Symmetric and Anti-Symmetric Wavefunctions

[Ed Quanta]

I am not sure if my title to this thread is appropriate for the question I am about to ask, but it is what we are currently studying in my Quantum Mechanics class so here it goes.

Two non-interacting particles with mass m, are in 1-d potential which is zero along a length 2a and infinite elsewhere

a) What are the values of the values of the 4 lowest energies of the system?
b) What are degeneracies of these energies if two particles are
i)identical, with spin 1/2
ii)not identical but both have spin 1/2
iii)identical with spin 1

What I am having trouble with is part a, but once I am helped with that can someone point me in the right direction to solve part b which is where I guess the anti-symmetric and symmetric wave functions come into play. I know spin 1/2 corresponds to fermions, while spin 1 correspnds to bosons. But I am still not sure how to this. Thanks and peace.

 

[dextercioby]

There's only one advice:solve the SE for point a).The energy spectrum should be degenerate.

Daniel.

 

[Ed Quanta]

I know, but how do you solve the Schroding equation for 2 particles?

 

[dextercioby]

What is the Hamiltonian?

Daniel.

 

[Gokul43201]

Dex, I think you scared Ed away !

Ed, it might help to start with the classical hamiltonian.

 

[Ed Quanta]

ok, E=(n1^2 + n2^2)E1

So the 4 lowest energy levels would be when n1=1 n2=1
1,2
2,1
2,2

So we have 2E1,5E1,5E1,8E1.

Now can someone help me with part b if part a is correct here? I am unsure how to apply the various spin situations to this to find degeneracies in the energy.

 

[dextercioby]

Point b)#1.I say 2.What do you say?

Daniel.

 

[Ed Quanta]

I am sorry. I am not clear what you mean. Since in this case, there are two electrons or fermions, it seems to me that the degenerate energy level would be 5E1 since this is the only possible energy level of the 4 that I listed. n1,and n2 cannot have the same value in the case of fermions. The other two parts to b involve bosons though so I am not sure what to do with these.

 

[Gokul43201]

For a spin-S particle, what are the allowed values of $m_s$ ?

 

[Ed Quanta]

1/2,-1/2? I don't think I learned this.

 

[Gokul43201]

Correct (for spin 1/2).

What about for a spin-1, spin-3/2 or spin-2 particle ? And in general, for a spin-S particle ?

 

[Ed Quanta]

Spin S particle has 2s + 1 possible states?

 

[QMrocks]

Then in answer to part (b)...

2 non-identical spin S particles... degneracy=(2S+1)(2S+1)
2 identical spin S particles... degneracy=(2S+1)(2S)

 

[Ed Quanta]

Yeah but how does this pertain to degeneracies in energy values?

 

[QMrocks]

These degeneracies are total possible of states (characterized by quantum numbers say [n,ms] in this case) that gives rise to same E. Only identical particle have to subject to Pauli principal which forbids 2 particles having same set of [n,ms].

 

[QMrocks]

Sorry... it should be [n1,n2,ms] not [n,ms]

 

[dextercioby]

...make that fermions and Pauli principle (sic).

Daniel.

 

[Ed Quanta]

Ok,that makes sense to me. Where I wrote down the four lowest energy values, I wrote 5E1 down twice, since this energy could be obtained where n1=1 and n2=2 and vice versa. Thus this energy level is doubly degenerate. So if I were to calculate the total degeneracy of this level in the case of let's say two identical spin 1/2 particles, would the total number of degeneracies increase by 2? What about for two non identical spin 1/2 particles?

And I didn't think the Pauli exclusion applied to identical spin 1 particles since they are bosons.

 

[dextercioby]

Yeah,spin degeneration is 2s+1...For each level.So 2 identical 1/2 particles should have distinct spin eigenvalues.For different 1/2 particles could have the same spins states simultaneously.

Daniel.

 

[QMrocks]

Hmm... i supposed Pauli exclusion applies only to Fermions (identical particles whose total wave functions have to be anti-symmetric). So if your questions says its Boson, then we can ignore Pauli exclusion..

 

[dextercioby]

identical fermions.If they're not identical,then they can be treated as independent particles (see the H atom:electron & nucleus,both fermions,yet they're different).

Daniel.

 

[QMrocks]

identical fermions.If they're not identical,then they can be treated as independent particles (see the H atom:electron & nucleus,both fermions,yet they're different).

Daniel.

i supposed the electron and nuclues in H are not called identical... though they are both Fermions.

 

[QMrocks]

does anyone knows why Fermions always have spin which are half integer, while Boson must have full integer spins?

 

[dextercioby]

Yes,it's the theorem spin-statistics of Schwinger,Pauli & Lüders.An essential result in QFT.

Daniel.

 

[Gokul43201]

This is a consequence of Pauli's spin-statistics theorem.

Edit : Oops ! Guess I had that page open too long without refreshing. Yeah, what Dexter said.

 

[rainbowings]

Comments:
1. the potential well is not, I presume "infinite elsewhere" - it is infinite only at the boundaries [lxl = a, with origin symmetrically chosen]. More clearly
V = 0, lxl not= a,
V = infinity, lxl = a

2. Once that is stated clearly, the hamiltonian is evidently

H = (-hbar^2/2m)[(grad1)^2 + (grad2)^2], lxl not= a

3. there is nothing classical about the hamiltonian!

4. since the particles are non -interacting, the energy eigenvalues of the total hamiltonian are just the sums of the energy eigenvalues for the individual hamiltonia.
E = E_1 + E_2
= [(n_1)^2 + (n_2)^2] E_0 , where E_0 = hbar^2/2m(2a)^2
n_1, n_2 = 1, 2, 3, ...

So the lowest allowed values of E are
2E_0, 5E_0, 8E_0, 10E_0

5. These correspond to (n_1, n_2 ) = (1, 1), [(1, 2), (2, 1)], (2, 2), [(1, 3), (3, 1)]
From here, the degenerecies can easily be enumerated, where necessary, including spin.

6. If the paricles are non identical, they won't usually have the same mass, so the E_0 term (which contains the mass) will be different for each particle now - so this has to be taken into account while adding to get the energy.

 

[Dream Light]

can you help me ?

1. Time-independent perturbation theory – a degenerate case involving angular momentum. An isotropic two-dimensional harmonic oscillator has Hamiltonian
H(0)=(p^2)/2m+(r^2)m(w^2)/2=(1/2m)[(Px)^2+(Py)^2]+(mw^2)(x^2+y^2)/2 [noticen the equation left, the '(0)' of H(0) is a superscript]

question:
a) We can find the energy eigenstates of this system by performing a separation of variables: looking for solutions of the form (x, y) = (x)  (y). Perform this separation and find the energy eigenvalue spectrum. (Use the results from the solution of the one-dimensional oscillator rather than re-deriving them. Introduce two sets of raising and lowering operators, one set for the x-motion, a†x and ax, and one set for the y-motion, a†yand ay. Eigenvalues and states thus carry two quantum numbers.) What is the degeneracy of each level?

b) The first excited eigenstate has a two-fold degeneracy. Now add to the Hamiltonian a perturbing term of the form H(1) = gL that lifts this degeneracy, where g 0 and L = xpy - ypx is the angular momentum in two dimensions (it only has one component; if the particle is charged then the z component of the magnetic field produces such a term, as we will see later). Obtain the matrix elements of L among the two degenerate first excited states that you obtained in part a). One convenient way to do this is to write L in terms of the raising and lowering operators. Find the good linear combinations of these states that are the eigenstates of this two-by-two matrix.

c) For these two states, what are the energy eigenvalues (with H = H(0) + gL) to first order in perturbation theory?

 

[Dream Light]

Comments:
1. the potential well is not, I presume "infinite elsewhere" - it is infinite only at the boundaries [lxl = a, with origin symmetrically chosen]. More clearly
V = 0, lxl not= a,
V = infinity, lxl = a

2. Once that is stated clearly, the hamiltonian is evidently

H = (-hbar^2/2m)[(grad1)^2 + (grad2)^2], lxl not= a

3. there is nothing classical about the hamiltonian!

4. since the particles are non -interacting, the energy eigenvalues of the total hamiltonian are just the sums of the energy eigenvalues for the individual hamiltonia.
E = E_1 + E_2
= [(n_1)^2 + (n_2)^2] E_0 , where E_0 = hbar^2/2m(2a)^2
n_1, n_2 = 1, 2, 3, ...

So the lowest allowed values of E are
2E_0, 5E_0, 8E_0, 10E_0

5. These correspond to (n_1, n_2 ) = (1, 1), [(1, 2), (2, 1)], (2, 2), [(1, 3), (3, 1)]
From here, the degenerecies can easily be enumerated, where necessary, including spin.

6. If the paricles are non identical, they won't usually have the same mass, so the E_0 term (which contains the mass) will be different for each particle now - so this has to be taken into account while adding to get the energy.

SOS!can anyone help me analysis these problems?

1. Time-independent perturbation theory – a degenerate case involving angular momentum. An isotropic two-dimensional harmonic oscillator has Hamiltonian
H(0)=(p^2)/2m+(r^2)m(w^2)/2=(1/2m)[(Px)^2+(Py)^2]+(mw^2)(x^2+y^2)/2 [noticen the equation left, the '(0)' of H(0) is a superscript]

question:
a) We can find the energy eigenstates of this system by performing a separation of variables: looking for solutions of the form (x, y) = (x)  (y). Perform this separation and find the energy eigenvalue spectrum. (Use the results from the solution of the one-dimensional oscillator rather than re-deriving them. Introduce two sets of raising and lowering operators, one set for the x-motion, a†x and ax, and one set for the y-motion, a†yand ay. Eigenvalues and states thus carry two quantum numbers.) What is the degeneracy of each level?

b) The first excited eigenstate has a two-fold degeneracy. Now add to the Hamiltonian a perturbing term of the form H(1) = gL that lifts this degeneracy, where g 0 and L = xpy - ypx is the angular momentum in two dimensions (it only has one component; if the particle is charged then the z component of the magnetic field produces such a term, as we will see later). Obtain the matrix elements of L among the two degenerate first excited states that you obtained in part a). One convenient way to do this is to write L in terms of the raising and lowering operators. Find the good linear combinations of these states that are the eigenstates of this two-by-two matrix.

c) For these two states, what are the energy eigenvalues (with H = H(0) + gL) to first order in perturbation theory?

 

[Neo_Anderson]

I am not sure if my title to this thread is appropriate for the question I am about to ask, but it is what we are currently studying in my Quantum Mechanics class

You're correct; this thread does not belong in this forum. Please take it to the appropriate forum, which is located here:
https://www.physicsforums.com/forumdisplay.php?f=152
Thank you in advance.