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To find the perturbation correction (fine structure) in the case of a degenerate energy $E_n^0$, we can diagonalize the operator $W_f^n$, the restriction of $W_f$ to the eigen-space associated to $E_n^0$.

According to C. C. Tanoudji, since $W_f$ doesn't depend on the proton spin, it's possible to divide by 2 the dimension of the problem ($\frac{g_n}{2} X \frac{g_n}{2}$ matrix instead of $g_n X g_n$ matrix) and diagonalize a sub-matrix. Why?

I would like to have a mathematical proof please.

Is the perturbation $W_f$ hermitian?

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    $\begingroup$ Answers like "math is just a tool, it's not the nature, etc." are not helpful. People who can't help, please don't disturb. $\endgroup$
    – aayyachi
    Aug 17, 2015 at 21:16

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So, you want to diagonalize an operator on a subspace.

Specifically you want diagonalize $W_f^n$, the restriction of $W_f$ to the eigen-space associated to $E_n^0$.

[Since] $W_f$ doesn't depend of the proton spin, it's possible to divide by 2 the dimension of the problem ($\frac{g_n}{2} X \frac{g_n}{2}$ matrix instead of $g_n X g_n$ matrix).

If an eigenvector in $E_n^0$ looks like a linear combination of $\{\Psi_n\otimes|+\rangle,\Psi_n\otimes|-\rangle\}$ where the $|\pm\rangle$ is the spin of the proton then we care about how $W_f$ acts on an arbitrary vector in the span of $\{\Psi_n\otimes|+\rangle,\Psi_n\otimes|-\rangle\}.$

Now supposed we find some specific functions $(\Phi_1, \Phi_2, \dots, \Phi_k)$ that are orthogonal and such that $W_f\Phi_i\otimes|+\rangle=\alpha_i\Phi_i\otimes|+\rangle$ then the operator is diagonal (this is the dimension half version of the problem). If we can solve it (i.e. find the $(\Phi_1, \Phi_2, \dots, \Phi_k)$ that satisfy $W_f\Phi_i\otimes|+\rangle=\alpha_i\Phi_i\otimes|+\rangle$) then since the perturbation didn't depend on the proton spin we also get $ W_f\Phi_i\otimes|-\rangle=\alpha_i\Phi_i\otimes|-\rangle$ this means that $\{\Phi_k\otimes|+\rangle,\Phi_k\otimes|-\rangle\}$ is a set of orthogonal eigenvectors (orthgonality follows since $(\Phi_1, \Phi_2, \dots, \Phi_k)$ are orthogonal).

So we have a bunch of orthogonal eigenvectors if there are enough of them, we are done, we've diagonalized the matrix. So the span of $\{\Psi_n\otimes|+\rangle,\Psi_n\otimes|-\rangle\}$ was of dimension $g_n$ so $\{\Psi_n\otimes|+\rangle\}$ is $g_n/2$ dimensional. So if this smaller problem can be diagonalized then we are fine.

Is it obvious it is diagonalizable? It isn't even obvious it is an operator, you can restrict your domain but to be an operator the range needs to be in the domain (or maybe the closure of the domain if you like unbounded operators.) You can project onto the domain to force it or be an operator. (For the projection composed with the function to be an operator.) Then you still have to ask whether it is diagonalizable. So we need to know if there are enough eigenvectors in the reduced space.

Is the perturbation $W_f$ hermitian?

If so, that solves the above problem. If the goal is to approximate a hermitian operator then having hermitian perturbations is reasonable, but it is not obvious that it has to be hermitian. However we can argue that if the perturbation operator is hermitian in the larger space then we can get that restriction to the smaller problem to be diagonalizable.

For that, we want to show the diagonalization can happen and has real eigenvalues. If we know the diagonalization happens (with real eigenvalues) in the larger space then any eigenvector $A\otimes(\alpha_+|+\rangle\alpha_-|-\rangle)$ with eigenvalue $a\in\mathbb R$ implies the state $A\otimes(\alpha_+|+\rangle-\alpha_-|-\rangle)$ is also an eigenvector with the same eigenvalue and so is the whole space spanned since it only differs by proton spin. And therefore so is $A\otimes|+\rangle$ and $A\otimes|-\rangle.$ Then we might as well replace all the other eigenvector with the same eigenvalue (if any) with one orthogonal to both of those (restrict to the subset of the eigenspace that is orthogonal to those two and diagonalized that, it is diagonalizable on the whole eigenspace it is a a multiple of the identity on that eigenspace). So we keep doing that (using Zorn's lemma if needed) until we have eigenvectors that are all in the space spanned by $\{\Psi_n\otimes|+\rangle\}$ or in the space spanned by $\{\Psi_n\otimes|-\rangle\}.$ There were $g_n$ eigenvectors in larger space each was replaced with a vector in the span of $\{\Psi_n\otimes|+\rangle\}$ or in the span of $\{\Psi_n\otimes|-\rangle\}.$ (Either directly or when one of the other ones was place in one but the point is that we had a diagonalized where all the $g_n$ eigenvectors are in span of $\{\Psi_n\otimes|+\rangle\}$ or in the span of $\{\Psi_n\otimes|-\rangle\}.$)

Since the eigenvectors in the larger space were orthogonal and nonzero they were linearly independent. So the ones in the span of $\{\Psi_n\otimes|+\rangle\}$ are linearly independent and the ones in the span of $\{\Psi_n\otimes|-\rangle\}$ Ade linearly independent. Since those spaces are of dimension $g_n/2$ (assuming $g_n$ is finite so the division makes sense) then there is at most $g_n/2$ vectors in each. If you tried to put $m$ fewer in one then the other needs have $g_n/2+M$ so that they add up to $g_n$ vectors but that would exceed $g_n/2$ for the other one so there must be $g_n$ exactly in each one. That's exactly enough to have the restricted one be diagonalizable. But these all have real eigenvalues so they are hermitian.

That means when we looked for eigenvectors of the smaller problem we could find $g_n/2$ orthogonal eigenvectors with real eigenvalues, it is hermitian, so it could be done.

why could we subdivide the matrix when it's still not diagonalized?

I'm saying that if you diagonalized the $g_nxg_n$ matrix then you can block diagonalise according to eigenspaces then for each eigenspace find a vector V then its projections onto the two $g_n/2$ dimensional spaces will be in the same eigenspace and any two orthogonal vectors in that eigenspace are eigenvectors. So we can pick one, project it onto both the $g_n/2$ dimensional spaces and if one of them gives zero take the other one and replace the $|\pm\rangle$ with $|\mp\rangle$ so we end up with two vectors down, then pick a random vector in the eigenspace that is orthogonal to all the ones we have so far and repeat. So you get your eigenvectors for the big matrix to be all vectors in the two $g_n$ dimensional spaces. Thus the restriction to the smaller space is an operator and is diagonalizable.

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  • $\begingroup$ I didn't understand the 2d part. Anyway, why could we subdivide the matrix when it's still not diagonalized? Thank you. $\endgroup$
    – aayyachi
    Aug 17, 2015 at 23:32
  • $\begingroup$ If you expand the matrix in the basis Timaeus provided, you'll see that it's block diagonal, e.g. it looks like $$\begin{pmatrix} A & 0 \\ 0 & A \end{pmatrix}$$ That is, it's block diagonal. When we subdivide, we're just throwing away the extra copy of $A$. $\endgroup$
    – knzhou
    Aug 18, 2015 at 2:03

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