Exploring Wave-Particle Duality in Physics

[R34p3r]

I have a question about an expirement that i saw yesturday in my physics class. They were shooting electrons one at a time at a wall on a video. They watched it and it showed a wave pattern on the sheet and it said that even when one electron goes throught it still goes through both slits and so on. Now when they observed it though it behaved like it was solid matter like a marble that made a two slit pattern on the wall. Now why does the observer change the situation for the electron to act so differently. I asked my physics teacher and he said that it required a 400 lvl class and about three weeks. So basically he didnt know. I was wondering if any of you knew why this happened i don't need to understand it right now i just want to know how that happened that the observer changed it. Thankyou tell me if you need it to be clearer.

 

[LnGrrrR]

Reaper, good luck understanding that one...I don't know if anyone understands exactly "why" it happens. :) We know it does though, and there are different theories/explanations regarding it.

Others who are more-versed (and actual physicists) will give more informational words than I could though, I'm sure. :)

 

[koantum]

...why does the observer change the situation for the electron to act so differently... I was wondering if any of you knew why this happened i don't need to understand it right now i just want to know how that happened that the observer changed it.

The behavior of the electrons does not depend on observers. It depends on the setup, regardless of the presence of observers. If the setup makes it possible to learn through which slit each electron went, then each electron went through a single slit, either left (L) or right (R). If the setup doesn't permit this, then each electron went through both slits.

Note that saying "the electron went through both slits" cannot have the same meaning as the conjunction "the electron went through L and the electron went through R". For if the electron goes through both slits (as a whole, without being thereby divided into parts that go through different slits) then "the electron went through L" and "the electron went through L" are each neither true nor false. These propositions are then meaningless. So what can saying "the electron went through both slits" possibly mean? Think about it!

As for understanding, that is in a completely different league...

 

[R34p3r]

and for the going throught both slits is actually exaculy what happens it goes through both until you add the observer then it goes through one. and i need to know I am just starting to become obbsessed so if anyone out there knows using mahamtics words whatever i don't care if i don't understand it ill learn to understand it.

 

[-Job-]

I like to think of a photon as a tiny particle, or some amount of energy, being displaced by some wave. Like a grain of rice on top of a water wave. The grain of rice only goes through one slit, but the wave goes through both slits. The water wave will interfere with itself at the region of the slits. Because the grain of rice is displaced by the water wave and the water wave interferes with itself, the final destination of the rice will exhibit interference. The reason why adding an observer at one of the slits might prevent the interference pattern might be because the wave doesn't go through slit A at the same time as it goes through slit B. Even a slight difference, considering that this wave is propagating at the speed of light could mean no interference. But if a quantum computer works then this can't be right, there must be superposition.
[edit] On a second thought why wouldn't a quantum computer work in this scenario? The effect is the same. I have to ask my teacher this.

 

[Hurkyl]

If we're given the fact we saw the electron pass through the left slit, then its subsequent behavior should be as if it passed through the left slit.

If we're given the fact we saw the electron pass through the right slit, then its subsequent behavior should be as if it passed through the right slit.

However, if there's nothing to determine which slit the electron passes through, then its subsequent behavior should be as if it passed through both slits (and thus you get interference).


(The following is based on my preferred interpretation of QM, which I will shamlessly plug with vanesch's favorite phrase: it's the one that takes QM seriously!)


Let's first analyze why interference happens: the state of the electron is the sum of what happens when it passes through the left slit, and what happens when it passes through the right slit. At some points P, the "electron strikes point P" parts of the left and right slits are out of phase, and they cancel each other out. Therefore, it's impossible to strike at point P.

However, when you have a device capable of "observing" through which slit the electron passes, the electron becomes entangled with the device. Then, at the point P, we have that the "device says left and electron strikes point P" and "device says right and the electron strikes point P are still completely out of phase. However, now they are different states (they're distinguished by the result of the measuring device), so they cannot cancel each other out!

(For the same reason "electron strikes point P" and "electron strikes point Q" don't cancel each other out when P and Q are different points)

 

[-Job-]

However, if there's nothing to determine which slit the electron passes through, then its subsequent behavior should be as if it passed through both slits (and thus you get interference).

If there were multiple slits in random positions, and we didn't know which way the photon/electron was fired, would its behavior still be as it if had passed through all slits? It's not like photon knows about the slits.

 

[jtbell]

While the photon is propagating, you can't think of it as being like a tiny pellet, localized in space at any given instant and following a well-defined classical trajectory. Its propagation is determined by a wave that spreads over a large volume of space, and is influenced by whatever obstacles are in that volume.

 

[-Job-]

I understand that that is the view adopted by Quantum Mechanics. I'm wondering if there is a classical view that explains why the photon/electron exhibits its strange behavior. Just seeing if there can be an explanation there is more familiar, and consitent with experience.

 

[Hurkyl]

Just seeing if there can be an explanation there is more familiar, and consitent with experience.

I'm not sure if you meant it that way, but I still feel the need to point out that QM is consistent with (most people's) experience.

In fact, since most people have zero experience with the microscopic world, just about any theory would be consistent with their experience.

What you meant, I think, is that you want to extrapolate the experience you do have into this domain where you have no experience.

 

[-Job-]

I suppose.

 

[koantum]

If there were multiple slits in random positions, and we didn't know which way the photon/electron was fired, would its behavior still be as it if had passed through all slits? It's not like photon knows about the slits.

The same interference fringes have been observed with http://thisquantumworld.com/images/c60a.jpg" and diffraction gratings containing thousands of slits.

 

[koantum]

… if there's nothing to determine which slit the electron passes through, then its subsequent behavior should be as if it passed through both slits …

If it looks like a duck, walks like a duck, and quacks like a duck, then it is duck. In science we can never say more than "it's as if", so we may as well drop the "as if".

… my preferred interpretation of QUANTUM MECHANICS .. the one that takes QM seriously!

Before one can take quantum mechanics seriously, one has to have an interpretation!

… the state of the electron is the sum of what happens when it passes through the left slit, and what happens when it passes through the right slit.

The "state of the electron" is a probability algorithm. According to this algorithm, the probability of electron detection at the backdrop is either the sum of the absolute squares of two amplitudes (http://thisquantumworld.com/2rules.htm" ), depending on the setup.

… when you have a device capable of "observing" through which slit the electron passes, the electron becomes entangled with the device.

Entanglement, like interference, is not a physical state or process but a mathematical feature of the quantum-mechanical probability algorithm. Without interference (Rule A) we have p = |A₁|²+|A₂|², with interference (Rule B) we have p = |A₁|²+|A₂|² + an "interference term". We speak of constructive interference if the interference term is positive and of destructive interference if it is negative. We speak of entangled systems if the probability algorithm for a composite system does not factorize into separate algorithms for the component systems.

Then, at the point P, we have that the "device says left and electron strikes point P" and "device says right and the electron strikes point P are still completely out of phase. However, now they are different states (they're distinguished by the result of the measuring device), so they cannot cancel each other out!

If you treat the device as capable of indicating the slit taken by the electron, then there is no question of phases – you have one outcome, not two. As long as the phases of the amplitudes (associated with the alternatives) matter, the appropriate probability algorithm is a superposition, and in this case the device cannot indicate anything.

For the same reason "electron strikes point P" and "electron strikes point Q" don't cancel each other out when P and Q are different points

The proper definition of an alternative is a sequence of measurement outcomes.

• The actual outcome of the initial measurement determines how probabilities are assigned to subsequent outcomes.
• The possible outcome of the final measurement is the one whose probability we are calculating.
• Alternatives that may interfere differ neither in the initial nor in the final outcome. They differ in the possible outcomes of intermediate measurements.
• Rule A applies if the intermediate measurements are made (or if it is possible to infer from other measurements what their outcomes would have been if they had been made).
• Rule B applies if the intermediate measurements are not made (and if it is not possible to infer from other measurements what their outcomes would have been).

 

[woodsy2k]

I was led to believe that the electron has with it an associated wave of probability. And it is this that passes through the slits. Thus there is a "quantum jump" (Ref - The New Quantum Universe. Hey & Walters, Cambridge university) after the slits, whereby the electrons' function of probability suddenly takes a specific form and the electron is detected once only at the detector.

 

[Hurkyl]

Before one can take quantum mechanics seriously, one has to have an interpretation!

The interpretation is "QM is right, let's take it seriously" -- you view the world as quantum. :tongue: In particular, there is no need for the collapse postulate, since its purpose in life is to get classical answers out of a quantum theory.

the state of the electron is the sum of what happens when it passes through the left slit, and what happens when it passes through the right slit.

The "state of the electron" is a probability algorithm.

Fine -- allow me to revise my comment to say:

The ket denoting the state of the electron is the sum of the ket denoting the state we'd get if we knew the electron passed through the left slit, and the ket denoting the state we'd get if we knew the electron passed through the right slit.

Entanglement, like interference, is not a physical state or process

It's certainly not a state -- "entangled" is an adjective that describes states.

I don't see why you would say "entanglement" is not a process: surely the verb "to entangle" applies to any interaction that took unentangled states as input and produced entangled states as output?


And, incidentally, the state here does not factor. The probability that the electron strikes a particular point on the screen is not independent of the outcomes of the measuring device.

In other words:

P(|left slit, point X>)

is not equal to

p(|right slit, point X>)

If you treat the device as capable of indicating the slit taken by the electron, then there is no question of phases – you have one outcome, not two. As long as the phases of the amplitudes (associated with the alternatives) matter, the appropriate probability algorithm is a superposition, and in this case the device cannot indicate anything.

Yes, it's in a superposition! An unnormalized ket denoting the state would look like:

|left slit>|left amplitudes> + |right slit>|right amplitudes>

Where |left amplitudes> is the ket that denotes the state of the electron if we knew it had passed through the left slit.

Your (rule A) is then derived by taking the partial trace to discard the information provided by the measuring device.


(for clarity, I'm going to assume that the distribution on the screen is discrete, not continuous.)

The whole point I'm trying to make is this: if |left amplitudes> looks like:

|left amplitudes> = ... + |P> + ...

and |right amplitudes> looks like:

|right amplitudes> = ... - |P> + ...

then without a measuring device in the middle, the ket denoting the state of the electron looks like:

|left amplitudes> + |right amplitudes> = ... + 0|P> + ...

which completely kills off the |P> component.

But with the measuring device, it looks like:

|left slit>|left amplitudes> + |right slit>|right amplitudes>
= ... + |left slit>|P> - |right slit>|P> + ...

and since |left slit>|P> and |right slit>|P> are different outcomes, they cannot interfere and cancel each other out.

 

[koantum]

I was led to believe that the electron has with it an associated wave of probability. And it is this that passes through the slits. Thus there is a "quantum jump" (Ref - The New Quantum Universe. Hey & Walters, Cambridge university) after the slits, whereby the electrons' function of probability suddenly takes a specific form and the electron is detected once only at the detector.

What do you think is a "wave of probability"? A wave is a physical entity that has an amplitude and a phase at every point of space and every instant of time (within some region of spacetime). Is a probability something that exists at every point of space and every instant of time (within some region of spacetime)? Is it something that has an amplitude or a phase? Is it something that can pass through slits? Get real, dude! Don't buy all that….

The quantum formalism is an algorithm for calculating the probabilities of possible measurement outcomes on the basis of actual outcomes. The calculation involves complex numbers called amplitudes, and these are functions of space and time in the same sense that probabilities are functions of space and time. The probability of detecting a particle in a region of space monitored by a detector is not something that is located in that region or that gets there after passing through the slits. The probability of obtaining a given outcome in a measurement performed at the time t is not something that exists at the time t. The dependence of quantum-mechanical probabilities on space is a dependence on the location of detectors, and the dependence of quantum-mechanical probability assignments on time is a dependence on the time at which a measurement is performed.

 

[koantum]

The interpretation is "QM is right, let's take it seriously" -- you view the world as quantum. In particular, there is no need for the collapse postulate, since its purpose in life is to get classical answers out of a quantum theory.

Thanks for the explanation that "viewing the world as quantum" means that there are no collapsible states or wave functions. In this sense, I happen to view the world as quantum.

The ket denoting the state of the electron is the sum of the ket denoting the state we'd get if we knew the electron passed through the left slit, and the ket denoting the state we'd get if we knew the electron passed through the right slit.

The state we'd get if we knew the electron passed through the left slit is not the ket |L> but the projector |L><L|. The state we'd get if we knew the electron passed through the right slit is not |R> but |R><R|. The state we'd get if we could know through which slit the electron passed but don’t care to is |L><L| + |R><R|. The state we'd get if it's impossible to know through which slit the electron passed is not (proportional to) |L>+|R> but (proportional to) (|L>+|R>)(<L|+<R|).

I don't see why you would say "entanglement" is not a process: surely the verb "to entangle" applies to any interaction that took unentangled states as input and produced entangled states as output?

The "input" is a joint probability algorithm that factors, the "output" is a joint probability algorithm that doesn’t. What transforms that input into this output? Why not simply admit that we don’t know?

You see, it is by definition impossible to find out by experiment what happened between one measurement and the next. Any story that tells you what happened between consecutive measurements is just that – a story. You may tell any story you like about the transformation of the initial algorithm into the final algorithm, but it has nothing to do with physics. What transforms the initial algorithm into the final algorithm is a computation performed in accordance with laws that correlate measurement outcomes. If you want to think of this computation as a physical process and call it "getting entangled" (or whatever), be my guest. I only want to point out that we have precious little hope of ever understanding the relation of the quantum formalism to the real world if we consistently fail to draw the distinction between a computation and a physical process.

And, incidentally, the state here does not factor.

? That's what I wrote. (I said: "We speak of entangled systems if the probability algorithm for a composite system does not factorize into separate algorithms for the component systems.")

Yes, it's in a superposition!

It makes no sense to say that a physical system it is in a superposition. It only makes sense to say that the probability algorithm associated with a physical system is a superposition.

An unnormalized ket denoting the state would look like:
|left slit>|left amplitudes> + |right slit>|right amplitudes>
Where |left amplitudes> is the ket that denotes the state of the electron if we knew it had passed through the left slit.

As I said, the state of the electron if we knew it had passed through the left slit is not denoted by a ket. Your following argument below is based on a misconception.

Your (rule A) is then derived by taking the partial trace to discard the information provided by the measuring device.

Surely this is not what you mean to say. The partial trace discards phase information because it takes into account that a measurement has occurred (without taking into account the outcome of the measurement).

 

[woodsy2k]

The square of the wavefunction gives the electron a probability of appearing anywhere on the screen. Perhaps i was wrong in saying that it has a "wave of probability" that is a physical entity, but there is certainly an element of probability associated with an electron when traveling throught he slits.

 

[koantum]

The square of the wavefunction gives the electron a probability of appearing anywhere on the screen. Perhaps i was wrong in saying that it has a "wave of probability" that is a physical entity, but there is certainly an element of probability associated with an electron when traveling throught the slits.

The wave function is the absolute square of the amplitude. Integrated over a region R, it gives the probability of detecting the electron in R.
Everything in the quantum world is described in terms of probabilities! Put in a nutshell, QM says that everything is possible, and it allows you (in principle) to calculate the probability of it. If this turns out to be zero, then it's impossible.

 

[Hurkyl]

The state we'd get if we knew the electron passed through the left slit is not the ket |L> but the projector |L><L|.

That's not a state, it's a projector. :tongue:

Density matrices, kets, positive linear functionals, points in projective Hilbert space -- they're all just different ways of denoting the same thing. Each one is just as good as the other. (Though some may be easier to use in certain contexts)

I like kets because that's what I'm used to, and they have the nice notational advantage that it's easy to give them verbose labels.

But in any case, your missing the main point of my whole exposition: you ought to look not just at the electron, but the whole system.

In the Hilbert space formalism, I could have just said:

If there's no detector, then the resulting (unnormalized) ket is:

(... + |P> + ...) + (... - |P> + ...) = ... + 0|P> + ...

but if there's a detector, the resulting (unnormalized) ket is:

... + 2|P> + ...

But that begs the question: "Why?" -- and since we can actually answer this question, we might as well, by looking at the system comprised of the electron and the detector.


You could do the same in the density matrix formalism, can't you? When there's a detector, you should write it as:

(|left, L> + |right, R>)(<left, L| + <right, R|)

and if you take the partial trace, you're left with

|L><L| + |R><R|

We don't have to present it as some mystical reason why we do one thing when there's a detector, and another thing when there's not a detector: we can actually give a theoretical reason why there's a difference!

 

[koantum]

That's not a state, it's a projector.

Since you seem to know that the ket |L> is nothing but a shorthand notation for the projector |L><L|, you should also know that (pure) quantum states are 1-dimensional projectors.

I like kets because that's what I'm used to, and they have the nice notational advantage that it's easy to give them verbose labels.

Kets can be terribly misleading since they contain spurious information. The phase of a ket has no physical significance whatsoever. Since it does not appear in the corresponding projector, it's much safer to work with projectors.

In the Hilbert space formalism, I could have just said:
If there's no detector, then the resulting (unnormalized) ket is:
(... + |P> + ...) + (... - |P> + ...) = ... + 0|P> + ...
but if there's a detector, the resulting (unnormalized) ket is:
... + 2|P> + ...

I'm sorry. Can't make sense of your notation. Does anyone else know what he is trying to say?

We don't have to present it as some mystical reason why we do one thing when there's a detector, and another thing when there's not a detector: we can actually give a theoretical reason why there's a difference!

Mystical = whatever you don’t understand?
Have a look at my recent post in the "pure and mixed" thread. There is a list that shows how the axioms of quantum mechanics can be derived. I haven’t mentioned it explicitly there, but the two rules (add absolute squares of amplitudes versus add amplitudes and take the absolute square of the sum) emerge naturally and are thereby fully explained.

 

[khellendros]

Multpile Worlds Theory

While you may argue about the conventional posibilities about why the photon does what it does, you might wish to contemplate de fact that it is this experiment that suporters of the many worlds theory claim proofs theyre ideas: that the photon passing thru BOTH slits means that in reality it actually DID pass both of them, at the same time, and as a consecuence a parallel universe is created, where the photon took the opposite slit as it did in ours. Just something to think about.

 

[Hurkyl]

Since you seem to know that the ket |L> is nothing but a shorthand notation for the projector |L><L|, you should also know that (pure) quantum states are 1-dimensional projectors.

That's exactly what I did not say. I am denying your stance that the density matrix has some special status as the "right" way to do things.

Kets can be terribly misleading since they contain spurious information. The phase of a ket has no physical significance whatsoever. Since it does not appear in the corresponding projector, it's much safer to work with projectors.

Yes, kets have disadvantages. That does not mean we should never use kets. :tongue: The reason we have different ways of doing the same thing is because each way has advantages in certain circumstances. One of the advantages of kets is that they're just plain simpler than the other methods of which I know.

(And, for the record, it is awfully difficult to talk about projectors unless you first adopt a space of kets for them to act upon. )

Incidentally, your objection doesn't apply to the other methods I mentioned: defining states as positive linear functionals on the "operator" algebra, or as points in the projective Hilbert space.

I'm sorry. Can't make sense of your notation. Does anyone else know what he is trying to say?

I had thjought it was clear that the "..." meant all the other stuff that was irrelevant to the problem. *sigh* Allow me to slightly reformulate the problem so that there are only two places the electron can strike after passing through the slit.

In this reformulation, the (unnormalized) ket arising from going through the left slit is:
|L> = |P> + |Q>
and through the right slit is
|R> = -|P> + |Q>

So if the electron passes through either slit with equal amplitude, the result is:
|L> + |R> = (|P> + |Q>) + (-|P> + |Q>) = 0|P> + 2|Q> = 2|Q>

in other words, the |P> term is killed off.

But if we have a detector, the result is:
|left>|L> + |right>|R> = (|left>|P> + |left>|Q>) + (-|right>|P> + |right>|Q>)
= (|left> - |right>)|P> + (|left> + |right>)|Q>

so the |P> term is not killed off. And if we take the partial trace to discard the information about the detector (and fiddle with the constant out front), it reduces to
|P> + |Q>

Mystical = whatever you don’t understand?

Here, I'm using "mystical" to refer to the fact that you could have explained why we use your rule A and rule B... but you chose instead to simply decree that we use those rules.

(By analogy, a magician could tell you exactly how he does what he's doing, but instead he prefers to hide them and present it as "magic")

 

[Hurkyl]

While you may argue about the conventional posibilities about why the photon does what it does, you might wish to contemplate de fact that it is this experiment that suporters of the many worlds theory claim proofs theyre ideas: that the photon passing thru BOTH slits means that in reality it actually DID pass both of them, at the same time, and as a consecuence a parallel universe is created, where the photon took the opposite slit as it did in ours. Just something to think about.

I'm having a hard time parsing your sentence -- but I've been attempting to present myself in an interpretation-independent way. MWI people still talk about probabilities too, but it has a different meaning than someone who prefers to get classical answers out of a problem.

 

[koantum]

I am denying your stance that the density matrix has some special status as the "right" way to do things.

I didn’t say that. Just that it's the safer way. Kets are shorthands for pure density matrices.

That does not mean we should never use kets.

Again, I didn't say that.

And, for the record, it is awfully difficult to talk about projectors unless you first adopt a space of kets for them to act upon.

Not awfully difficult but impossible.

Allow me to slightly reformulate the problem so that there are only two places the electron can strike after passing through the slit. In this reformulation, the (unnormalized) ket arising from going through the left slit is:
|L> = |P> + |Q>
and through the right slit is
|R> = -|P> + |Q>

Using unnormalized kets is even more potentially misleading than using kets, but never mind. Adding and subtracting your two equations yields
|Q> = (|L> + |R>)/2 and |P> = (|L> - |R>)/2.
The ket |P> stands for detection at P, the ket |Q> for detection at Q. Since detection at P rules out detection at Q (and vice versa) we have that <Q|P>=0. It follows that
<P|L> + <P |R>=0 and <Q|L> - <Q |R>=0 or
<P|L> = -<P |R> and <Q|L> = <Q |R>.
In other words, the amplitude for propagation from the left slit to P is minus the amplitude for propagation from the right slit to P, whereas the amplitude for propagation from L to Q equals the amplitude for propagation from R to Q. Consequently, the probability of getting from L to P is the same as the probability of getting from R to P, and the probability of getting from L to Q is also the same as the probability of getting from R to Q. Which means that both P and Q and equidistant from L and R. Which means that P and Q are the same point. Do you now see that something is wrong with your handling of kets?

 

[Hurkyl]

I didn’t say that. Just that it's the safer way. Kets are shorthands for pure density matrices.

Not explicitly. You're saying it in the way that every way I write a ket that's living in the Hilbert space, you say that I really meant it as a short-hand notation for a density matrix living in my operator algebra.

And I'm entirely baffled how you can, on the one hand, say that a ket is just shorthand for density matrices, but on the other hand admit that you need some vector space of kets before you can even speak about projectors.

Do you now see that something is wrong with your handling of kets?

First off, surely it was clear that I was simplifying the problem to illuminate the concept I was trying to set forth?

Secondly, you got your criticism wrong:

Which means that both P and Q and equidistant from L and R. Which means that P and Q are the same point.

It's easy to find a pair of distinct points P and Q each of which is equidistant both from L and from R. (In fact, I can make all 4 distances equal)

e.g. let L = (-1, 0, 0), R = (1, 0, 0), P = (0, -1, 3), Q = (0, 1, 3)

 

[koantum]

every way I write a ket that's living in the Hilbert space, you say that I really meant it as a short-hand notation for a density matrix living in my operator algebra.

If you have |A> +|B> , it's a shorthand notation for (|A> +|B>)(<A|+<B|) but not (of course) for |A><A| +|B><B|. What's your problem? It's well known that the phase of |A> +|B> is irrelevant. This vector therefore contains the same information as that operator. (The relative phases of |A> and |B> in |A> +|B> do of course matter.)

And I'm entirely baffled how you can, on the one hand, say that a ket is just shorthand for density matrices, but on the other hand admit that you need some vector space of kets before you can even speak about projectors.

For the dummies amongst us (myself not excluded), it's easier to think in terms of Hilbert space. In this context I'm saying that the vectors in this space have no physical meaning. Only the subspaces (or the corresponding projectors) have. A vector has meaning only as a shorthand for a 1 dimensional projector. For the experts, J.M. Jauch (Foundations of Quantum Mechanics,1968) has shown that

1. the lattice of propositions about a physical system ("proposition system") with continuous observables is a complete, ortho-complemented, separable, atomic, and weakly modular lattice;
2. every proposition system with these properties is a unique direct union of irreducible propositions systems;
3. an irreducible proposition system can always be represented as the set of subspaces of a Hilbert space.

It's easy to find a pair of distinct points P and Q each of which is equidistant both from L and from R.

In 3 dimensions, of course. But the dimension parallel to the slits and the interference fringes is irrelevant in this context. Your points P and Q belong to the same interference fringe, which could be a minimum or a maximum. If it's a minimum, the amplitudes <P|L> and <P|R> as well as the amplitudes <Q|L> and <Q|R> are out of phase, and if it's a maximum, <P|L> and <P|R> as well as <Q|L> and <Q|R> are in phase, whereas you implied that <P|L> and <P |R> are out of phase and that <Q|L> and <Q |R> are in phase, which means they do not belong to the same interference fringe.

 

[Hurkyl]

If you have |A> +|B> , it's a shorthand notation for (|A> +|B>)(<A|+<B|) but not (of course) for |A><A| +|B><B|. What's your problem?

That you're wrong.

And we can already see some of the notational problems with what you say. If |A> really means, as you say, |A><A|, and |B><B| really means |B><B|, then |A> + |B> ought to mean |A><A| + |B><B|. But, you want it to mean (|A> + |B>)(<A| + <B|)!

And, of course, if |A> is merely shorthand for |A><A|, then |A><A| is really shorthand for |A><A|<A|, which is really shorthand for |A><A|<A|<A| which is really shorthand for... and you get into all sorts of silliness.

It's well known that the phase of |A> +|B> is irrelevant.

Wrong. If you were to go on and say "is irrelevant for denoting a physical" state you'd be correct -- it is well known that |A> and k|A> denote the same physical state... or at least denote physical states that are completely indistinguishable by experiment.

But the phase is not irrelevant. For example, it is of crucial importance to the Hilbert space structure -- unless k = 1, or |A> = 0, the two kets |A> and k|A> must be different... otherwise it's not a Hilbert space!

This vector therefore contains the same information as that operator.

Wrong. When you go from vector to operator, you have irrevokably lost its phase -- there's no way to recover the original vector. Therefore, vectors have more information than projection operators. Or, we can instead count independent components.

But yes, the vector and operator do have the same capability for describing physical states. (up to indistinguishability)


But what is the point? Sure, the vector |A> denotes the same physical state as |A><A| (up to indistinguishability), but that does not mean that |A> means |A><A|! |A> is a vector that denotes some physical state, and |A><A| is an operator that denotes some physical state. |A> is still a vector, and |A><A| is still an operator, even if we say they denote the same thing.

(note: when I talk about indistinguishability, I do not mean to assert that there can exist distinct physical states that are indistinguishable -- I just cannot bring myself to be unrigorous and ignore that possibility)

In this context I'm saying that the vectors in this space have no physical meaning.

(Given that you believe the density matrices have physical meaning) Then you're wrong -- we can interpret the kets as having physical meaning. One way to do it would be to interpret the ket |A> as denoting the same physical thing as the operator |A><A|. And, of course, the ket k|A> would denote the same thing too.

This particular interpretation simply results in a many-to-one correspondence between vectors and physical states.

But the interpretation doesn't say anything about the objects mathematical properties. No matter how you want to interpret the vectors, mathematically speaking they're still vectors living in a Hilbert space and have properties such as phase.

A vector has meaning only as a shorthand for a 1 dimensional projector.

I won't even bother talking about the problems with this view, and I'll just make the point I want to make:

However you want to interpret your vectors, they're still vectors. The interpretation of a mathematical object does not change its mathematical identity.

And you need to have honest-to-goodness vectors to do things (at least in the formalism you seem to use). You like to talk about projectors. Well, you have to have an honest-to-goodness vector space to talk about them. You even like to write your projectors like |A><A| -- notation that doesn't make any sense unless you have a vector space of kets and a covector space of bras! So even the very way you write your projectors is absolutely meaningless unless you use kets to denote vectors and bras to denote covectors.


If the notion of working with vectors as vectors is really that offensive to you, then you should try your hand at some other formalism. For example:

(1) A projective Hilbert space does nothing more than discard the phase -- it does not go extra steps to try and reinterpret vectors as operators. Things like the Schrödinger equation can be homogenized and work on the projective Hilbert space directly. There is still an irrelevant phase at work, but it's merely in our choice of coordinates, as opposed to being a part of a point in projective Hilbert space.

(2) You could use a more algebraic approach. You have a C*-algebra of "operators". I put operators in quotes because we don't think of them as operators on Hilbert space, but as simply being an algebra in its own right. We can then talk about idempotents: things satisfying PP=P, and they would play the same role as projectors. States are then introduced as something called a "positive linear functional" -- intuitively, that's just a function that assigns an expected value to each element of our algebra.

(3) Another algebraic approach is via Clifford algebra. CarlB would be happy to talk about that, if you asked him.

 

[reilly]

It's good, now and then again, to remember that the Schrodinger Eq, is a wave equation. So, Schrodinger predicts an interference pattern from electrons, say, passing through a multiple slit screen. And, in fact, one of the early quantum puzzles, the Davisson-Germer experiment, was beautifully explained by the Schrodinger Eq. Waves are waves -- exp(i K.x - wt) -- whether generated by Maxwell or Schrodinger, and thus necessarily show interference and diffraction phenomena. (And, remember, it's the phenomena that drove and drives QM, not the other way around. QM is odd because Nature is odd.

If you restrict two slits to just one, of course the interference pattern will vanish -- it's a totally different problem. And, the only classical analogue worth a damn is optical, Young's experiment, interference. Also remember that virtually all of particle physics, many-body and solid state physics are, in many regards, conceptual children of Davisson-Germer and the two-slit experiment. The support for QM's superposition of states, is critical for most of quantum physics, and, so far, it's held up very well. (My thesis concerned radiative corrections to electron-proton scattering. Scattering, emission and absorbtion of radiation all are fundamentally dependent on superposition of states, on interference phenomena. There's no controversy in most of the trade about such matters.)

There are literally thousands of pages of theory and experiments based on QM; the theory of QM has stood up with spectacular success. After more than 50 years of trying, Bohm and his followers are lagging behind in the dust, still looking for new physics, for confirmation that hidden variables have something to offer. So far they don't. Which, of course, is why so many physicists pay little or no heed to Bohm's hidden variable work -- not true for the extremely clever work by Ahronhov and Bohm on the physical "existence" of the vector potential.

Again, it is experiment that forces us to deal with the two slit problem for electrons. Standard QM does a very nice job -- understandable as well -- of dealing with the two slit problem. What's missing here?

Regards,
Reilly Atkinson

 

[koantum]

Hurkyl,
Do you really think I am as dumb as you seem to think or as you make me seem? I fully agree with you. Of course vectors are vectors and projectors are projectors as far as the maths is concerned.
What I was saying is that the standard quantum formalism represents possible measurement outcomes as projectors on a Hilbert space, and it represents the algorithm by which probabilities are assigned by a density operator W. If this a projector (necessarily 1 dimensional) then we call it a pure state, otherwise it can be written as a probability distribution over pure states, and we call it a mixed state. The probability of outcome P equals Tr(WP). But precisely because the quantum formalism attaches significance only to the projectors and the density operator, we can use unit vectors are shorthand notations for the corresponding 1 dimensional projectors. This implies, among other things, that the overall phases of state vectors lack physical significance. So the state |A> is physically equivalent to the state e^ia|A> because they both are shorthand notations for |A><A|, but the state |A> + |B> is not physically equivalent to the state e^ia|A> + |B> because the former is a shorthand notation for
(|A> + |B>)(<A| + <B|)=|A><A| +|B><B| +|A><B| +|B><A|
whereas the latter is short for
(e^ia|A> + |B>)(e^-ia <A| + <B|)=|A><A| +|B><B| +e^ia|A><B| +e^-ia|A><B|.

 

[Hurkyl]

Do you really think I am as dumb as you seem to think or as you make me seem? I fully agree with you. Of course vectors are vectors and projectors are projectors as far as the maths is concerned.

No, which is why I'm baffled that you keep telling me that vectors are just projectors.

But precisely because the quantum formalism attaches significance only to the projectors and the density operator

The formalism doesn't attach significance to anything. It's the physical interpretation that attaches significance to things. And quite frankly, this is the first time I've ever seen anyone advocate a physical interpretation that happily attaches meaning to the density operators but not to the kets.

More commonly, when using the Hilbert space formalism, people will interpret kets as one way of denoting physical states, and they will interpret the density operators as another way of denoting physical states, and we have the property that |A> and |A><A| denote the same physical state.

 

[reilly]

Since you seem to know that the ket |L> is nothing but a shorthand notation for the projector |L><L|, you should also know that (pure) quantum states are 1-dimensional projectors.

(I'm saying many of the same things as hurkyl has. He is definitely on the right track.)

koantum
I suggest you revisit basic QM, and, best of all, go back to review Dirac's work in his QM book. Your statement is incorrect; states and projection operators are different.

Kets can be terribly misleading since they contain spurious information. The phase of a ket has no physical significance whatsoever. Since it does not appear in the corresponding projector, it's much safer to work with projectors.

Misleading?
Phases can occur in a projection operator, particularly in complicated angular momentum problems, compund states defined by means of 3-j symbols for example(Multipole radiation, for example.(We are, of course, talking relative phases. What spurious info is in a ket? If that's what you think, then the wavefunction has exactly the same problem

Tell me how to do the hydrogen atom with projectors. If you can get the hydrogen spectrum without states, but with projection operators, then I'll reconsider. In fact, I'll bet if you can do it, you will find yourself lecturing at CERN, Harvard, MIT, Stanford,... Such a calculation would astonish most of us -- unless it is one of the standard approaches in disguise, imitating Schrodinger or Heisenberg.

Perhaps I'm naive, even though I learned QM from a master, J. H. VanVleck. That, as is the case in most linear vector space matters, the wave function W(x) can be interpreted as a vector in Hilbert space. Dirac simply formulated a new notation, very clever indeed, very useful indeed, to aid in the basic QM. A projection operator is just that, an operator, which acts on state vectors And anyway, this is as basic as anything in QM -- see Landau and Lifschitz, Dirac, Kemble, Schiff, Cohen-Tannoudji, Messiah, Weinberg (Field theory), Condon and Shortley, Mott and Massey, Golberger and Watson, Schwinger, Feynman, Pauli, Wesskopf and on and on. You might even say that there's an overwhelming consensus, since the early 1930s on the mathematical structure of QM -- states and operators, that's all she wrote. And most of us who are, or have been in the physics trade use Dirac's ideas and notation -- like it's the best game in town.

I greatly regret that I must conclude from your confusion over operators and states, that you are just plain wrong, both the math and the physics. Even if you have taken a QM course, you should go back and review in detail that course. If your Prof suggested the nonsense that a state and a projection operator are the same thing, then that Prof. ought to be chastized. If you are going to criticize QM, at least understand QM well before you start.

I have a feeling that you might be dealing with a "sorta density" matrix approach. Does not change things at all. Also, in field theory we do use operators to create states, but, still, a state is a state, and an operator is an operator

Mystical = whatever you don’t understand?
Have a look at my recent post in the "pure and mixed" thread. There is a list that shows how the axioms of quantum mechanics can be derived. I haven’t mentioned it explicitly there, but the two rules (add absolute squares of amplitudes versus add amplitudes and take the absolute square of the sum) emerge naturally and are thereby fully explained.

Unless I'm badly mistaken, the sum of squares never was considered in QM, at least not by the founders. (The impetus for the square of the wave function to be interpreted (Born) as a probability density follows from general notions about wave intensity -- freshman physics, optics-- and or Poynting's Thrm -- E&M. Jackson's E&M text does a very nice job with semi-classical quantum E&M, in which a direct connection between (E*E + B*B)/2 as an energy density, and photons with energy h*Frequency. You say:
The "state of the electron" is a probability algorithm.
A state is not an algorithm -- at best it can be determined with an algorithm.

You say: Entanglement, like interference, is not a physical state or process but a mathematical feature of the quantum-mechanical probability algorithm.

How then do you explain entanglement in classical systems?

I learned my QM some 50 years ago, and taught QM 40 years ago. What I learned is still valid (cf, Schiff's and Kemble's texts). I could use my lecture notes today with minimal modification. Lot's of things have not changed. And, little has changed in the criticisms of QM. The plain fact is that much of the discussion in this thread could easily have taken place 25 or 30 years ago. The "anti-QMs" have brought little new to the table, and nothing that makes QM easier to use. The basic anti-QM arguments have not changed much over the past 50 years. On the other hand, the past 50 years have shown an explosion of successes -- from superconductivity to quarks. Telling isn't it.

I think that standard QM is here to stay. Even if it is superseded by a deeper theory, our QM will still be used to deal with atomic spectra, scattering, solid state, ... However, I recognize that QM might eventually be found wanting. Whoever makes that claim will, of necessity, be extensively familiar with QM theory and applications -- from atomic spectra to high energy particle physics -- not necessarily an encyclopedic knowledge, but enough to understand and communicate the problems and triumphs in the many fields in which QM is used -- get any of that wrong, and the game is over. QM is a huge subject, slit experiments compose a very small portion of QM.(For example, if you have problems with superposition, check out the K-meson system, in which the role of superposition is crucial.This system is of far more consequence than a two slit experiment.)Secondly it is of the highest importance to write with simplicity and to use precise rather than sloppy language.

To be very honest, my response was in no small measure due to the poor writing -- states=algorithm? The problem for you is that given such mistakes in writing, it is hard to take your ideas seriously. (I've said such things to my students, and to those who worked for me in the consulting business.) The problems generated by sloppy writing are compounded when the topic takes exception to standard notions.

That being said, I think your ideas are potentially interesting. While I think there's a lot of jive that goes in this Forum, I do not apply that epithet to you. If you can write your ideas in a compellingly clear fashion, then I'll be among the first to give you a fair hearing.

Once a professor, always a professor.
Regards, Reilly Atkinson

 

[koantum]

No, which is why I'm baffled that you keep telling me that vectors are just projectors.

But I don't say that, for heaven's sake!

this is the first time I've ever seen anyone advocate a physical interpretation that happily attaches meaning to the density operators but not to the kets.

On which planet do you live? Read J.M. Jauch's Foundations of Quantum Mechanics (Addison-Wesley, Reading, MA, 1968) for starters, or A.M. Gleason's seminal paper "Measures on the closed subspaces of a Hilbert space,'' Journal of Mathematics and Mechanics 6, 885-94, 1957. (The original proof of Gleason's theorem is reproduced in V.S. Varadarajan, Geometry of Quantum Mechanics (Springer, 1985). Another proof is due to I. Pitowsky. Simpler proofs using POVMs instead of PVMs have been given by (i) P. Busch and (ii) Caves, Fuchs, Manne, and Rennes.

More commonly, when using the Hilbert space formalism, people will interpret kets as one way of denoting physical states, and they will interpret the density operators as another way of denoting physical states, and we have the property that |A> and |A><A| denote the same physical state.

What else am I saying?

 

[koantum]

states and projection operators are different.

They are different mathematically, but when you use a state to assign probabilities to possible outcomes, you may as well use the corresponding density operator, and when you use a state to represent a possible outcome, you may as well use the corresponding projector. This is basic quantum mechanics.

Phases can occur in a projection operator, particularly in complicated angular momentum problems, compound states defined by means of 3-j symbols for example (Multipole radiation, for example.(We are, of course, talking relative phases.

Well, exactly. It's the only relevant phase information because it's the only one that affects the probabilities of measurement outcomes.

What spurious info is in a ket?

Everything that has no effect on the probabilities of measurement outcomes is spurious.

Tell me how to do the hydrogen atom with projectors.

If you want to understand quantum mechanics deeply, you look for the one rule that is at the heart of it all. And that, IMHO, is Gleason's trace rule, according to which probabilities are assigned (i) by density operators and (ii) to projectors representing outcomes. So what ultimately matters, if you want to understand quantum mechanics deeply, is density operators and projectors. (I suppose Hurkyl is right in that C* operator algebras or Clifford algebras can also be used, but I am not sufficiently familiar with them.) I never denied that it is a whole lot easier to work with kets when calculating the hydrogen spectrum and such.

Misleading?

OK, I take back the "terribly misleading". Not misleading when calculating the hydrogen spectrum. Misleading only if you think of a wave function or state vector as representing the ontological state of the universe, for then you assign physical significance to the overall phase of the wave function. I know you don’t do that, so I need not caution you.

the wave function W(x) can be interpreted as a vector in Hilbert space.

Of course.

Dirac simply formulated a new notation, very clever indeed, very useful indeed, to aid in the basic QM. A projection operator is just that, an operator, which acts on state vectors. And anyway, this is as basic as anything in QM… You might even say that there's an overwhelming consensus, since the early 1930s on the mathematical structure of QM -- states and operators, that's all she wrote. And most of us who are, or have been in the physics trade use Dirac's ideas and notation -- like it's the best game in town.

I absolutely agree with what you say!

A state is not an algorithm -- at best it can be determined with an algorithm.

Do you mean determined by an algorithm? Please clarify, for I want to get back to this.

How then do you explain entanglement in classical systems?

What do you mean? I thought it was common knowledge that entanglement is a uniquely quantum phenomenon. Please give me an example of classical entanglement.

I think that standard QM is here to stay.

That's precisely what I wrote https://www.physicsforums.com/showpost.php?p=952106&postcount=40"!

if you have problems with superposition, check out the K-meson system, in which the role of superposition is crucial.

I don’t. And thanks for the reminder of this fascinating system.

Secondly it is of the highest importance to write with simplicity and to use precise rather than sloppy language.

Once again I fully agree. Unfortunately it often takes much more than a post of acceptable size to forestall all possible misunderstandings.

That being said, I think your ideas are potentially interesting. While I think there's a lot of jive that goes in this Forum, I do not apply that epithet to you. If you can write your ideas in a compellingly clear fashion, then I'll be among the first to give you a fair hearing.

Thank you for that. I would greatly appreciate your feedback on my http://thisquantumworld.com" . I know there is much scope for improvement.

Regards,
Ulrich Mohrhoff (aka koantum)

 

[Hurkyl]

What else am I saying?

You keep saying |A> is shorthand for |A><A|!

When you say that, it doesn't sound like you're merely saying that |A> and |A><A| denote the same physical state.

It sounds like you're saying that |A> is a projector. It's the projector |A><A|, to be precise, but we used the shorthand notation |A> because it's shorter and more convenient.