The mathematics of QM

[BlackHole]

Both QM and relativity (which uses mainly diffential geometry and tensors) require a thorough backround in CM and Electromagnetism (including classical electrodynamics).

 

So Quantum Mechanics (non-relativistic) uses linear Algebra (especially complex numbers, matrix algebra and linear transformations), functional Analysis (Hilbert spaces), advanced calculus & diffential equations (ordinary diffential equations & partial differential equations). QM also uses Harmonic analysis using Fourier series & transforms. What else? Can someone give a little more information?

 

Edit: So basically modern QM is a combination of Werner Heisenberg's matrix mechanics with Erwin's wave mechanics using Dirac's linear transformations.

[Tom Mattson]

What else? Can someone give a little more information?

 

You've got most of it. I'd add group theory and various algebras (Lie, Grassman, Clifford,...)

[Johnny5]

You can do quantum mechanics using a partial differential equation, shown by Erwin to be equivalent to the matrix approach.

 

When you make something too abstract, you get yourself confused.

 

And you can get a fourth quantum number, if you use a different Laplacian operator on the wavefunction.

[Tom Mattson]

You can do quantum mechanics using a partial differential equation' date=' shown by Erwin to be equivalent to the matrix approach.

[/quote']

 

You can't take quantum mechanical spin into account with any differential equation. You have to rely on the Lie algebra that defines spin operators:

 

[math]

[s_i ,S_j]=i \hbar \epsilon _{ijk} S_k

[/math]

 

edit: OK, who broke LaTeX???

[Johnny5]

You can't take quantum mechanical spin into account with any differential equation. You have to rely on the Lie algebra that defines spin operators.

 

Is there any chance that you are wrong?

[Tom Mattson]

Is there any chance that you are wrong?

 

Nope. I should put a finer point on my statement though: You can't take quantum mechanical spin into account with differential operators (as you can with position, momentum, and energy to get the Schrodinger equation). You can of course use Heisenberg's equation of motion to predict the time evolution of spin operators, but this is of course a differential equation in the operators themselves.

 

The observation of phase changes of spin-1/2 particles under 2pi rotations is proof positive that you can't use a function of coordinates to describe spin. If you try, you will be forced to use functions that are multiple valued functions of position, which are physically meaningless.

[Johnny5]

 

The observation of phase changes of spin-1/2 particles under 2pi rotations is proof positive that you can't use a function of coordinates to describe spin. If you try' date=' you will be forced to use functions that are multiple valued functions of position, which are physically meaningless.[/quote']

 

In the simplest way imaginable, can you explain this?

[Tom Mattson]

In the simplest way imaginable, can you explain this?

 

Yes, you take a beam of spin-1/2 particles, prepared in some quantum state |a>. I'm leaving it as an abstract vector for the time being because my whole point is that it can't be projected into physical 3-space to make a wavefunction.

 

Now you do an experiment in which you split the beam and rotate the particles in one beam by 2p radians (say we're changing the f coordinate, for concreteness), while leaving the other beam unchanged. Upon bringing the beams together you will find that there is complete destructive interference at the point at which they meet. This indicates that the altered beam is now in the state -|a>.

 

This is precisely the reason why there cannot be a "spin wavefunction". If we project our state vector into 3-space, we have:

 

<r|a>=a(r,t)

 

But to be physically meaningful, the function a should not take on 2 different values when f=0 and when f=2p. So on these grounds, we determine that spin is not a function of the coordinates.

 

On more general grounds, if you look at the differential operator for orbital angular momentum (a physical quantity you can have a differential equation for), you will find that it gives nonsensical results when l (the orbital quantum number) is anything other than an integer.

[Johnny5]

Yes' date=' you take a beam of spin-1/2 particles, prepared in some quantum state |a>.

 

Now you do an experiment in which you split the beam and rotate the particles in one beam by 2p radians (say we're changing the f coordinate, for concreteness), while leaving the other beam unchanged. Upon bringing the beams together you will find that there is complete destructive interference at the point at which they meet. This indicates that the altered beam is now in the state -|a>.

 

Are you referring to the Stern-Gerlach experiment?

[Tom Mattson]

Are you referring to the Stern-Gerlach experiment?

 

No, I'm referring to neutron interferometry experiments that involve 2p rotations.

[Johnny5]

 

The observation of phase changes of spin-1/2 particles under 2pi rotations is proof positive that you can't use a function of coordinates to describe spin. If you try' date=' you will be forced to use functions that are multiple valued functions of position, which are physically meaningless.[/quote']

 

Can you explain to me what a "phase change of a spin 1/2 particle under 2p rotation" means?

[Tom Mattson]

I'll edit this once LaTeX is fixed, but for now here it is in regular font.

 

To rotate a state vector about a unit normal vector n by an angle f you have to use the rotation operator, which is:

D(n,f)=exp(-in^.Sf/(h-bar))

 

where S is the spin operator.

 

Take a state vector for an arbitrary spin-1/2 particle:

 

|a>=c₊|+>+c_-|->

 

and let D act on it. For concreteness, let n=k and let f=2p.

 

D(k,2p)=exp(-2piS_z)|a>

 

D(k,2p)=exp(-2piS_z)c₊|+>+exp(-2piS_z)c_-|->

D(k,2p)=exp(-ip)c₊|+>+exp(+ip)c_-|->

D(k,2p)=-c₊|+>-c_-|->

D(k,2p)=-|a>

 

The state vector picks up a negative sign, which is equivalent to a phase change of p radians.

[Johnny5]

Take a state vector for an arbitrary spin-1/2 particle:

 

|a>=c₊|+>+c_-|->

 

 

How do you get the RHS' date=' admittedly it's foreign to me. I don't know how to interpret it. I don't know what I'm looking at.

 

How did you derive the state vector for a spin 1/2 particle?

 

What is c[sub']+[/sub]... probability something is spin up state? Is the RHS just the sum of the probability that some arbitrary spin 1/2 particle is in a spin up state, with the probability that it is in a spin down state, given that it must be one or the other, something like that? Bits and pieces are coming back to me.

[Tom Mattson]

You don't derive it, it's just an arbitrary vector. The vectors |+> and |-> are defined as follows:

 

s₃|+>=(1/2)|+>

s₃|->=(-1/2)|->

 

where s₃ is the 3rd Pauli spin matrix. |+> and |-> are the only eigenstates available, so an arbitrary spin state can be expressed as a linear combination of them, which is what I did.

[Johnny5]

, so an arbitrary spin state can be expressed as a linear combination of them, which is what I did.

 

So c+,c- are just arbitrary constants?

 

I'm reading this now, and this .

 

That last thing is what you are trying to show me isn't it.

[Johnny5]

Is "spinor" short for spin operator?

 

Now I'm reading this

 

Masochism really. In all my ignorance, I would just like to say, there has got to be a better way to do this.

 

I read a few things so far:

 

1. Spin operator is a vector observable.

2. If you rotate ket vector through 2pi radians, you do not get back the original ket vector, you get back - ket. You have to rotate a ket vector through 720 degrees to get back the original ket vector.

 

They say this is a classically unexpected result.

 

Lepton (fermion)-generic name for any spin 1/2 particle which does not feel the strong nuclear force.

[Johnny5]

Let me go all the way back in time, to the reason anyone ever added a fourth quantum number. Wasn't the reason QM was seen as incomplete due to the observation of the Zeeman effect?

 

I have done the experiment myself, and witnessed the zeeman effect.

 

$25,000 SPEX spectrometer was used.

 

Hydrogen light entered the spex, and the computer plotted peaks which corresponded to the different photon wavelengths entering.

 

Then a huge magnet (1-5 Teslas 50 lbs, big sucker) was placed between the light source (excited hydrogen gas) and the SPEX (diffraction grating inside).

 

Then there was a splitting of the spectral lines observed. The number of hydrogen lines increased... the Zeeman effect.

 

Isn't the Zeeman effect the reason why spin quantum number was added to the model of hydrogen atoms?

 

This is actually the most helpful thing I've read so far, and yes of course I've went through this before, but that was long ago.

 

Electron spin

[Tom Mattson]

So c+' date='c- are just arbitrary constants?

[/quote']

 

Not completely arbitrary, there is a constraint. Because of normalization, we must have:

 

|c₊|²+|c_-|²=1

 

I'm reading this now, and this .

 

That's a good start.

 

That last thing is what you are trying to show me isn't it.

 

Yes.

[Tom Mattson]

Is "spinor" short for spin operator?

 

No' date=' it's not an operator. A spinor is an eigenvector of the Pauli spin matrices.

 

Now I'm reading this

 

Masochism really. In all my ignorance, I would just like to say, there has got to be a better way to do this.

 

Eye of the beholder, and all that. I think it's beautiful.

 

I read a few things so far:

 

1. Spin operator is a vector observable.

 

True.

 

2. If you rotate ket vector through 2pi radians, you do not get back the original ket vector, you get back - ket. You have to rotate a ket vector through 720 degrees to get back the original ket vector.

 

Not true in general. That is only the case for spin-1/2 particles. For spin-1 particles, a 2p rotation gets you right back to "ket".

 

They say this is a classically unexpected result.

 

It is. Take any object in your house. Now rotate it through 360 degrees, and see if you don't get back to where you started.

 

Lepton (fermion)-generic name for any spin 1/2 particle which does not feel the strong nuclear force.

 

That's right for lepton, but not for fermion. Quarks are fermions, and they are subject to both nuclear forces.

[Johnny5]

It is. Take any object in your house. Now rotate it through 360 degrees' date=' and see if you don't get back to where you started.

[/quote']

 

Stay with me here...

 

We are not rotating a material object in real space.

 

The rotation is of a ket vector, a ket vector isn't a solid object now is it?

[Tom Mattson]

We are not rotating a material object in real space.

 

We are indeed. If you turn a spin-1/2 particle through a complete revolution' date=' its physical state changes in just the way I described.

 

The rotation is of a ket vector, a ket vector isn't a solid object now is it?

 

A ket is not a solid object, but the spin-1/2 particle it describes is.

[Johnny5]

We are indeed. If you turn a spin-1/2 particle through a complete revolution' date=' its physical state changes in just the way I described.

 

 

 

A ket is not a solid object, but the spin-1/2 particle it describes is.[/quote']

 

Electron is an example of a spin 1/2 particle. Originally, an electron was modelled as a 3D object. You can visualize such a thing. A particle has no size. So which is it?

 

Is an electron a point particle, or a 3D object, or neither...

 

I might as well add, I cannot visualize a 360 rotation of a 3D object, and not get back to where I started, I will never be able to visualize such a thing, no one can.

 

It is not clear to me, that you are correctly interpreting the math.

 

Exactly what is electron spin anyways?

 

I can imagine a tiny sphere, rotating about an axis in some frame. It's spinning. It has some tangential speed in the frame.

 

Then I can imagine its charge density distributed evenly throughout its volume.

 

Then there is a magnetic moment which points, say upwards, as the 3D particle spins say counter-clockwise.

 

Now you want to 'rotate' this particle which is already spinning, through 360 degrees. No matter what you do to this thing, I cannot see how it will not be back to the way it was before you rotated it.

[Tom Mattson]

Electron is an example of a spin 1/2 particle. Originally' date=' an electron was modelled as a 3D object. You can visualize such a thing. A particle has no size. So which is it?

 

Is an electron a point particle, or a 3D object, or neither...

[/quote']

 

All we know is that our theories which treat the electron as a point particle work exceptionally well. But I should note that these 2p rotation experiments are typically done with neutrons, which have a known radius.

 

I might as well add, I cannot visualize a 360 rotation of a 3D object, and not get back to where I started, I will never be able to visualize such a thing, no one can.

 

That does not preclude its occurance, of course.

 

It is not clear to me, that you are correctly interpreting the math.

 

It would be, if you knew more QM.

 

Exactly what is electron spin anyways?

 

I can imagine a tiny sphere, rotating about an axis in some frame. It's spinning. It has some tangential speed in the frame.

 

That's not what electron spin is. Its spin is not analogous to anything classical, such as a spinning top. The spin of the electron is just an intrinsic quantity of angular momentum that the electron is born with.

[Johnny5]

 

That's not what electron spin is. Its spin is not analogous to anything classical' date=' such as a spinning top. The spin of the electron is just an intrinsic quantity of angular momentum that the electron is born with.[/quote']

 

This is so highly abstract, do you regard this as knowledge?

 

I would like to ask an important question.

 

Exactly what is done to a spin-1/2 particle, in order to rotate it 360 degrees.

 

Let me guess, an external B field is applied?

[Tom Mattson]

This is so highly abstract, do you regard this as knowledge?

 

Yes, I do regard it as knowledge. It is a well-confirmed prediction of a deductively valid scientific theory.

 

I would like to ask an important question.

 

Exactly what is done to a spin-1/2 particle, in order to rotate it 360 degrees.

 

Let me guess, an external B field is applied?

 

You guessed right.