would allow you to derive the expected number of photons from the characteristics of the used laser light source. Plug this expected number as parameter ##\lambda## into the Poisson distribution to get the distribution of the number of photons, and use this together with the predictions from the measurement postulate to predict a statistics of measured images the can be compared with the images actually recorded by the CCD detector. (It is possible with current technology to get the expected number of photons below 1, but then you also need a very good shielding against background noise, which is of course no problem either.)
Using a CCD detector for the Stern-Gerlach experiment is not such a great idea, because the silver atoms would make sensitive surfaces dirty and quickly degrade your CCD detector. So you better just use some film where the silver atoms can deposit themselves. If you want some temporal resolution, then you could move the film at an appropriate speed in the appropriate direction, or use "fast blankers" and a step-move strategy for the film.
Even so the described experiments could be done in the described form, the experiments themselves are a bit too trivial to justify this effort. But related less trivial experiments (or rather measurements) are sometimes done in such ways.

ARoyC said:
Then is the postulate only for known |ψ〉?
Well, the effect on the density matrix reduces to the effect on |ψ〉via an Eigen-decomposition of the density matrix.

ARoyC said:
Could you please elaborate a bit more?
What exactly would you like to have elaborated? The notion and role of an "idealized measurement"? The notion and role of "a set of postulates"? The notion and role of being "limited in its ability"?

dextercioby
ARoyC said:
Thank you. Will it be possible for you to give an example? Like what an experimentalist would do during a measurement step-by-step (mathematically, not experimental nitty-gritty)?
And here are examples from a Quantum Chemistry textbook, which describe measurements that can be done both in theory and practice, but which are not related in any simple way to typical sets of postulates for quantum measurement.

And to complement those non-examples of quantum measurements, let me also quote some of my opinions on measurement vs preparation. In Jul 2021:
gentzen said:
The point is that you don't need the projection postulate (and probably not even the Born rule in any form whatsoever) to derive the statistics of the thermal state or prepare a system in it. To prepare a system in it, you probably just have to control the thermodynamical degrees of freedom and keep them constant long enough for the system to settle/converge sufficiently to the thermal state. So you can get away with the much weaker assumptions that the thermodynamical degrees of freedom can be controlled and measured. At least I guess that this assumption is weaker than assuming the existence of a classical description plus some version of the Born rule.
In Feb 2023:
gentzen said:
My answer would be that preparation does not just determine the initial state, but also the Hamiltonian.

And in general, I guess that the failure to distinguish between preparation and measurement is responsible for some of the confusion with QM and its interpretation. Using measurement to emulate preparation seems so convenient and straightforward, just like an additional quantum symmetry. But you risk a totally unnecessary circularity in this way.
gentzen said:
Well, the quest for unification is not my quest. I guess my quest is just to be able to communicate (about physics), without too much appeal to authority.

I sort of get why preparation and measurement are closely related. For example, if I prepare atoms by shielding them and waiting long enough until they nearly all relaxed to their ground state, then I know that they are in their ground state. And because I know it, I can claim that I somehow measured it, because what else is measurement than knowing some specific properties. But ... I would prefer to measure properties which were there before I measured, and prepare states which will be there after I prepared.

ARoyC said:
TL;DR Summary: What is a measurement in Quantum Mechanics? Is the outcome definite or probabilistic if I apply a particular measurement operator? How is it different from Classical measurements?

Hi.

This is Annwoy Roy Choudhury. I have just completed my first-year undergraduate studies in Physics. I am new to Quantum Mechanics. There are certain confusions I have regarding Quantum Measurements. It would be really kind of you to help me out.

Postulate 3 states,

An example is,

Let's come to my questions.

1. For known |ψ〉, say if I apply a Measurement Operator Mm, then will I always get the projection of |ψ〉on |m〉? Or will it be probabilistic? Can the measurement outcome be a projection on some other basis state? What will I actually get from the measurement? An example would be beneficial.
The formulation of the postulate is a bit inaccurate. It's not a measurement that's described by a self-adjoint operator in Hilbert space but an observable. A set of observables ##A_j## is called compatible, if the self-adjoint operators, ##\hat{A}_j##, representing them are commuting, and it's called complete, if the then existing subspaces of common eigenvectors for each combination of possible eigenvalues ##(a_1,\ldots,a_n)## is one-dimensional. It's called a minimal complete set of compatible observables, if none of the corresponding operators is a function of the others.

Then, if the system is prepared in a pure state, described by a normalized vector ##|\psi \rangle##, the probability for finding the values ##(a_1,\ldots,a_n)## when measuring a minimal complete set of observables ##A_j## is given by
$$P(a_1,\ldots,a_n = |\langle a_1,\ldots,a_n|\psi \rangle|^2.$$

Whether or not the system is projected to the corresponding state ##|a_1,\ldots,a_n \rangle##, depends on the specific way you do the measurement. This kind of measurements is called a "von Neumann filter measurement". Often it's not realized in standard ways to measure things. E.g., if you register a photon, usually it's absorbed by the detector.

The correct statement of the postulate is that if the system is prepared in a pure state, described by
ARoyC said:
2. What happens for unknown |ψ〉?
If you don't know ##|\psi \rangle## you cannot calculate the probabilities in this way, of course.
ARoyC said:
3. What is essentially the difference between Classical Measurements and Quantum Measurements?
None. A measurement is performed with some apparatus in the lab.

The difference is in the meaning of the state. In classical mechanics all observables always take determined values, and measuring an observable just gets this value of the measured observable. In principle the value of the observable is determined even before the measurement.

In quantum theory the complete determination of the state, i.e., the preparation of the system in a pure quantum state, implies only the probabilities for the outcome of measurements (for a complete minimal set of observables, as described above) are known. Before the measurement the values of observables in general are not determined (that's the case only if the state vector ##|\psi \rangle## is an eigenvector of the corresponding observable).

ARoyC said:
3. What is essentially the difference between Classical Measurements and Quantum Measurements?
The thing to realize is that measurement always disturbs the system, but in the macroscopic world you have two main ways to handle it.

The first way is to say that the act of measuring is negligible, think of a thermometer in a pool. Yes, technically, the thermometer disturbs the system, but it's such a small percentage, we can just say that what temperature we measure is the true value for the pool.

The second way is to say that yes, the act of measurement disturbs the system a lot, but we can account for it. Think of mixing two chemicals together. We know doing such a thing is very chaotic, but, we can account for what happens, and we can predict the result correctly.

So in both cases, we acknowledge that the measurement will disturb our system, but we can still predict the individual results correctly.

This is not true for quanta. When we do experiments that deal with quanta, the first ideal flies out the window. The measurement process here is not negligible at all. The second case you may think saves us. Unfortunately, due to doing experiments with quanta, the act of measurement results in individual events being random.

Random isn't really "random" in the sense that it can be ANY value. Electron spin [from the stern-gerlach experiement] can only be 2 values, and this is where quantum shines. We can't know prior to measurement what spin our electron has, but we CAN predict their statistical distributions! So in the quantum world, you throw out the value of predicting individual events, and move onto predicting the average outcome.

TL;DR: Macroscopic measurements exists in a deterministic, causal world. Knowing the state at one time lets me predict any physical phenomenon associated with the state, which includes individual events. Microscopic measurement exists in a statistically deterministic, causal world. Knowing the state at one time lets me predict the average outcome of any physical phenomenon associated with the state, which means I can't predict individual events.

vanhees71
Demystifier said:
No.Yes.YesOne of the states from this other basis.You can measure spin in direction ##x##, or you can measure spin in direction ##y##.You still know the basis in which you measure, so the result will be one of the states from this basis. But you cannot compute the probability of each possible state in the basis.Contextuality, namely the fact that quantum measurement changes the properties of the measured system. For example, suppose that you measure the spin of the same particle twice. You can first measure spin in ##x##-direction, and then in ##y##-direction. Or you can measure in ##y##-direction twice. The results of the second measurement in ##y##-direction will be different in the two cases.
Thank you. I got it.

Demystifier
gentzen said:
And here are examples from a Quantum Chemistry textbook, which describe measurements that can be done both in theory and practice, but which are not related in any simple way to typical sets of postulates for quantum measurement.

And to complement those non-examples of quantum measurements, let me also quote some of my opinions on measurement vs preparation. In Jul 2021:

In Feb 2023:
Thanks for the insight.

vanhees71 said:
The formulation of the postulate is a bit inaccurate. It's not a measurement that's described by a self-adjoint operator in Hilbert space but an observable. A set of observables ##A_j## is called compatible, if the self-adjoint operators, ##\hat{A}_j##, representing them are commuting, and it's called complete, if the then existing subspaces of common eigenvectors for each combination of possible eigenvalues ##(a_1,\ldots,a_n)## is one-dimensional. It's called a minimal complete set of compatible observables, if none of the corresponding operators is a function of the others.

Then, if the system is prepared in a pure state, described by a normalized vector ##|\psi \rangle##, the probability for finding the values ##(a_1,\ldots,a_n)## when measuring a minimal complete set of observables ##A_j## is given by
$$P(a_1,\ldots,a_n = |\langle a_1,\ldots,a_n|\psi \rangle|^2.$$

Whether or not the system is projected to the corresponding state ##|a_1,\ldots,a_n \rangle##, depends on the specific way you do the measurement. This kind of measurements is called a "von Neumann filter measurement". Often it's not realized in standard ways to measure things. E.g., if you register a photon, usually it's absorbed by the detector.

The correct statement of the postulate is that if the system is prepared in a pure state, described by

If you don't know ##|\psi \rangle## you cannot calculate the probabilities in this way, of course.

None. A measurement is performed with some apparatus in the lab.

The difference is in the meaning of the state. In classical mechanics all observables always take determined values, and measuring an observable just gets this value of the measured observable. In principle the value of the observable is determined even before the measurement.

In quantum theory the complete determination of the state, i.e., the preparation of the system in a pure quantum state, implies only the probabilities for the outcome of measurements (for a complete minimal set of observables, as described above) are known. Before the measurement the values of observables in general are not determined (that's the case only if the state vector ##|\psi \rangle## is an eigenvector of the corresponding observable).
Thank you.

romsofia said:
The thing to realize is that measurement always disturbs the system, but in the macroscopic world you have two main ways to handle it.

The first way is to say that the act of measuring is negligible, think of a thermometer in a pool. Yes, technically, the thermometer disturbs the system, but it's such a small percentage, we can just say that what temperature we measure is the true value for the pool.

The second way is to say that yes, the act of measurement disturbs the system a lot, but we can account for it. Think of mixing two chemicals together. We know doing such a thing is very chaotic, but, we can account for what happens, and we can predict the result correctly.

So in both cases, we acknowledge that the measurement will disturb our system, but we can still predict the individual results correctly.

This is not true for quanta. When we do experiments that deal with quanta, the first ideal flies out the window. The measurement process here is not negligible at all. The second case you may think saves us. Unfortunately, due to doing experiments with quanta, the act of measurement results in individual events being random.

Random isn't really "random" in the sense that it can be ANY value. Electron spin [from the stern-gerlach experiement] can only be 2 values, and this is where quantum shines. We can't know prior to measurement what spin our electron has, but we CAN predict their statistical distributions! So in the quantum world, you throw out the value of predicting individual events, and move onto predicting the average outcome.

TL;DR: Macroscopic measurements exists in a deterministic, causal world. Knowing the state at one time lets me predict any physical phenomenon associated with the state, which includes individual events. Microscopic measurement exists in a statistically deterministic, causal world. Knowing the state at one time lets me predict the average outcome of any physical phenomenon associated with the state, which means I can't predict individual events.
Thank you for the explanation.

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