Question

can anybody explain electron spin by using concepts of quantum mechanics?

Answer 1

I'm not sure how deep of an answer you want here, but you probably need a more specialized site than OpenStudy to get a good answer. I would suggest this one:
http://www.physicsforums.com/

Answer 2

I suggest you read about the Stern–Gerlach experiment. I can also try explain some of it my self, but it is rather heavy stuff, but the explanation of spin can be found when you look at how Dirac combined quantum mechanics with special relativity. So you should actually be looking into relativistic quantum mechanics.

Answer 3

I'm gonna try do my best here and hope @TurningTest can correct me if I say anything wrong:
What we are going to be looking at is the rotation in three dimensions, the particle on a sphere.
We start by setting up the Schrödinger equation:
\[\hat{H}=-\frac{ \hbar }{ 2m } \nabla^2 + V\]
Where:
\[\nabla^2=\frac{ \partial ^{2} }{ \partial x ^{2} }+\frac{ \partial ^{2} }{ \partial y ^{2} }+\frac{ \partial ^{2} }{ \partial z ^{2} }\]
For the particle confined to a spherical surface, V=0 wherever it is free to travel and the radius, r, is a constant. The wavefunction is therefor a function of the colatitude, theta, and the azimuth, psi, and we write it as:
\[\psi(\theta,\phi)\]
The Schrödinger equation therefor become:
\[-\frac{ \hbar }{ 2m } \nabla ^{2} \psi=E \psi\]
We are now going to consider the sphere as a stack of rings and the equation above have the solution:
\[\psi(\theta,\phi)=\Theta(\theta) \Phi(\phi)\]
Where capital theta, is a function of only little theta and capital phi only a function of little phi, is the solution for a particle on a ring. You can show this using separation of variables*, where you will also see that the solutions are specified by the quantum numbers l and mi restricted to the values:
\[l=0,1,2.....\]
\[m _{l}=l,l-1,...,-l\]
Remember that the orbital angular momentum quantum number, l, is non-negative and, for a given value of l, there are 2l+1 permitted values of the magnetic quantum number.

* sence this is what we want to show in the end we show now how this is true:
To take advantage of this problem we use the symmetry of the sphere and use the fact that r is a constant for a particle on a sphere. I like to do this with spherical polar coordinates, so we now define:
The radius r; the colatitude delta and the azimuth phi, where:
\[x=r \sin(\theta) \cos(\phi)\]
\[y=r \sin(\theta) \sin (\phi)\]
\[z=r \cos(\theta)\]
You can show with some trouble that the laplacian in spherical polar coordinates is:
\[\nabla ^{2}=\frac{ \partial ^{2} }{ \partial r ^{2} }+\frac{ 2 }{ r } \frac{ \partial }{ \partial r }+\frac{ 1 }{ r ^{2} } \Lambda ^{2}\]
Where the legendrian, Lambda squared, is:
\[\Lambda ^{2}=\frac{ 1 }{ \sin ^{2}\theta }\frac{ \partial ^{2} }{ \partial \phi ^{2} }+\frac{ 1 }{ \sin \theta }\frac{ \partial }{ \partial \theta }\sin \theta \frac{ \partial }{ \partial \theta }\]
to be continued....

Answer 4

CORRECTION: everytime I wrote the Dirac constant please square it... a very stupid mistake of me.

Answer 5

Because r is constant, we acn discard the part of the laplacian that involves differentiation with respect to r and we can therefor write the Schrödinger equation as:
\[-\frac{ \hbar ^{2} }{ 2mr ^{2} } \Lambda ^{2}=E \psi\]
We notice that the moment of inertia had appeared:
\[I=mr ^{2}\]
This expression can be rearranged into:
\[\Lambda ^{2}\psi=-\epsilon \psi\]
Where:
\[\epsilon=\frac{ 2IE }{ \hbar ^{2} }\]
We verify this expression is separable by substituting our previously solution into the legendrian:
\[\Lambda ^{2}\Theta \Phi=\frac{ 1 }{ \sin ^{2}\theta }\frac{ \partial ^{2}(\Theta \Phi) }{ \partial ^{2} \phi }+\frac{ 1 }{ \sin \theta }\frac{ \partial }{ \partial \theta }\sin \theta \frac{ \partial (\Theta \Phi) }{ \partial \theta }=-\epsilon \Theta \Phi\]
We use the fact that capital Theta and capital Phi are each functions of only 1 variable so our partial derivatives become complete derivatives:
\[\frac{ \Theta }{ \sin ^{2}\theta }\frac{ d ^{2} \Phi }{ d \phi ^{2} }+\frac{ \Phi }{ \sin(\theta) }\frac{ d }{ d \theta }\sin \theta \frac{ d \Theta }{ d \theta }=-\epsilon \Theta \Phi\]
Divide with Theta and Phi, multiply with sin^2(theta) and rearrange it into:
\[\frac{ 1 }{ \Phi }\frac{ d ^{2} \Phi }{ d \phi ^{2} }+\frac{ \sin \theta }{ \Theta } \frac{ d }{ d \theta } \sin \theta \frac{ d \Theta }{ \theta }+\epsilon \sin ^{2} \theta=0\]
Now we notice that the first left term only depends on phi and the remaining terms depend on theta. We can now argue for that the each term is equal to a constant (not gonna show that) so we now say:
\[\frac{ 1 }{ \Phi }\frac{ d ^{2} \Phi }{ d \phi ^{2} }=-m _{l}^{2}\]
\[\frac{ \sin \theta }{ \Theta } \frac{ d }{ d \theta } \sin \theta \frac{ d \Theta }{ \theta }+\epsilon \sin ^{2} \theta=m_{l}^{2}\]

Answer 6

Now the first equation is not unknown as it is used in rotation in two dimensions; a particle on a ring and know its solution. the 2. equation is not my best, by mathematicians would be smiling and see the solutions as associated Legendre functions (this is where I am no longer worth anything).
The cyclic boundary condition for the matching of the wavefunction at phi=0 and 2π restricts the magnetic quantum number to positive and negative integer values (including 0). The additional requirement that the wavefunction also match on a journey over the poles results in the introduction of the orbital angular momentum quantum number, l, with non-negative integer values. However, the presence of the magnetic quantum number in the second equation implies that the ranges of the two quantum numbers are linked, and it turns out (just as we expected) that the orbital angular momentum quantum number, l, can take the values 0, 1, 2 and so on and for a given value of l the magnetic quantum number ranges in integer steps from -l to +l.

Answer 7

Now all this is good.... most of the way, but when Stern and Gerlach did their experiment they saw that two bands on their measure instrument meaning electrons would have two possible fates when exposed to a inhomogenous magnetic field.
This observation conflicts with one of the predictions from quantum mechanics, because the orbital angular momentum quantum number give rise to 2l+1 orientations, which is only equal to 2 if l =1/2, which is against what we predicted for l that most be a integer.

But it have be found that all the fermions are particles with half-integral spin and the particles with integral spin (including 0) are the bosons.

Answer 8

This is my limitations to quantum mechanics so far.