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Griffiths' explanation of degenerate perturbation theory

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Niles
#1
Feb4-09, 02:50 PM
P: 1,863
Hi all.

I'm reading about time-independent perturbation theory for degenerate states in Griffiths' Introduction to QM.

I have a question on the things he writes in chapter 6.2, page 269. What does he mean by the so-called "good" linear combinations?

I hope you can shed some light on this. Thanks in advance.

Best regards,
Niles.
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Niles
#2
Feb5-09, 06:37 AM
P: 1,863
Can I get a moderator to move this thread to the homework section, preferably the "Advanced Physics"-section? Thanks in advance.
Niles
#3
Feb5-09, 07:26 AM
P: 1,863
Ok, this is as far as I've come.

We want W to be a diagonal matrix, because then we can solve the problem just like we would solve the unpertubated problem. But in order for W to be a diagonal matrix, we would have to write it in the basis spanned by its eigenvectors.

Normally, W is given in the basis of H0, i.e. the space spanned by the unpertubated wave functions.

And this is where I can't go any further.

ice109
#4
Feb5-09, 12:37 PM
P: 1,705
Griffiths' explanation of degenerate perturbation theory

i started writing a response to this yesterday and quit half way because though i've just suffered through trying to interpret griffith on this as well i still don't get it.

here's what i got:

when we write down the matrix representation of H' in the eigenbasis of H0 it's not diagonal, like H0 is in its eigenbasis. but we don't need it to be if the eigensystem of H0 isn't degenerate, the eigenvalues are distinct. perturbation theory still works. if the eigensystem of H0 is degenerate then perturbation theory breaks down, ( in the denominator of some of the terms where there's a difference , a minus sign ).

what the breakdown actually is is a zero in the denominator which is an infinity. it occurs when you consider two different eigenstates in that same degenerate part of the eigenbasis, therefore they have the same energy eigenvalue but different wavefunctions. these are of the off diagonal part of the matrix representation of H'. we fix this break down making the numerator zero, on the occasion that the denominator is zero. this equates to making the off diagonals zero which itself equates to making the basis in which H' is represented orthogonal. this is diagonalization.

i doubt that helps but post some response question and maybe we can figure it out together.

the whole argument about good and bad combinations i don't understand though what he means by 'good' combinations are the new diagonalized eigenstates in that degenerate subspace.
Niles
#5
Feb5-09, 01:20 PM
P: 1,863
First, thanks for replying. It is very comforting to know that there are other people who believe that this part of Griffiths' book is poorly written.

Try and take a look at this thread (especially posts 3 and 4): http://www.physicsforums.com/showthread.php?t=65773

What it all boils down to is that we wish to be able to use non-degenerate perturbation theory: But in order to do this, the W-matrix has to be diagonal (that's what Griffiths concludes on page 271). W is a diagonal matrix when it is expressed in its eigenspace, and to find this diagonal matrix, we must find the eigenvectors of W and use:

Mdiagonal matrix = S-1eigenvectors * W * Seigenvectors.

When we have the W-matrix in this form, we can go ahead and use: <\psi_i |H'| \psi_i>, where \psi_i are the eigenvectors of H0 as expressed in the eigenspace of W. But in Griffiths' example, he does not use this approach, rather he just goes ahead and finds the eigenvectors and their corresponding eigenvalues of W in the eigenspace of H0. This is where I cannot interpret Griffiths' procedure anymore. I'm not sure where to go from here. Does the above thread give you any ideas?

Best regards,
Niles.
ice109
#6
Feb5-09, 02:40 PM
P: 1,705
well the eigenvalues you compute end up being <\psi_i |H'| \psi_i> in the diagonalized basis of W, i mean they're constructed as such. is that what you're having a problem with? what exactly don't you understand about his prescription? the stuff about the commuting operator? that's just the change of basis operator which i can say with a cavalier attitude but don't expect me to guess the eigensystem just by looking at the symmetries of the system, which is what i think my professor expects from us.
Niles
#7
Feb5-09, 02:43 PM
P: 1,863
Quote Quote by ice109 View Post
well the eigenvalues you compute end up being <\psi_i |H'| \psi_i> in the diagonalized basis of W
That is what I do not understand. I believe that above statement is only true if \psi_i are the basisvectors of the eigenspace of H' (and thus W), but they are not. They are the basisvectors of H0.
ice109
#8
Feb5-09, 02:46 PM
P: 1,705
Quote Quote by Niles View Post
That is what I do not understand. I believe that above statement is only true if \psi_i are the basisvectors of the eigenspace of H' (and thus W), but they are not. They are the basisvectors of H0.
no they're not, they're linear combinations of the basis vectors of H0.

p.s.

that cleared up for me my last concern. thanks.
Niles
#9
Feb5-09, 03:11 PM
P: 1,863
Quote Quote by ice109 View Post
no they're not, they're linear combinations of the basis vectors of H0.

p.s.

that cleared up for me my last concern. thanks.
I'm glad to hear that.

The wavefunctions \psi0i that we use to find the first order corrections are eigenfunctions of H0. This is on page 269 (the beginning of the subsection).
ice109
#10
Feb5-09, 03:25 PM
P: 1,705
Quote Quote by Niles View Post
I'm glad to hear that.

The wavefunctions \psi0i that we use to find the first order corrections are eigenfunctions of H0. This is on page 269 (the beginning of the subsection).
yes but the one we are considering is [itex]\psi^0 = \alpha \psi_a^0 + \beta \psi_b^0 [/itex]
where [itex] \psi_a^0 [/itex] and [itex] \psi_b^0 [/itex] are eigenfunctions of H0

hopefully you're referring to what i think you are, eqns 6.16-6.18 in 6.3. my pages are numbered different from yours.
Niles
#11
Feb5-09, 03:37 PM
P: 1,863
Quote Quote by ice109 View Post
hopefully you're referring to what i think you are, eqns 6.16-6.18 in 6.3. my pages are numbered different from yours.
Yeah, the pages are numbered differently then. But we agree: It is a linear combination.

Ok, I think I should take this from the top. I think I am confusing myself.

We know that [itex]<\psi_i |H'| \psi_i>[/itex] will give us the eigenvalues of W when both H' (i.e. W) and [itex]\psi_i[/itex] are given in the eigenspace of W. But [itex]\psi_i[/itex] is given in the basis of H0, since it is a linear combination: [itex]
\psi^0 = \alpha \psi_a^0 + \beta \psi_b^0.
[/itex]

Now H and [itex]
\psi^0 = \alpha \psi_a^0 + \beta \psi_b^0.
[/itex] must be in the same basis. Which one do we choose?


This is my problem.
ice109
#12
Feb5-09, 04:23 PM
P: 1,705
you are confusing yourself

Quote Quote by Niles View Post
Yeah, the pages are numbered differently then. But we agree: It is a linear combination.

Ok, I think I should take this from the top. I think I am confusing myself.

We know that [itex]<\psi_i |H'| \psi_i>[/itex] will give us the eigenvalues of W when both H' (i.e. W) and [itex]\psi_i[/itex] are given in the eigenspace of W.
this is true

from here on you lose me
Quote Quote by Niles View Post
But [itex]\psi_i[/itex] is given in the basis of H0, since it is a linear combination: [itex]
\psi^0 = \alpha \psi_a^0 + \beta \psi_b^0.
[/itex]

Now H and [itex]
\psi^0 = \alpha \psi_a^0 + \beta \psi_b^0.
[/itex] must be in the same basis. Which one do we choose?


This is my problem.
you're over complicating things. here's the prescription

we have perturbation theory

it doesn't work when the eigensystem of the unperturbed hamiltonian has a degenerate subspace

in a degenerate subspace of an eigenspace linear combinations of vectors are also eigenvectors:

if x and y are eigenvectors of A with eigenvalue z, for both then A(2x+3y)=A3x + A2y = 3Ax+2Ay = 3zx+2zy = z(3x+2y)

hence we are allowed to choose any linear combination of vectors in that subspace when we do w/e we want with it. hence we can choose such a combination of the original eigenfunctions that diagonalizes H' which allows regular perturbation thry to work again.

so we diagonalize H' and apply perturbation theory but when you do that you're doing

[itex]<\psi^0|H'|\psi^0> [/itex] where now [itex]\psi_0 = \alpha \psi_a^0 + \beta \psi_b^0 [/itex] but that's an eigenstate of H' and hence you get it's eigenvalue which you already calculated.


there's two different diagonalization floating around here. H0 is diagonal in it's own eigenbasis representation. H' a priori isn't diagonal in that same basis, nor does it need to be because vanilla perturbation theory doesn't need H' to be diagonal but vanilla perturbation does need the eigenfunctions of the unperturbed hamiltonian. the point here is that a linear combination of eigenfunction in a degenerate subspace are still eigenfunctions and therefore all the stuff they derived in terms of them still works, just that now it doesn't blow up around H' too.
Niles
#13
Feb6-09, 02:28 AM
P: 1,863
I assume you mean "degenerate perturbation theory" when you say "vanilla perturbation theory"? Nonetheless, thank you for your description. I can see that I was on the right track before, but I was over-complicating stuff.

Now only one question remains, which is quite important actually: How are we sure that there will always exist a linear combination of eigenstates of H0 that can diagonalize H'?
ice109
#14
Feb7-09, 12:53 PM
P: 1,705
Quote Quote by Niles View Post
I assume you mean "degenerate perturbation theory" when you say "vanilla perturbation theory"? Nonetheless, thank you for your description. I can see that I was on the right track before, but I was over-complicating stuff.

Now only one question remains, which is quite important actually: How are we sure that there will always exist a linear combination of eigenstates of H0 that can diagonalize H'?
cause you can. cause what you're actually doing is orthonormalizing the basis for the degenerate eigenspace which you can always do. there are in fact several algorithms for it. more on this later
Niles
#15
Feb7-09, 03:34 PM
P: 1,863
Ok, I will look forward to it.

Thanks for helping with all the stuff above.


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