# Measurement and preparation

1. Feb 8, 2016

### naima

Is there a difference between measurement and preparation?
When we have two Young slits we prepare a superposition of position. The apparatus which can measure this state is a copy of these slits near them with detectors outside them. If they never click you have build the correct measurement set up.
So measurement would be filtering.

2. Feb 8, 2016

3. Feb 8, 2016

### A. Neumaier

Yes, there is a very big difference.

In a preparation, you create a source producing an ensemble of quantum systems behaving in a prescribed quantum statistical manner, described by its quantum state (density matrix). You measure nothing.

In a measurement, you read some numbers or events from a macroscopic device. The measured object may or may not be destroyed in the process, but what remains of it after the measurement is not of interest.

A filter transforms a source into another source, thereby changing its state and preparing a new ensemble. There are two main kinds of filters. A filter may be unitary; in this case it just transforms the input source into an output source without leaving any trace in the environment. (Example: an ideal rotator for a photon beam.) Or a filter is dissipative. In this case it leaves an in principle measurable trace in the environment and transforms the input source into an output source that depends on what was deposited in the environment. Thus a dissipative filter involves also a measurement. (Examples: a polarizer or a double slit.) There are also other filters that consume energy to operate properly.

In quantum experiments that involve multiple times, each intermediate time where some classical object is passed acts as a filter, and depending on whether the filter leaves information in the environment one can read off a measurement from it. The state of the input source is transformed into an output state, and this output state is a (partially) collapsed version of the input state. In sufficiently idealized cases, it is a collapse of the form postulated in the Copenhagen interpretation.

Last edited: Feb 8, 2016
4. Feb 8, 2016

### naima

A screen with two slits is used to prepare states. It may be used as an apparatus to measure the prepared state. You put it behind the original slits. If it has the correct position and the correct distance between the slits (this is your number)
you can have an electronic device which clicks when the particle have passed through the 2 screens.
when you have prepared a state you can describe it to an experimentalist . He will build a set up which will give him a known output (eigen value).
What is for you the set up in the cas of Young slits?

5. Feb 8, 2016

### A. Neumaier

How do you use the screen to measure the prepared state??
Where do you get the measurement results from??

The distance between the slits is not a measurement of a state. It is the same number no matter whether something or nothing goes through the slit. How can it measure the state?

The way you set up the experiment, the second screen is just a dissipative filter that changes the input state (a spherical wave centered at the first slit, or a superposition of several if there were multiple slits in the first screen) into the output state (a spherical wave centered at the second slit), with a loss in intensity proportional to the energy absorbed. The amount of energy absorbed depends on where you put the second screen, hence the energy absorbed indirectly measures the energy density of the input field.

The amount of energy absorbed is the only thing one could measure if one would try hard enough, though you don't in your setting.

Eigenvalues don't play a role in the whole argument.

6. Feb 8, 2016

### naima

I can understand you when say that my setup is not an measurement apparatus. What is yours?

7. Feb 8, 2016

### A. Neumaier

I didn't describe a particular setup, but explained the difference between a preparation and a measurement, as answer to the question you had posed in post #186. My explanation applies to any experimental setting in which states are prepared, modified, or measured in a sequential fashion.

8. Feb 8, 2016

### naima

I repeat my question: do you know an apparatus which could give an absolutely certain result, knowing the distance between the slits, the energy and so on?

What i said is that just like measurement can be repeated, the preparation setup can be used to verify that it select twice the same state and that it acts just like a measurement device. A measurement apparatus is not always something with a ground state which interact withe the thing to be measured and jump to the output.
Take a steelyard balance. It gives you the output once you have displaced the weight at the correct position. That is what i did with my second screen with variable distances and position. I have to put it so that it reproduces the caracteristics of the initial preparation device.

9. Feb 8, 2016

### A. Neumaier

I never needed to know such an apparatus. I am not a friend of highly idealized settings.

And I know that the state of any sufficiently simple stationary source can be determined arbitrary well by appropriate measurements. $n^2-1$ of them suffice for an $n$ level state, if the measurements are carefully chosen and ideally performed. So why should I bother about a very specific instance?

10. Feb 8, 2016

### naima

Why have we measurement devices in QM? The theory tells us that there are operators on a hilbert space. An hermitian operator has eigenvectors with eigenvalues. Such an eigenvector is something which can be measured. ONE measurement with a chosen measurement apparatus will give you the eigen value. If you have a given spin in a direction you will verify immediatly its value by measuring it along that direction. What you say is that it is enough to have multiple measurements along x and y and z. It is not the way i see things.

11. Feb 8, 2016

### A. Neumaier

In orthodox quantum mechanics we don't have measurement devices, only quantum systems. They are prepared and measured by classical devices, whose behavior is outside the scope of QM. Indeed, once a measurement device is treated as a quantum system, it immediately loses its status as a measurement device and becomes something that itself must be measured to find anything out about it.

Measuring the state of a quantum source is a standard procedure in quantum detection and estimation theory. The density matrix has $n^2-1$ independent degrees of freedom, and it is not difficult to find this many independent observables such that, from the measurement of their mean (by repeated observation of individual systems emitted by the source), the density matrix can be reconstructed by solving a linear system of equations. That Wikipedia doesn't know about this doesn't change the fact that this is a standard result. No eigenvalues are needed to understand or perform these computations.

12. Feb 8, 2016

### naima

What i say is that a prepared system has an answer to one question. If you ask the good question you will always get the same good answer. If you ask n^2 - 1
"bad" guestion you will get random answers and you will need mean values, changes of basis and the algebraic machinary (that i know).

13. Feb 8, 2016

### drvrm

At the moment our best description is that the electron/microparticle is an excitation of a quantum field. Using quantum field theory allows us to calculate the behaviour of electrons whether they happen to be involved in particle-like or wave-like interactions. This doesn't mean that the electron is a quantum field, and we may almost certainly replace quantum field theory by some even more complicated e.g. some future development of string theory.If one wishes to describe electron as in Bohr's theory then a particle representation demands a collapse of the wave function-actually the representation of the reality demands the collapse-as the measured particle must be found to be there.
as all experiments are being performed in a classical world which is 'deterministic' leads to such formalism-and the question about collapse can be answered only by an intelligent system which knows 'its being measured'- there is opinion in general the collapse is related to 'decoherence' but what is 'decoherence'

14. Feb 8, 2016

### A. Neumaier

Only if it is not used for anything useful. In practice, systems are prepared because they deliver many similar items that can be used in many different ways, not just to answer one question where the answer is known beforehand.

15. Feb 8, 2016

### drvrm

The job of decoherence is to bring a quantum system into an apparently classical state. What especially differentiates a quantum system from a classical system is the concept of a superposition of states. In the classical realm of physics, we would say that a particle is at a position (x,y,z). In the quantum realm, the formalism allows us to state that a particle is in a superposition of positions (say, and [PLAIN]http://www.physics.drexel.edu/~tim/open/main/img7.png). [Broken] However, as the canon of quantum mechanics postulates, when we actually measure the position of this particle, we will find it at either of the two positions, that is, we will have collapsed the wave function''into one or the other state. A classical mind (which we all have) would ask, what does this mean? This is the so called measurement problem''.
Decoherence does not generate actual wave function collapse. It only provides an explanation for the observation of wave function collapse, as the quantum nature of the system "leaks" into the environment. That is, components of the wavefunction are decoupled from a coherent system, and acquire phases from their immediate surroundings. A total superposition of the global or universal wavefunction still exists (and remains coherent at the global level), but its ultimate fate remains aninterpretational issue. Specifically, decoherence does not attempt to explain themeasurement problem. Rather, decoherence provides an explanation for the transition of the system to a mixture of states that seem to correspond to those states observers perceive. Moreover, our observation tells us that this mixture looks like a properquantum ensemble in a measurement situation, as we observe that measurements lead to the "realization" of precisely one state in the "ensemble".

Last edited by a moderator: May 7, 2017
16. Feb 9, 2016

### gxu

Please excuse me for so abruptly jumping in - simply because I found this thread just now. Isn't this type of unitary operator based model essentially along the same line of the decoherence theory? So eventually the system is unitarily transformed onto the preferred orthogonal basis, giving all the same predictions as by applying the mixed state density.

17. Feb 9, 2016

### gxu

My understanding is that, the decoherence theory resolved the "preferred-basis problem" (probably for most physicists), but not the "definite-outcome problem" (presumably the essence of collapse problem). Here is a reference of the comprehensive discussion on this subject:

Decoherence, the measurement problem, and interpretations of quantum mechanics, Maximilian Schlosshauer, 2003

where Schlosshauer in his concluded remark stated that, "within the standard interpretation of quantum mechanics, decoherence cannot solve the problem of definite outcomes in quantum measurement: We are still left with a multitude of (albeit individually well-localized quasiclassical) components of the wave function, and we need to supplement or otherwise to interpret this situation in order to explain why and how single outcomes are perceived."

Last edited by a moderator: May 7, 2017
18. Feb 9, 2016

### drvrm

<decoherence cannot solve the problem of definite outcomes in quantum measurement:>......gsu ...i agree with you and will look up the reference given by you ..in my submission also this point has been raised!

19. Feb 9, 2016

### stevendaryl

Staff Emeritus
Just for clarification: Decoherence solves the preferred basis problem only after we've split the universe up into the system of interest plus the environment. Tracing out the environmental degrees of freedom leaves a mixed state for the system of interest.

20. Feb 9, 2016

### zonde

I don't think that Schlosshauer's approach is consistent.
Right at the start he quotes Zurek:
"The idea that the “openness” of quantum systems might have anything to do with the transition from quantum to classical was ignored for a very long time, probably because in classical physics problems of fundamental importance were always settled in isolated systems."
But later on he forgets about that idea (maybe it's not exactly his fault) and considers a model where we can speak about well defined Hilbert space of environment and well defined Hilbert space of measurement apparatus $H_E \otimes H_A$. If the measurement apparatus is open quantum system then the split between environment and system is floating, say at time $t_0$ the split is $H_{E0} \otimes H_{A0}$ and at a later time $t_1$ the split is $H_{E1} \otimes H_{A1}$ . And $H_{A1}$ compared to $H_{A0}$ is the same from classical perspective but quite different from QM perspective.