# Probably easy perturbation theory question (quantum)

1. Feb 22, 2007

### quasar987

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.

1. The problem statement, all variables and given/known data

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 $\vec{E}=-2Cz\hat{z}$ derivable from a potential $V=Cz^2$. 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?

2. Relevant equations

Perturbation theory: Given an hamiltonian $H=H_0+W$ and a g-fold degenerate energy $E_n^0$ of H_0, the first order corrections due to the presence of W are the eigenvalues of the gxg matrix $<\phi_n^i|W|\phi_n^{i'}>$, the $\phi_n^i$'s being the g eigenstates of eigenvalue $E_n^0$.

Wigner-Eckart Theorem: $T_Q^{(K)}$ (Q=-K,...,K) are the components of an irreducible tensorial operator of rank K if and only if the matrix elements of $T_Q^{(K)}$ 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 $<J', K; M', Q|J,M>$ via a proportionality constant independent of M. More precisely, iff

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

Theorem: If {|n,J,M>} is a standard basis of the state space, then the matrix element $<n,J,M|T_Q^{(K)}|n',J',M'>$ is zero as soon as $M\neq Q+M'$

3. The attempt at a solution

Here, W=qCz² and in the formalism of the W-E thm, we shall write W as $W_0^{(0)}$ 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 $$E_{n^*l^*}$$. According to perturbation theory then, we have that the first order corrections $\epsilon_1^{m_{l^*}}$, $$(m_l^*=-l^*,...,l^*)$$ are the eigenvalues of the matrix

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

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

$$\epsilon_1^{m_{l^*}}=<n^*,l^*,m_{l^*}|W_0^{(0)}|n^*,l^*,m_{l^*}>$$

Or, according to W-E,

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

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 $j_1=l^*$ and other the other is $j_2=0$. In this case, J can only take the value l* and if M=$m_{l^*}$, we get

$$|l^*,m_{l^*}>=\sum_{m_1=-l^*}^{l^*}<l^*,0;m_1,0|l^*,m_{l^*}>|l^*,m_1>|0,0>$$

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

$$|l^*,m_{l^*}>=<l^*,0;m_{l^*},0|l^*,m_{l^*}>|l^*,m_{l^*}>|0,0>$$,

i.e.,

$$<l^*,0;m_{l^*},0|l^*,m_{l^*}>=1$$

So there we have it:

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

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!

Last edited: Feb 22, 2007
2. Feb 22, 2007

### Physics Monkey

Hi quasar,

$$W = qCz^2$$ is not a spherical tensor of rank 0! Rank 0 spherical tensors are scalars, but $$z^2$$ is not a rotational scalar. You need to figure out what the spherical tensor components of $$z^2$$ are before you go any further. Hint: look at spherical harmonics for a clue.

Hope this helps.

3. Feb 22, 2007

### quasar987

Your post was the first time I heard "spherical tensor". I wikipediated it and got redirected to two sites that talk a bit about that...

http://electron6.phys.utk.edu/qm2/modules/m4/wigner.htm

http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/TensorOperators.htm

but neither help demistifying your

Are you asserting that to any operator, there corresponds spherical tensors components or something?

What I got out of my lectures is that a set of three scalar operators are the components of a spherical tensor of rank 1 (or vector operator) if they satisfy a set of commutation relation with J_z similar to the ones the spherical harmonics have with J_z. (Or more generally, one can define a spherical tensor of rank K in a similar manner).

You said z² is not of rank 0. So I suppose I want to find what rank z² is and what are its spherical components. How can one do that?

Last edited: Feb 22, 2007
4. Feb 22, 2007

### quasar987

I may have squeezed something out of your spherical harmonics hint there.

I noticed that z=r²cos²O, and that

$$Y_{2,0}=\sqrt{\frac{5}{16\pi}}(3\cos^2\theta-1) \ \ \Leftrightarrow \ \ \frac{1}{3}\left(\sqrt{\frac{16\pi}{5}}Y_{2,0}+1\right)=\cos^2\theta$$

So

$$z^2=\frac{r^2}{3}\sqrt{\frac{16\pi}{5}}Y_{2,0}+\frac{r^2}{3}$$

And if it weren't for that aditionnal r²/3, this would have behaved under commutation with J exactly like a spherical harmonic.. :/

Last edited: Feb 22, 2007
5. Feb 22, 2007

### Physics Monkey

Sorry for the confusion, "spherical tensor" is just a more technical name for what you've called a tensor operator.

You know what these things are already, I assure you. A scalar (1 component) is a tensor operator of rank 0. J^2 is the basic example of a scalar. A vector (3 components) is a tensor of rank 1. J is the basic example of a vector. But one must be careful about what one means by components. In this language vectors are not specified by their cartesian components but rather by their "spherical" components (hence the name spherical). For example, we use J+ and J- instead of Jx and Jy, while Jz is the same in both. You can think of J+ as the M = 1 component of the rank L = 1 tensor J. Similarly, J- is the M = -1 component while Jz is the M = 0 component. Notice the complete analogy with the L = 1 angular momentum states.

What are the L = 2 spherical tensors? For such a tensor M can range from -2 to 2, so we need 5 components. You can convince yourself with a little counting that such tensors correspond to the symmetric traceless part of 3x3 matrices (the trace is a scalar and the antisymmetric part is a vector, count the components).

Now hopefully you can begin to see what this has to do with z^2. This quantity (which isn't a scalar or a vector) can be thought of as the bottom right component of a matrix given by [ xx xy xz ; yx yy yz ; zx zy zz ]. The trace of this matrix is simply x^2 + y^2 + z^2 which is a rotational scalar as I claimed. Your job is to figure out how to write z^2 in terms of spherical harmonics and the radius r. As warm up you might try working out what x or y or z are in terms of spherical harmonics.

6. Feb 22, 2007

### Physics Monkey

Yes, you've got it! Remember that J doesn't do anything to r since r is a rotational scalar.

7. Feb 22, 2007

### quasar987

Forgive me for being slow but why was it my job to figure out how to write z² in terms of spherical harmonics and the radius r? I wanted to prove that z² was some component of a tensorial operator. How do I know what component of a tensor of what rank z² is, knowing that

$$z^2=\frac{r^2}{3}\sqrt{\frac{16\pi}{5}}Y_{2,0}+\frac{r^2}{3}$$

Last edited: Feb 22, 2007
8. Feb 22, 2007

### Physics Monkey

It's quite alright. Since you've written z^2 ~ A Y20 + B Y00 (Y00 is just a constant), and since the spherical harmonics are eigenstates of angular momentum, the correspondence I mentioned earlier suggests that z^2 is made up of L = 2 M = 0 and L = 0 M = 0 spherical tensor pieces.

9. Feb 22, 2007

### quasar987

So we can write

$$W=qCz^2=qCT_0^{(2)}+qCT_0^{(0)}$$

and then apply W-E.

Thank you so much Physics Monkey.