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quasar987

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Note that the post is long but only because I wanted to make the content cristal clear. The same post could easily have been 10 lines long.

A spinless particle of charge q is in a spherically symetric potentiel V(r). The energy levels depend on l but not on m_l. The system is now perturbed by a weak electric field [itex]\vec{E}=-2Cz\hat{z}[/itex] derivable from a potential [itex]V=Cz^2[/itex]. For a subspace of a given value of l, what can we say about the first order corrections of the energy? (Use Wigner-Eckart's theorem). Is the degeneracy completely raised?

[tex]<n,J,M|T_Q^{(K)}|n',J',M'>=\frac{1}{\sqrt{2J+1}}<n,J||T^{(K)}||n',J'>[/tex]

Here, W=qCz² and in the formalism of the W-E thm, we shall write W as [itex]W_0^{(0)}[/itex] because it is an tensorial operator of rank 0.

Next, following the question, let us fix l to, say, l=l*. Let us fix n also to, say, n=n* in order to concentrate on the corrections to the particular energy [tex]E_{n^*l^*}[/tex]. According to perturbation theory then, we have that the first order corrections [itex]\epsilon_1^{m_{l^*}}[/itex], [tex](m_l^*=-l^*,...,l^*)[/tex] are the eigenvalues of the matrix

[tex]<n^*,l^*,m_{l^*}^{(1)}|W_0^{(0)}|n^*,l^*,m_{l^*}^{(2)}>, \ \ \ \ \ \ \ \ m_{l^*}^{(1)},m_{l^*}^{(2)}=-l^*,...,l^*[/tex]

And according to the theorem stated above, these matrix elements are all 0 unless possibly those on the diagonal, because here, Q=0, M=[itex]m_{l^*}^{(1)}[/itex], and M'=[/itex]m_{l^*}^{(2)}[/itex], so the only (possibly) non nul matrix elements are those for which [itex]m_{l^*}^{(1)}=0+m_{l^*}^{(2)}[/itex], i.e the diagonal elements. But the eigenvalues of a diagonal matrix are precisely the elements of its diagonal. So we have directly,

[tex]\epsilon_1^{m_{l^*}}=<n^*,l^*,m_{l^*}|W_0^{(0)}|n^*,l^*,m_{l^*}>[/tex]

Or, according to W-E,

[tex]\epsilon_1^{m_{l^*}}=\frac{<n^*,l^*||W^{(0)}||n^*,l^*>}{\sqrt{2l^*+1}}<l^*,0;m_{l^*},0|l^*,m_{l^*}>[/tex]

Alright, but these Clebschs do not figure in the tables because they are kind of the "trivial Clebsch" in that they arise when you want to express the states of a two-particle system one of which whose angular momentum is [itex]j_1=l^*[/itex] and other the other is [itex]j_2=0[/itex]. In this case, J can only take the value l* and if M=[itex]m_{l^*}[/itex], we get

[tex]|l^*,m_{l^*}>=\sum_{m_1=-l^*}^{l^*}<l^*,0;m_1,0|l^*,m_{l^*}>|l^*,m_1>|0,0>[/tex]

But all the Clebsch that don't satisfy [tex]m_{l^*}=m_1+0[/tex] are null, so the whole expansion shrinks to

[tex]|l^*,m_{l^*}>=<l^*,0;m_{l^*},0|l^*,m_{l^*}>|l^*,m_{l^*}>|0,0>[/tex],

i.e.,

[tex]<l^*,0;m_{l^*},0|l^*,m_{l^*}>=1[/tex]

So there we have it:

[tex]\epsilon_1^{m_{l^*}}=\frac{<n^*,l^*||W^{(0)}||n^*,l^*>}{\sqrt{2l^*+1}}[/tex]

Each first order correction to the energy is equal and the degeneracy remains intact.

Have I goofed somewhere? The answer does not depend in any way in the form (W=qCz²) of the perturbation, which I found extremely dubious!

## Homework Statement

A spinless particle of charge q is in a spherically symetric potentiel V(r). The energy levels depend on l but not on m_l. The system is now perturbed by a weak electric field [itex]\vec{E}=-2Cz\hat{z}[/itex] derivable from a potential [itex]V=Cz^2[/itex]. For a subspace of a given value of l, what can we say about the first order corrections of the energy? (Use Wigner-Eckart's theorem). Is the degeneracy completely raised?

## Homework Equations

__Perturbation theory:__Given an hamiltonian [itex]H=H_0+W[/itex] and a g-fold degenerate energy [itex]E_n^0[/itex] of H_0, the first order corrections due to the presence of W are the eigenvalues of the gxg matrix [itex]<\phi_n^i|W|\phi_n^{i'}>[/itex], the [itex]\phi_n^i[/itex]'s being the g eigenstates of eigenvalue [itex]E_n^0[/itex].__Wigner-Eckart Theorem:__[itex]T_Q^{(K)}[/itex] (Q=-K,...,K) are the components of an irreducible tensorial operator of rank K if and only if the matrix elements of [itex]T_Q^{(K)}[/itex] in a standard basis of the state space {|n,J,M>} (i.e. an orthonormal set of eigenstates common to both J² and J_z) are proportional to the Clebsch-Gordan coefficients [itex]<J', K; M', Q|J,M>[/itex] via a proportionality constant independent of M. More precisely, iff[tex]<n,J,M|T_Q^{(K)}|n',J',M'>=\frac{1}{\sqrt{2J+1}}<n,J||T^{(K)}||n',J'>[/tex]

__Theorem:__If {|n,J,M>} is a standard basis of the state space, then the matrix element [itex]<n,J,M|T_Q^{(K)}|n',J',M'>[/itex] is zero as soon as [itex]M\neq Q+M'[/itex]## The Attempt at a Solution

Here, W=qCz² and in the formalism of the W-E thm, we shall write W as [itex]W_0^{(0)}[/itex] because it is an tensorial operator of rank 0.

Next, following the question, let us fix l to, say, l=l*. Let us fix n also to, say, n=n* in order to concentrate on the corrections to the particular energy [tex]E_{n^*l^*}[/tex]. According to perturbation theory then, we have that the first order corrections [itex]\epsilon_1^{m_{l^*}}[/itex], [tex](m_l^*=-l^*,...,l^*)[/tex] are the eigenvalues of the matrix

[tex]<n^*,l^*,m_{l^*}^{(1)}|W_0^{(0)}|n^*,l^*,m_{l^*}^{(2)}>, \ \ \ \ \ \ \ \ m_{l^*}^{(1)},m_{l^*}^{(2)}=-l^*,...,l^*[/tex]

And according to the theorem stated above, these matrix elements are all 0 unless possibly those on the diagonal, because here, Q=0, M=[itex]m_{l^*}^{(1)}[/itex], and M'=[/itex]m_{l^*}^{(2)}[/itex], so the only (possibly) non nul matrix elements are those for which [itex]m_{l^*}^{(1)}=0+m_{l^*}^{(2)}[/itex], i.e the diagonal elements. But the eigenvalues of a diagonal matrix are precisely the elements of its diagonal. So we have directly,

[tex]\epsilon_1^{m_{l^*}}=<n^*,l^*,m_{l^*}|W_0^{(0)}|n^*,l^*,m_{l^*}>[/tex]

Or, according to W-E,

[tex]\epsilon_1^{m_{l^*}}=\frac{<n^*,l^*||W^{(0)}||n^*,l^*>}{\sqrt{2l^*+1}}<l^*,0;m_{l^*},0|l^*,m_{l^*}>[/tex]

Alright, but these Clebschs do not figure in the tables because they are kind of the "trivial Clebsch" in that they arise when you want to express the states of a two-particle system one of which whose angular momentum is [itex]j_1=l^*[/itex] and other the other is [itex]j_2=0[/itex]. In this case, J can only take the value l* and if M=[itex]m_{l^*}[/itex], we get

[tex]|l^*,m_{l^*}>=\sum_{m_1=-l^*}^{l^*}<l^*,0;m_1,0|l^*,m_{l^*}>|l^*,m_1>|0,0>[/tex]

But all the Clebsch that don't satisfy [tex]m_{l^*}=m_1+0[/tex] are null, so the whole expansion shrinks to

[tex]|l^*,m_{l^*}>=<l^*,0;m_{l^*},0|l^*,m_{l^*}>|l^*,m_{l^*}>|0,0>[/tex],

i.e.,

[tex]<l^*,0;m_{l^*},0|l^*,m_{l^*}>=1[/tex]

So there we have it:

[tex]\epsilon_1^{m_{l^*}}=\frac{<n^*,l^*||W^{(0)}||n^*,l^*>}{\sqrt{2l^*+1}}[/tex]

Each first order correction to the energy is equal and the degeneracy remains intact.

Have I goofed somewhere? The answer does not depend in any way in the form (W=qCz²) of the perturbation, which I found extremely dubious!

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