Spin Up & Spin Down in Quantum Mechanics

[trosten]

Its frequently written in my QM book that there can be two electrons in every state, one spin up and one spin down. Where does this come from? My book seems to take it as an axiom.

If we have a He atom with its electrons in the singlet state. Then as I understand it the wave function will be something like this

U = f(u,v) |0 0> where f is symmetric since |0 0> isnt.

Since electrons are fermions I take it that this must imply that f(r,r) = 0 for all r? If this is the case, is it known what order of zero f(r,r) is?
eg. f(u,v) = g(u,v)*(u-v)^n what then is n in the He case if g(r,r)<>0.

 

[selfAdjoint]

If you look at any particular line in space, the fermion space has two states with respect to it, just as you can go two ways along a road. These two orientations are arbitrarily labled "spin up" and "spin down" along that axis. The odd thing is that this completely determines the spin state; spin around another line can be obtained by a simpe change of basis.

So if you look at some electron in a shell, there are those two possibilities; it could have its spin "up" along some line you specify, or it could have it "down" along that line. Since these are different configurations, they escape from Pauli's exclusion principle and you can have two electrons in the shell, each with one of the two spin states.

 

[trosten]

yes that's what my book aslo says. But I assume that if we make a meassurement we can't find 2 electrons at the same place ?

 

[dextercioby]

yes that's what my book aslo says. But I assume that if we make a meassurement we can't find 2 electrons at the same place ?

Define "at the same place"...In what space??The R^{6},the Hilbert space of states,which space?Or you meant 2 electrons in the same quantum state??According to Mr.Pauli,you can't have two identical fermions in the same quantum state.

Daniel.

 

[Norman]

forgot the last part

It might be usefull to explain to trosten what is meant by a quantum state. What is meant by a quantum state is ALL the quantum numbers that can describe the given particle. So, for an electron, it can be described by its "principal" quantum number n, the "azimuthal" quantum number l, the "magnetic" quantum number m_l, and the spin quantum number s_z. So since the electron is a fermion (with s=1/2), Pauli told us that only one electron can be in any given quantum state. So for any given n,l,m_l there can be only 2 electrons- one with spin up, s_z=+\frac{1}{2} and one with spin down s_z=-\frac{1}{2}.
Hope that helps.
Cheers

 

[trosten]

I ask again

Perhaps I am not that god at explaining my question. My question is essentially this

Can we or can we not meassure two electrons at the same infinitesimal volume in space (in theory)?? (the space where the experiment takes place) If we have two electrons in the singlet state the pauli principle doesn't say anything about this since the anti-symmectric part gets caught in the spinpart.

For a detailed explanation of what I mean see below.

Take the example of the Helium atom where the two electrons are in the lowest energy state, the same state, represented by the same numbers n,l,m. This state can be described by some function f of six positions variables multiplied by their common spin state |0 0> ((1/sqrt(2))(up|down - down|up)). I've not seen a solution to this problem I just assume it exists.

Now say we want to find the probability of finding both electrons within the same small volume element around the positions r1 and r2 (r1=[x,y,z],r2=[u,v,w]).

The|0 0> drop out since its normalized (?).

This will then be f*(r1,r2)f(r1,r2)(d3r1)(d3r2) or
f*([x,y,z],[u,v,w]) f([x,y,z],[u,v,w]) dxdydz dudvdw

This will be the thing to integrate to find the probability of a measurement finding both particles at some volumes. Now if we try to find both particles within the same volume, and we let that volume go to a infinitesimal volume the probability will simply be
f*([x,y,z],[u,v,w]) f([x,y,z],[u,v,w]) dxdydz dudvdw where x=u, y=v and z=w.
Now can we or can we not meassure two electrons at the same infinitesimal volume in space (in theory) ??

 

[dextercioby]

Perhaps I am not that god at explaining my question. My question is essentially this

Can we or can we not meassure two electrons at the same infinitesimal volume in space (in theory)?? (the space where the experiment takes place) If we have two electrons in the singlet state the pauli principle doesn't say anything about this since the anti-symmectric part gets caught in the spinpart.

For a detailed explanation of what I mean see below.

Take the example of the Helium atom where the two electrons are in the lowest energy state, the same state, represented by the same numbers n,l,m. This state can be described by some function f of six positions variables multiplied by their common spin state |0 0> ((1/sqrt(2))(up|down - down|up)). I've not seen a solution to this problem I just assume it exists.

Now say we want to find the probability of finding both electrons within the same small volume element around the positions r1 and r2 (r1=[x,y,z],r2=[u,v,w]).

The|0 0> drop out since its normalized (?).

This will then be f*(r1,r2)f(r1,r2)(d3r1)(d3r2) or
f*([x,y,z],[u,v,w]) f([x,y,z],[u,v,w]) dxdydz dudvdw

This will be the thing to integrate to find the probability of a measurement finding both particles at some volumes. Now if we try to find both particles within the same volume, and we let that volume go to a infinitesimal volume the probability will simply be
f*([x,y,z],[u,v,w]) f([x,y,z],[u,v,w]) dxdydz dudvdw where x=u, y=v and z=w.
Now can we or can we not meassure two electrons at the same infinitesimal volume in space (in theory) ??

Yes,the spin part part really doesn't matter,because it's normalized and it would not yield infinite when integrated on the space of the spin variables.

The probability density to detect 2 electrons within an element of volume dV of the 6 dimensional space of 'positions' is indeed given by (assume stationary states)
dP_{e_{1},e_{2}}=\psi^{*}(\vec{r}_{1})\psi(\vec{r}_{1})d^{3}\vec{r}_{1} \psi^{*}(\vec{r}_{2})\psi(\vec{r}_{2}) d^{3}\vec{r}_{2}

As u can see this function is defined in a 6-dim.space.
dP:R^{3}\otimes R^{3}\rightarrow L^{2}(R^{3}\otimes R^{3})

It cannot be computed in 3 dimensions.We cannot measure electrons,we can detect them with a certain probability density in one point,or with a certain probability over a FINITE volume in its own space.


Daniel.

 

[trosten]

ok thanks for your answer, I asked this question because of when we have for example two identical fermions in stationary states f and g, then the antisymmetri operation gives the state

f(u)g(v) - g(u)f(v) where u and v are position vectors.

now as you all can see if u = v = r then f(r)g(r)-g(r)f(r) = 0 that is we then have zero probability density that the electron are detected at same point. But when the electron are in the singlet state there no antisymmetri and we can have a nonzero probability density of the electrons being detected at the same point.

Now I suspect that the schroedinger equation always will hold some sort of potential that always goes to infinity when the distance between two fermions goes to zero now that would yield two things either the energy also goes to infinity or psy goes to zero faster. What do you think about this?

 

[dextercioby]

ok thanks for your answer, I asked this question because of when we have for example two identical fermions in stationary states f and g, then the antisymmetri operation gives the state

f(u)g(v) - g(u)f(v) where u and v are position vectors.

now as you all can see if u = v = r then f(r)g(r)-g(r)f(r) = 0 that is we then have zero probability density that the electron are detected at same point. But when the electron are in the singlet state there no antisymmetri and we can have a nonzero probability density of the electrons being detected at the same point.

Nope,the antisymmetry is always present and is required by the Pauli principle.You can't by-pass it.The spin part of the wave function will always be antisymmetric and if you take two electrons to be in the same quantum state,it will always yield zero.

Now I suspect that the schroedinger equation always will hold some sort of potential that always goes to infinity when the distance between two fermions goes to zero now that would yield two things either the energy also goes to infinity or psy goes to zero faster. What do you think about this?

Depends on the system.In general,Schroedinger equation does not go with electron (fermion) description,as it is not suitable for spin description.A thorough analysis should be using Dirac's equation.Dirac's eq.uses potentials as well,and the Coulomb potential is one example of potential with 'good' behavior,meaning that one can find wavefunctions square modulus integrable.Describing systems of systems of fermions is done actually in QFT,where the spin-statistics theorem is essential.The problem becomes really very complicated.It makes the Schroedinger's equation look ridiculously simple.
So let's not mix spin,fermions and Schroedinger equation.

Daniel.

 

[trosten]

this is all a bit confusing, I am going to take one more course in QM and then I am going to study QFT. Thanks for your explanations.