Showing that antisymmetry in a two nucleon wave function implies that L + S + T = odd

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Homework Statement


Show that the antisymmetry of the two nucleon wave function in an oscillator model implies that L + S + T = odd. Secondly would this condition change if one worked in a more general single particle model?

T = isospin
S = intrinsic spin
L = orbital angular momentum

The Attempt at a Solution



Using the Slater determinant gives us the antisymmetric wave function for a two nucleon system such that.

[tex]\psi(1,2) = 1/\sqrt(2)*[\varphi_{j1 m1} (1)\varphi_{j2 m2} (2) - \varphi_{j2 m2} (1)\varphi_{j1 m1} (2)][/tex]

Which can be rephrased as...

[tex]\psi(1,2) = <j_1 m_1 j_2 m_2|JM>\varphi_{j1 m1} (1)\varphi_{j2 m2} (2) + <j_2 m_2 j_1 m_1|JM> \varphi_{j2 m2} (1)\varphi_{j1 m1} (2)[/tex]

Using the transformation property of the Clebsch Gordon co-efficients we can substitute the below into the above.

[tex]<j_2 m_2 j_1 m_1|JM> = (-1)^{j_1+j_2-J}<j_1 m_1 j_2 m_2|JM> [/tex]

[tex]\psi(1,2) = <j_1 m_1 j_2 m_2|JM>\varphi_{j1 m1} (1)\varphi_{j2 m2} (2) + (-1)^{j_1+j_2-J<j_1 m_1 j_2 m_2|JM>\varphi_{j2 m2} (1)\varphi_{j1 m1} (2)[/tex]

According to the clebsch gordons of the two nucleon problem we should find the following.

[tex]\psi(1,2) = 1/\sqrt(2)*[1-(-1)^{j_1+j_2-J}]\varphi_{j1 m1} (1)\varphi_{j2 m2} (2)[/tex]

Rephrasing in terms of L & S...

[tex]\psi(1,2) = 1/\sqrt(2)*[1-(-1)^{l_1+s_1+l_2+s_2-L-S}]\varphi_{j1 m1} (1)\varphi_{j2 m2} (2)[/tex]

We know for a two fermion system the spin's will each be a half thus S = 1, we also know that for a two nucleon system isospin will be a half for each nucleon therefore T = 1;

So the above equation now reads...

[tex]\psi(1,2) = 1/\sqrt(2)*[1-(-1)^{l_1+l_2-L}]\varphi_{j1 m1} (1)\varphi_{j2 m2} (2)[/tex]

Two things to note here from the above argument.

If [tex] l_1 +l_2 - L = even [/tex] then the wavefunction = 0 however if it = odd then the wave function is non-zero and therefore possible.

L + T + S = L + 2.

I would like someone to please help me figure out how L can be odd in order for L + T + S to make sense /Or possibly point out a flaw in my logic/ better way to do this problem.

This is my first time posting so I'm not sure if I've got this in the right section. Please feel free to move it.

Thanks
Darryl Fleming
 

Answers and Replies

  • #2
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Is this too confusing? What have I done wrong here? Have I not given this enough time? Should I send this to a higher up forum?
 
  • #3
Hao
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I probably don't understand the topic well enough, but how does the equality

[tex]
\varphi_{j1 m1} (1)\varphi_{j2 m2} (2) = \varphi_{j2 m2} (1)\varphi_{j1 m1} (2)
[/tex]

come about in moving from [tex]
\psi(1,2) = <j_1 m_1 j_2 m_2|JM>\varphi_{j1 m1} (1)\varphi_{j2 m2} (2) + <j_2 m_2 j_1 m_1|JM> \varphi_{j2 m2} (1)\varphi_{j1 m1} (2)
[/tex]

to [tex]
\psi(1,2) = 1/\sqrt(2)*[1-(-1)^{j_1+j_2-J}]\varphi_{j1 m1} (1)\varphi_{j2 m2} (2)
[/tex]

should both wavefunctions represent different states, ie. [tex]
\varphi_{j1 m1}(1) \neq \varphi_{j2 m2} (1)
[/tex]?

Additionally, how does the final answer demonstrate antisymmetry with respect to swapping the positions of particles 1 and 2?
 
  • #4
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Applying the Clebsch Gordon transformation allows us to do [tex]

\varphi_{j1 m1} (1)\varphi_{j2 m2} (2) = \varphi_{j2 m2} (1)\varphi_{j1 m1} (2)

[/tex].

Yes the wave functions do represent different states.

The antisymmetry of the two nucleon system is intrinsic to the system. So the final answer shouldn't demonstrate the antisymmetry, as it's given. More like the antisymmetrical nature of the two nucleon system should help you to find that L + S + T = odd. I might have misinterpreted your question.

I have chatted with two different lectures in my department. The super clever one that no one understands led me down this route. However after chatting with the other one he showed me a more intuitive and less threatening way of going about this.

I'll post it soon.

Darryl Fleming
 
  • #5
4
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A wavefunction can be described as follows.

[tex]
\psi_{total} = \psi_L \chi_S \chi_T
[/tex]

Where [tex]\psi_{total}[/tex] refers to the total wave function and [tex]\psi_L
,\chi_S \mbox{ and } \chi_T [/tex] refer to the orbital angular momentum,
the spin and the isospin parts of the wave function.

Some of the things we know at the outset are as follows

The total wavefunction is antisymmetric
[tex]
T = 0 ,1 \mbox{ The possible isospin values for two nucleons as }t_1 = 1/2,t_2 = 1/2
[/tex]
[tex]
S = 0 , 1 \mbox{ The possible spin values for two fermions as}s_1 = 1/2 , s_2 = 1/2
[/tex]
L is an integer


I will now attempt to show how for every possible combination of T = 0,1 and S = 0,1 we find that L + S + T = odd.

The first case I will deal with is T = 0, S = 0.

If T = 0 then [tex]\chi_T[/tex] is odd and we know [tex]\psi_{total}[/tex] is odd then
that implies that [tex]\Pi(\psi_L \chi_S)[/tex] must be even.

Now if S = 0 then [tex]\chi_S[/tex] is odd but since [tex]\Pi(\psi_L \chi_S)[/tex] is
even that implies that L must be an odd integer (Integer due to 1.5).

We now find that T + S + L = 0 + 0 + odd integer = odd.

Thus it works for the T = 0, S = 0 case...

and so on and so forth for T = 0 S = 1, T = 1 S = 0 and T = 1 S = 1...

Thus in all cases we can see that T + L + S = odd.
 
  • #6
Hao
93
0


Ah, thanks. I think I see where I went wrong.

I was thinking about only the position space representation (radial + angular) of the wavefunction, but the wavefunction in question here is for the spin (spinor), isospin (spinor), and angular momentum (position space; angular) components.

The fact that you are using Clebsch Gordon coefficients should have clued me in.

As for antisymmetry, I was referring to [tex]\psi(1,2) = - \psi(2,1)[/tex] (positions).
 
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