The Naked Scientists

The Naked Scientists Forum

Poll

QM angular momentum question

abc
0 (0%)
def
0 (0%)

Total Members Voted: 0

Author Topic: Quantum Mechanics  (Read 2044 times)

Offline kolahalb

  • First timers
  • *
  • Posts: 1
    • View Profile
Quantum Mechanics
« on: 25/10/2007 17:29:50 »
After developing the concept of quantum mechanical angular momentum
[ I mean the relation L_z=m (h') where h' means h/2π
newbielink:http://hyperphysics.phy-astr.gsu.edu [nonactive]...um/qangm.html]

a book says:

"The quantum mechanical origin of these strange restrictions lies in the require-
ment that if either the particle or the laboratory is turned through a complete
rotation around any axis,the observed situation will be the same as before the
rotation.Because observables are related to the square of the wavefunction,the
wavefunction must turn into either plus or minus itself under a rotation by 2π
radians.Its sign remains unchanged if the angular momentum around the rotation
axis is an integer multiple of h(i.e.,forbosons)but changes if the angular momen-
tum around the rotation axis is a half-integer multiple of h (i.e.,forfermions).
Because of this difference in sign under 2π rotations,bosons and fermions each
obey a different type of quantum statistics"

I cannot exactly follow the book here.How can exp[i 2π] result in - of the same wave function?

can anyone please explain?


 

Offline lightarrow

  • Neilep Level Member
  • ******
  • Posts: 4586
  • Thanked: 7 times
    • View Profile
Quantum Mechanics
« Reply #1 on: 26/10/2007 08:21:29 »
After developing the concept of quantum mechanical angular momentum
[ I mean the relation L_z=m (h') where h' means h/2π
http://hyperphysics.phy-astr.gsu.edu...um/qangm.html]

a book says:

"The quantum mechanical origin of these strange restrictions lies in the require-
ment that if either the particle or the laboratory is turned through a complete
rotation around any axis,the observed situation will be the same as before the
rotation.Because observables are related to the square of the wavefunction,the
wavefunction must turn into either plus or minus itself under a rotation by 2π
radians.Its sign remains unchanged if the angular momentum around the rotation
axis is an integer multiple of h(i.e.,forbosons)but changes if the angular momen-
tum around the rotation axis is a half-integer multiple of h (i.e.,forfermions).
Because of this difference in sign under 2π rotations,bosons and fermions each
obey a different type of quantum statistics"

I cannot exactly follow the book here.How can exp[i 2π] result in - of the same wave function?

can anyone please explain?

Sincerely I don't know anything about spin;
anyway, if, mathematically, the wavefunction contained a term of the kind: e(iθS/h), then, in the case of fermions (S = 1/2 h) you would have eiθ/2 and when θ = 2π you have e = -1, while for bosons (S = nh) you would have ei2πn = +1.

(Can you explain me the poll?)
« Last Edit: 26/10/2007 08:23:49 by lightarrow »
 

The Naked Scientists Forum

Quantum Mechanics
« Reply #1 on: 26/10/2007 08:21:29 »

 

SMF 2.0.10 | SMF © 2015, Simple Machines
SMFAds for Free Forums