Collapse the wave function?

[Obnoxious]

I heard that phrase spammed many times on many phizzy books I've read, just what does it mean, and how would you go about collapsing the wave function?

[swansont]

I heard that phrase spammed many times on many phizzy books I've read, just what does it mean, and how would you go about collapsing the wave function?

 

A system may be in an indeterminate state, so the wave function that describes it is a superposition of states with various probabilities. But when you actually "look" at the particle, it can only be in one state. The wave function is said to collapse to that one state when you do a measurement that tells you what state the particle is in.

[Johnny5]

A system may be in an indeterminate state, so the wave function that describes it is a superposition of states with various probabilities. But when you actually "look" at the particle, it can only be in one state. The wave function is said to collapse to that one state when you do a measurement that tells you what state the particle is in.

 

And to add to what Dr. Swanson said, mathematically, this happens due to usage of the Dirac delta function.

 

See this link here: Measurement in quantum mechanics

 

In particular, read this:

 

The case of a continuous spectrum is more problematic, since the basis has aleph eigenvectors. These can be represented by a set of Delta functions. Since the delta function is in fact not a function, and moreover, doesn't belong to the Hilbert space of square-integrable functions, this can causes difficulties such as singularities and infinite values. In all practical cases, the resolution of any given measurement is finite, and therefore the continuous space may be divided into discrete segments. Another solution is to approximate any lab experiments by a "box" potential (which bounds the volume in which the particle can be found, and thus ensures a countable spectrum).

 

The usage of the mathematical theory of probability in quantum mechanics leads to certain "conceptual problems." These problems arise for any number of reasons.

 

Regards

[swansont]

And to add to what Dr. Swanson said' date=' mathematically, this happens due to usage of the Dirac delta function.

[/quote']

 

Note that it says that's for a continuous spectrum of states.

[Johnny5]

Note that it says that's for a continuous spectrum of states.

 

Yes, I saw that Dr. Swanson.

[swansont]

Yes, I saw that Dr. Swanson.

 

Since typically you use QM when you have discrete states, it's very important and means that "mathematically, this happens due to usage of the Dirac delta function." doesn't apply in most situations.

[Johnny5]

Since typically you use QM when you have discrete states, it's very important and means that "mathematically, this happens due to usage of the Dirac delta function.[/i']" doesn't apply in most situations.

 

Even the usage of discrete distribution functions, still leads to conceptual problems. Just a dumb question... how is one to know how many elements are in the sample space?

 

Regards

[swansont]

Even the usage of discrete distribution functions' date=' still leads to conceptual problems. Just a dumb question... how is one to know how many elements are in the sample space?

 

Regards[/quote']

 

You solve Schroedinger's equation for the boundary conditions involved.

[Johnny5]

You solve Schroedinger's equation for the boundary conditions involved.

 

In the case of a hydrogen atom, free of all external forces, you have an electron in a potential given by:

 

[math] U® = \frac{1}{4\pi \epsilon_0} \frac{Q_1Q_1}{r} [/math]

 

And Schrodinger's equation is:

 

[math] \frac{- \hbar^2}{2 \mu} \nabla^2 \Psi +U\Psi = i \hbar \frac{\partial \Psi}{\partial t} [/math]

 

Where mu is the reduced mass of the electron.

 

So we can perform the analysis in the center of mass frame, of the proton-electron system.

 

Now, suppose that an electron is in some energy state, above the ground state, and you seek to measure its position.

 

Is position a well defined observable quantity?

 

And if not, then you are going to use probability theory. So my question is, how do you go from the boundary conditions for a hydrogen atom in free space, to knowing the number of elements in the sample space, in order to justify the treatment of the electron's position as a random variable.

 

If that didn't make sense, then I will solve the TISE for Hydrogen, to make the question clearer.

 

You aren't going to exactly follow where I am going with this, so to be crystal clear:

 

How many possible places are there for an electron in hydrogenic atom in free space to be at the next moment in time? One, or more than one?

[BlackHole]

Johnny, do you mean the probability density applet. Take a lot at this:

 

http://www.phy.davidson.edu/StuHome/cabell_f/Density.html

http://scienceworld.wolfram.com/physics/HydrogenAtom.html

http://physics.ius.edu/~kyle/physlets/quantum/hydrogen.html

 

Btw, the wave-function collapse, how do we know it really happens? Does the state vector collapse to one of the eigenvectors of the observable (operator)?

 

http://linas.org/theory/experiment.html

[Johnny5]

Johnny' date=' do you mean the probability density applet. Take a lot at this:

 

http://www.phy.davidson.edu/StuHome/cabell_f/Density.html

http://scienceworld.wolfram.com/physics/HydrogenAtom.html

http://physics.ius.edu/~kyle/physlets/quantum/hydrogen.html

 

Btw, the wave-function collapse, how do we know it really happens? Does the state vector collapse to one of the eigenvectors of the observable (operator)?

 

http://linas.org/theory/experiment.html[/quote']

 

Yes, on a cursory glance, that is exactly what I was referring to.

 

Using any or all of the knowledge at these sites, what is the answer to the question which I posed?

 

How many possible places can the 'electron' in a hydrogenic atom be, at the next moment in time, according to quantum mechanics?

 

 

I was thinking of the expectation values of the radius in particular.

 

Here is the one I wanted first:

 

[math] <R> = \frac{A_0}{2Z}(3n^2 - l(l+1)) [/math]

 

Now, in the formula above, Z is the atomic number of the atom in question, and for Hydrogen Z=1, which simplifies to:

 

[math] <R> = \frac{A_0}{2}(3n^2 - l(l+1)) [/math]

 

A0 is the first Bohr radius, n is the principle quantum number, l is the angular momentum quantum number, and the angular momentum L is given by:

 

[math] L = \sqrt{l(l+1)} \hbar [/math]

 

And the z-component of the angular momentum is given by:

 

[math] L_z = m_l \hbar [/math]

 

where m_l is the magnetic quantum number, and comes from solving the azimuthal equation, one of three ODE's you have to solve, after you have used separation of variables on the time independent Schrodinger equation.

 

Ok so...

 

I really should just do it, because all the information comes pouring out on one go of it, but from memory, l can range from 0 to n-1.

 

At any rate, it is the expectation of the position of the electron node which I am interested in.

 

How many places can it be, at the next moment in time, according to conventional wisdom?

 

Anyone?

[swansont]

How many possible places can the 'electron' in a hydrogenic atom be' date=' at the next moment in time, according to quantum mechanics?

[/quote']

 

The position operator does not commute with itself over an elapsed time interval. Since they are not eigenstates, you don't collapse a wave function.

[Johnny5]

The position operator does not commute with itself over an elapsed time interval. Since they are not eigenstates, you don't collapse a wave function.

 

Dr swanson, can you show me this mathematically?

 

The momentum operator, from memory is:

 

[math] -i \hbar \frac{\partial}{\partial x} [/math]

 

Or do it in three dimensions.

 

Show me mathematically, how to say that the position operator does not commute with itself.

 

For the position of the electronal node in the hydrogenic rest frame, we have the variable r.

 

There is some neat symbolism, though I've forgotten it, to show this.

 

Something about Hermitian operators, and simultaneous observables.

 

It's coming back but very slowly.

 

It seems to me that you have something like this;

 

[r,r*]

 

for the position operator, and

[math]

[-i \hbar \frac{\partial}{\partial x},i \hbar \frac{\partial}{\partial x}] [/math]

 

For the momentum operator, and you are checking to see if the operators are commutative. I think in the case of the position operator r, it does commute with itself.

 

Yes or no?

[Johnny5]

Found something on it.

 

Hermitian operators

There is a special class of operators which are called Hermitian operators. They are of particular importance in quantum mechanics because they have the property that all of their eigenvalues are real (not complex). This is convenient for the measurement outcome of any experiment must be a real number. There are non-Hermitian operators, but they do not correspond to observable properties. All observable properties are represented by Hermitian operators (but not all Hermitian operators correspond to an observable property).

 

And here is the part about simultaneous observables:

 

Simultaneous observable quantities

 

 

There you see the notation I was talking about.

 

Anyways, I think the position operator r does commute with itself.

 

But i have to run through that notation again, to prove so.

[BlackHole]

Isn't there a surprising similarity between the quantum world and the Newtonian world?

 

http://www.walter-fendt.de/ph11e/bohrh.htm

 

This is also very informative. I'll have to work it through...

 

http://vergil.chemistry.gatech.edu/notes/quantrev/node25.html

[Johnny5]

This is also very informative. I'll have to work it through...

 

http://vergil.chemistry.gatech.edu/notes/quantrev/node25.html

 

That's the famous formula for energy quantization. It is equivalent to what Bohr got, using a planetary model.

 

Have you ever gone through the solution of the Schrodinger PDE before?

[BlackHole]

That's the famous formula for energy quantization. It is equivalent to what Bohr got' date=' using a planetary model.

 

Have you ever gone through the solution of the Schrodinger PDE before?[/quote']

 

So the planetary model is not physical?

 

If it is then i think we have a final theory. Matter (which is also energy) interacts with the CMBR (the ground state of the quantum vacuum) which causes motion. This motion emits radiation (the CMB photons).

 

Now if the sun is the nucleus and the planets are the quantum numbers, then we have a physical model of gravity. A world only of energy and interactions.

 

This is very hard to prove mathematically and i've never gone through PDEs but i will work on it and get back to you.

 

Here are some solutions to schroedinger equation.

 

Regards

[swansont]

I think in the case of the position operator r' date=' it does commute with itself.

 

Yes or no?[/quote']

Yes. But since I didn't claim otherwise, it's irrelvant. What I said was that it does not commute with itself over a time interval. In order to know r(t) from measuring r(0), you need to know the momentum.

 

Look here, equation 258

[swansont]

So the planetary model is not physical?

No. From your own link: You must not take the idea of electrons' date=' orbiting around the atomic nucleus, for reality.[/i']

[BlackHole]

No. From your own link: [i']You must not take the idea of electrons, orbiting around the atomic nucleus, for reality.[/i]

 

Why is Bohr's interpretation rejected?

 

If it's not correct, then it's not possible to combine gravity with quantum mechanics and we enter the problem of the many worlds interpretation of quantum mechanics. Combining gravity with magnetism also becomes an impossibility.

 

PS: I like the Copenhagen Interpretation better. It also makes more sense to me.

[swansont]

Why is Bohr's interpretation rejected?

 

Because the orbits aren't circular. QM came along and gave a much more complete picture.

[BlackHole]

Because the orbits aren't circular. QM came along and gave a much more complete picture.

 

I was wrong. Today we use Schroedinger's equation and the probability function.

 

I think this would make it impossible to combine gravity and magnetism.

[Severian]

I was wrong. Today we use Schroedinger's equation and the probability function.

 

I think this would make it impossible to combine gravity and magnetism.

 

I am 99.9% sure they will be combined one day.

[SHtRO]

Alright, I think this is where I belong...I'm new here so bear with me if I look like an idiot (I'm not). I've been a fan of Einstein since I was 9 and came to understand his General Theory of Relativity. As a relativist I have always had difficulty accepting the assertions of Schroedinger, Heisenberg, et al. I have only recently come to understand and accept the concepts such as entanglement and wave-collapse. However, I've always felt I had an "open understanding" of the nature of light. So now, on to my question/puzzlement...

 

Isn't wave-collapse just a transfer of momentum? Since photons, electrons, etc. travel at c (or near c) relativity seems to me to explain such experiments as delayed-choice. That is to say, from the relativistic perspective of the "wave/particle" (field?) time is practically stand-still, so there is no such thing as "delayed choice" with respect to the particle(?) frame of reference. I have read that because electrons are STL particles this perspective does not hold ("The Fabric of the Cosmos", Greene, 2004. p. 512, Ch. 7 footnote #4). However, I'm not in full agreement as it seems the relativistic speeds of quanta are so close to c so that practically relativistic logic holds.

 

Regardless, the puzzlement for *all* seems to focus on whether we are dealing with decoherence or something else with respect to wave-collapse. Is it not possible that the momentum of the wave function is transferred into our own (observer's material world) "entangled wave-function"? That is the interaction of coherent (big) matter, bound by all the usual forces, with energy fields such as photons and electrons is simply a transfer of momentum?

 

If I'm off-base, someone please direct me to experimental data if you can...

[Severian]

Alright' date=' I think this is where I belong...I'm new here so bear with me if I look like an idiot (I'm not). I've been a fan of Einstein since I was 9 and came to understand his General Theory of Relativity.

[/quote']

 

Claiming you understood general relativity at age 9 does indeed make you look like an idiot.

 

That is to say, from the relativistic perspective of the "wave/particle" (field?) time is practically stand-still, so there is no such thing as "delayed choice" with respect to the particle(?) frame of reference. I have read that because electrons are STL particles this perspective does not hold ("The Fabric of the Cosmos", Greene, 2004. p. 512, Ch. 7 footnote #4). However, I'm not in full agreement as it seems the relativistic speeds of quanta are so close to c so that practically relativistic logic holds.

 

Relativity tells us that only relative speed is important. Since the laws of physics are applicable in all frames of reference (all inertial frames for SR, all frames for GR) there is no notion of any particular frame being more special than any other. Since the elctron has mass and travels <c then it does have a rest frame, and your argument breaks down. It makes no difference how fast an electron actually goes since it is still infinitely far away from travelling at c (ie. you would have to put an infinite amount of energy in).

 

Secondly, this deosn't even work for phtons which have no mass and travel at c, since the collapse of the wavefunction is non-local. So the wavefunction collapses everywhere at once, quicker than you could send a photon across it. This sounds paradoxical, since it seems like some parts of the wavefunction actually collapse before you measure it (in certain frames) but is not when worked out in detail.