Quantum interference and superposition

  • #1
3,507
26
I would like to have a better grasp of the relation between quantum interference and state superposition that are linked in QM. For instance in the photon double-slit experiment the interference pattern is seen as a consequence of the superposition of the wave function(pure state), and when there is knowledge of wich slit must have been traversed i.e. by putting a detector the interference pattern is supposed to disappear switching to what one would expect in the case with defined trajectories without interference. But one slit experiments also give interference patterns if dimmer, so what seems to be happening is that the initial strong interference pattern is substituted by 2 one slit very weak diffraction patterns, wich makes sense if we think the interaction with the detector lowers the intensity of the pattern.
In short, instead of interference versus no interference as complementarity would demand, we can also interpret the outcome of the experiment as double-slit interference pattern vs. two adjacent single-slit degraded interferences patterns.
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,857
1,655
  • Like
Likes bhobba
  • #3
Matterwave
Science Advisor
Gold Member
3,965
326
Was there a question in there somewhere? Or are you just trying to make a point about something?
 
  • #4
130
0
I'd like to ask a question and don't want to start a new thread, so let me ask it here in this inactive thread of a related view..

What is wrong with the idea that the reason the particle tries out all paths in the double slit experiment is because fundamentally, there is no space (in the quantum world). Is there any serious work with regards to this and what would be the problems of this view?
 
  • #5
9,596
2,676
What is wrong with the idea that the reason the particle tries out all paths in the double slit experiment is because fundamentally, there is no space (in the quantum world).
Why do you think that no space (whatever that is supposed to mean) implies that?

For it to be an actual reason you would need to define the idea rigorously and show, again rigorously, Feynmans sum over histories follows. Just one of the things you will need to deduce is why it requires complex numbers - which is the reason the idea works to give part cancellation for most paths.

Thanks
Bill
 
  • #6
130
0
Why do you think that no space (whatever that is supposed to mean) implies that?

For it to be an actual reason you would need to define the idea rigorously and show, again rigorously, Feynmans sum over histories follows. Just one of the things you will need to deduce is why it requires complex numbers - which is the reason the idea works to give part cancellation for most paths.

Thanks
Bill
QM has complex numbers. General Relativity has complex numbers too.. what do their complex numbers have in common? Are their complex numbers fundamentally compatible in the calculations?
 
  • #7
9,596
2,676
General Relativity has complex numbers too
Here complex numbers means the following:
http://en.wikipedia.org/wiki/Complex_number

The question is how does your no space idea imply that?

That is even leaving aside the issue of what a path is, which resides in space, and what it can mean without space.

The path integral approach assigns a complex number to each point in the path and that number changes with time.

Thanks
Bill
 
Last edited:
  • #8
3,507
26
Was there a question in there somewhere? Or are you just trying to make a point about something?
[Sorry for the late reply, busy week-end]
I was trying to make the point that if the OP's description of interference is basically correct and well known (as confirmed in post #2), IOW if quantum interference-superposition is fundamentally different from classical wave interference-superposition, then why is the mathematical formalism the same concerning states as vectors and their space a vector space?
Quoting from Dirac's QM he basically warns against the analogies one can derive from this: "It is important to remember, however, that the superposition that occurs in quantum mechanics is of an essentially different nature from any occurring in the classical theory, as is shown by the fact that the quantum superposition principle demands indeterminacy in the results of observations in order to be capable of a sensible physical interpretation. The analogies are thus liable to be misleading."
But then in the next paragraph he goes on to use the mathematical formalism of the superposition principle he doesn't want to be confused with the quantum one anyway. i,e. : "We shall begin to set up the scheme by dealing with the mathematical relations between the states of a dynamical system at one instant of time, which relations will come from the mathematical formulation of the principle of superposition. The superposition process is a kind of additive process and implies that states can in some way be added to give new states. The states must therefore be connected with mathematical quantities of a kind which can be added together to give other quantities of the same kind. The most obvious of such quantities are vectors."
It may be the most obvious if one doesn't care much about misleading analogies but if "the quantum superposition principle demands indeterminacy in the results of observations in order to be capable of a sensible physical interpretation" due to observables not commuting (Heisenberg indeterminacy principle) and that makes superposition essentially different, then vectors don't seem the most obvious quantities to choose as states, do they? At least for dynamical systems.

In short, vectors are invariants in the mathematical formalism, states are not, due to Heisenberg indeterminacy demanding a preferred basis for anything having to do with observables (that's everything regarding physical systems) i.e. involving measurements or state preparations.

.
 
  • #9
atyy
Science Advisor
14,309
2,546
[Sorry for the late reply, busy week-end]
I was trying to make the point that if the OP's description of interference is basically correct and well known (as confirmed in post #2), IOW if quantum interference-superposition is fundamentally different from classical wave interference-superposition, then why is the mathematical formalism the same concerning states as vectors and their space a vector space?
The superposition is indeed mathematically analogous to the classical wave equation. However, the big difference is the probability interpretation. Here let me go with the orthodox Copenhagen-style interpretation, where a pure state is a ray in Hilbert space, and is the complete state of a single particle.

What in probability terms is a pure state? To understand it, let's go over to the density matrix formalism and compare pure and mixed ensembles. A pure ensemble is one in which every particle is in the same pure state. A mixed ensemble is one in which some fraction of the particles are in a particular pure state, while another fraction is in another pure state. So the key idea in a pure ensemble is that it is *not* a statistical mixture, and that every particle in a pure ensemble is identical to the limits of the theory.

This idea that a pure state is not a statistical mixture holds in both classical and quantum theory. The difference is that a classical pure state is not a superposition of other pure states, while quantum pure states are superpositions of other pure states. So the classical pure state is completely pure, while the quantum pure state, although not a statistical mixture is still "impure" in the sense that it is a superposition.

OK, above I distinguished pure state and pure ensemble, but they are so closely related that a we do not usually use different words for them, and I will now use the terms interchangeably. In technical terms you will find it said that the state space (here referring to the density matrix formalism) has pure states that are extremal points of a simplex in the classical case, but not the quantum case. This is why cartoons of classical state spaces are polygons, while the quantum state space is usually shown as a circle or something without pointy edges, eg. http://arxiv.org/abs/1303.2849 (Fig. 2, actually this is not the state space, but the correlations, but the idea is similar).
 
Last edited:
  • #10
3,507
26
The superposition is indeed mathematically analogous to the classical wave equation. However, the big difference is the probability interpretation.
The probability interpretation is different yes, but that distinction comes in a later stage, after the mathematical formulation is chosen. My question was why choose a mathematical analogy that in the words of Dirac is liable to be misleading. Yes, in practical terms it leads to a very good approximation to what we observe if one gets around the fact that it only gives probabilities in the dynamical picture. But on the other hand the obvious disconnect between the abstract formalism(basis-independent) and the physical theory(basis-dependent) makes people interested in interpreting the theory go into an endless loop and blocks any putative progress to better approachs.
 
  • #11
atyy
Science Advisor
14,309
2,546
The probability interpretation is different yes, but that distinction comes in a later stage, after the mathematical formulation is chosen. My question was why choose a mathematical analogy that in the words of Dirac is liable to be misleading. Yes, in practical terms it leads to a very good approximation to what we observe if one gets around the fact that it only gives probabilities in the dynamical picture. But on the other hand the obvious disconnect between the abstract formalism(basis-independent) and the physical theory(basis-dependent) makes people interested in interpreting the theory go into an endless loop and blocks any putative progress to better approachs.
Because the mathematical analogy is very good. For many particles the wave is in Hilbert space, but for one particle one can treat it as a wave in ordinary space. And the relationship between position and momentum is exactly the time-frequency uncertainty of classical Fourier analysis. All's fair in love, war and science :)
 
  • #12
3,507
26
Because the mathematical analogy is very good. For many particles the wave is in Hilbert space, but for one particle one can treat it as a wave in ordinary space.
So you would say the main justification is to keep the appearance of classical particles(defined as locally causal bodies as demanded by the distinction one/many)?

All's fair in love, war and science :)
:D
 
  • #13
atyy
Science Advisor
14,309
2,546
So you would say the main justification is to keep the appearance of classical particles(defined as locally causal bodies as demanded by the distinction one/many)?
Actually, I guess I don't know what you are asking. Most posters above, me included, think what you are saying is correct. As long as one agrees on the formalism and the predictions obtained, anyone including Dirac can use whatever partial analogy they find helpful. I was just saying the Scroedinger equation is a linear wave equation, so we can use whatever intuition we have for linear wave equations.
 
  • #14
3,507
26
Actually, I guess I don't know what you are asking. Most posters above, me included, think what you are saying is correct. As long as one agrees on the formalism and the predictions obtained, anyone including Dirac can use whatever partial analogy they find helpful. I was just saying the Scroedinger equation is a linear wave equation, so we can use whatever intuition we have for linear wave equations.
Ok, so I guess it is just me who sees some incongruence in having a physical theory with preferred-basis FAPP described by a math formalism that is basis-independent, wich actually all physical theories ought to be, if the notion of invariance in physics is to have any meaning. For a proof of the lack of invariance of pure states in QM in the dynamical case see http://arxiv.org/abs/1412.2701v1
 
  • #15
atyy
Science Advisor
14,309
2,546
Ok, so I guess it is just me who sees some incongruence in having a physical theory with preferred-basis FAPP described by a math formalism that is basis-independent, wich actually all physical theories ought to be, if the notion of invariance in physics is to have any meaning. For a proof of the lack of invariance of pure states in QM in the dynamical case see http://arxiv.org/abs/1412.2701v1
The preferred basis is put in by hand in quantum theory when a measurement is made, and is due to the free choice that the observer has to perform whatever measurement he wants. It can be argued that decoherence picks a preferred basis, but one still needs a measurement to get definite outcomes, ie. make the preferred basis a reality. So it is due to the classical/quantum cut, with the observer being not in the system, so to speak. If one removes the classical/qauntum cut, say using Bohmian Mechanics, one needs hidden variables. But the hidden variables are more complex eg. http://arxiv.org/abs/0711.4770, so it is (almost always) easier to calculate without hidden variables, and the quantum formalism is a terrific effective theory from this point of view.
 
  • Like
Likes vanhees71
  • #16
Simon Bridge
Science Advisor
Homework Helper
17,857
1,655
Footnote... the schrodinger equation is a special case of the helmholtz equation. This is why the analogy works.
Its not all that uncommon to have different physical effects modelled with the same equation.

You may like Marcella "quantum interference at slits" and critique by Rothman I think.
The second has some insights to your question, but you need the first for context... Marcella tries to avoid the analogous approach you, and many others, find problematic.
 
  • #17
3,507
26
The preferred basis is put in by hand in quantum theory when a measurement is made, and is due to the free choice that the observer has to perform whatever measurement he wants. It can be argued that decoherence picks a preferred basis, but one still needs a measurement to get definite outcomes, ie. make the preferred basis a reality.
Preparation procedures already pick up a preferred basis, just picking a representation space picks a preferred basis, and sure the observer is free to choose the observable and thus the basis, but that choice makes a difference wrt what is observed(observable by definition), due to the commuting relations the order of the observables matter, this doesn't happen in a truly basis-independent case in wich one can also pick a basis to represent a vector but that choice has no consequences at all wrt any outcomes(true invariance wrt basis).
So it is due to the classical/quantum cut, with the observer being not in the system, so to speak. If one removes the classical/qauntum cut, say using Bohmian Mechanics, one needs hidden variables. But the hidden variables are more complex eg. http://arxiv.org/abs/0711.4770, so it is (almost always) easier to calculate without hidden variables, and the quantum formalism is a terrific effective theory from this point of view.
If it is due to the cut it should be removed from interpretations with no cut(like BM), but BM has preferred basis, so that example seems to counter your argument, as Demystifier showed in the "QM thrashed by non-physicists" thread the preferred basis issue is interpretation independent.
 
  • #18
atyy
Science Advisor
14,309
2,546
If it is due to the cut it should be removed from interpretations with no cut(like BM), but BM has preferred basis, so that example seems to counter your argument, as Demystifier showed in the "QM thrashed by non-physicists" thread the preferred basis issue is interpretation independent.
Did you mean BM has a preferred basis or has no preferred basis?
 
  • #19
3,507
26
Footnote... the schrodinger equation is a special case of the helmholtz equation. This is why the analogy works.
Its not all that uncommon to have different physical effects modelled with the same equation.
Yes, I know why the analogy works, I'm just pointing out obvious problems of going to a special case, maybe going to a mathematically more general case could have been an option too.

You may like Marcella "quantum interference at slits" and critique by Rothman I think.
The second has some insights to your question, but you need the first for context... Marcella tries to avoid the analogous approach you, and many others, find problematic.
I'll take a look at this.
 
  • #20
3,507
26
Did you mean BM has a preferred basis or has no preferred basis?
Has position preferred basis.
 
  • #21
atyy
Science Advisor
14,309
2,546
Has position preferred basis.
Yes, that's what I think too. I thought you were complaining that the usual quantum formalism has no preferred basis until a measurement is done?
 
  • #22
3,507
26
Yes, that's what I think too.
But you said the preferred basis was due to the Heisenberg cut and that BM had no cut.

I thought you were complaining that the usual quantum formalism has no preferred basis until a measurement is done?
No, QM has preferred basis inherently due to observables non-commuting, or seen from the complementary perspective due to its quantum nature that identifies pure states with rays in Hilbert space. The abstract mathematical formulation, for instant Dirac's bra-ket notation hides this fact.
 
  • #23
atyy
Science Advisor
14,309
2,546
But you said the preferred basis was due to the Heisenberg cut and that BM had no cut.
No, QM has preferred basis inherently due to observables non-commuting, or seen from the complementary perspective due to its quantum nature that identifies pure states with rays in Hilbert space. The abstract mathematical formulation, for instant Dirac's bra-ket notation hides this fact.
Let's see if I can be more coherent. The state as a ray and unitary evolution hides the preferred basis, so that there is no preferred basis except at measurement. When the measurement occurs there is a preferred basis chosen, and the evolution becomes random. So the preferred basis is hidden by having two rules for time evolution - deterministic unitary evolution during which we hide the preferred basis, and random collapse of the wave function when we choose it. In contrast BM has only one time evolution, so it doesn't hide the preferred basis. This is why I'm contrasting standard QM where the preferred basis is hidden into one of two types of time evolution, with BM where the preferred basis is not hidden. So if it is standard QM versus BM whether the preferred basis is hidden or explicit, then it is the hidden variables that choose the basis explicitly. So my argument was that standard QM chooses to hide the preferred basis rather than use hidden variables, because of the additional complexity that hidden variables usually bring.
 
  • Like
Likes vanhees71
  • #24
3,507
26
Let's see if I can be more coherent. The state as a ray and unitary evolution hides the preferred basis, so that there is no preferred basis except at measurement. When the measurement occurs there is a preferred basis chosen, and the evolution becomes random. So the preferred basis is hidden by having two rules for time evolution - deterministic unitary evolution during which we hide the preferred basis, and random collapse of the wave function when we choose it. In contrast BM has only one time evolution, so it doesn't hide the preferred basis. This is why I'm contrasting standard QM where the preferred basis is hidden into one of two types of time evolution, with BM where the preferred basis is not hidden. So if it is standard QM versus BM whether the preferred basis is hidden or explicit, then it is the hidden variables that choose the basis explicitly. So my argument was that standard QM chooses to hide the preferred basis rather than use hidden variables, because of the additional complexity that hidden variables usually bring.
Ok, I see what you mean, but don't you think hiding the preferred basis in an abstract formalism leads to confusion and gets people into loops that have no way out within the theory-say like your discussions with vanhees71 over collapse ;) - unless one admits the preferred frame as a shortcoming of the theory that reflects its being an approximation to a truly basis independent theory?
I think this conclusion in unavoidable unless one is convinced before hand that what QM models is a truly stochastic reality, whatever that means, but that is a strong belief that should have some justification.
 
  • #25
atyy
Science Advisor
14,309
2,546
Ok, I see what you mean, but don't you think hiding the preferred basis in an abstract formalism leads to confusion and gets people into loops that have no way out within the theory-say like your discussions with vanhees71 over collapse ;) - unless one admits the preferred frame as a shortcoming of the theory that reflects its being an approximation to a truly basis independent theory?
I think this conclusion in unavoidable unless one is convinced before hand that what QM models is a truly stochastic reality, whatever that means, but that is a strong belief that should have some justification.
Well, whether it's hidden or not is a matter of interpretation :) In Copenhagen, the classical quantum cut associated with the need to decide from outside when a measurement occurs to fix the basis and get random outcomes and time evolution is stated right up front - it's in the operning pages of Landau and Lifshitz, and in Weinberg's coverage of Copenhagen. So what I like about Copenhagen is that it is extremely clear in stating its limitations.
 

Related Threads on Quantum interference and superposition

Replies
4
Views
720
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
2
Views
586
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
873
Replies
5
Views
726
Replies
3
Views
2K
Replies
3
Views
3K
Top