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Spectrum of linear operators

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Homework Statement


Let A be a linear transformation on the space of square summable sequences [itex]\ell[/itex]2 such that (A[itex]\ell[/itex])n = [itex]\ell[/itex]n+1 + [itex]\ell[/itex]n-1 - 2[itex]\ell[/itex]n. Find the spectrum of A.

2. The attempt at a solution
I see that A is self-adjoint, so its spectrum must be a subset of the real line. We also showed in class that it is bounded, and it's clearly the discrete analog of the second derivative, but I'm not sure how to use these facts. It seems to me that for any real λ one can construct a sequence [itex]\ell[/itex] such that (A - λ)[itex]\ell[/itex] = 0, since we then have the condition [itex]\ell[/itex]n+1 = -[itex]\ell[/itex]n-1 + (2 + λ)[itex]\ell[/itex]n, which can be used to recursively construct a sequence setting [itex]\ell[/itex]0 = a and [itex]\ell[/itex]1 = b for arbitrary a and b. I'm guessing that the problem with this is that such a sequence would only converge for certain values of λ, but I don't really know how I could show this. Any help would be greatly appreciated!
 

Answers and Replies

  • #2
Dick
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Try exploiting your hunch that it looks like a second derivative operator. Then what should eigensequences look like?
 
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  • #3
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Try exploiting your hunch that it looks like a second derivative operator. Then what should eigensequences look like?
Oh, I think I understand! This finite difference approximation has error of order h^3, so if the terms in my series are e^(kn) expanded to order n^2 I should have an eigensequence. But it seems like this sequence isn't square summable except of course for k = 0, because otherwise it will diverge at one end... so the point spectrum is just 0, and the rest of the real line is in the continuous spectrum?
 
  • #4
Dick
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Oh, I think I understand! This finite difference approximation has error of order h^3, so if the terms in my series are e^(kn) expanded to order n^2 I should have an eigensequence. But it seems like this sequence isn't square summable except of course for k = 0, because otherwise it will diverge at one end... so the point spectrum is just 0, and the rest of the real line is in the continuous spectrum?
I don't know, I haven't followed the whole argument out. I'm not really an expert, that was just supposed to be a hint. But I do think e^(kn) is an eigensequence if k<0 and it is square summable. What's the eigenvalue? You probably know more about this than I do.
 
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I don't know, I haven't followed the whole argument out. I'm not really an expert, that was just supposed to be a hint. But I do think e^(kn) is an eigensequence if k<0 and it is square summable. What's the eigenvalue? You probably know more about this than I do.
Oh, I don't think I mentioned that these are two-sided sequences, so e^(kn) for k<0 would diverge at -∞. I guess the problem is that I don't yet have a clear understanding of what a spectrum contains beyond eigenvalues. I seem to remember from class an example where an operator "seemed" to have an eigenvector, but the vector wasn't an element of the Hilbert space, so the value it corresponded to wasn't an eigenvalue, but was still an element of the spectrum. I believe the example given was the hydrogen atom potential, where the bound states are eigenvectors and the scattering states are part of the continuous spectrum since they aren't normalizable. It seems to me that this operator is similar, with the e^(kn)-like sequences acting like scattering states for all k (including k=0, I don't know what I was saying before). Can anyone clarify this?
 
  • #6
Dick
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Oh, I don't think I mentioned that these are two-sided sequences, so e^(kn) for k<0 would diverge at -∞. I guess the problem is that I don't yet have a clear understanding of what a spectrum contains beyond eigenvalues. I seem to remember from class an example where an operator "seemed" to have an eigenvector, but the vector wasn't an element of the Hilbert space, so the value it corresponded to wasn't an eigenvalue, but was still an element of the spectrum. I believe the example given was the hydrogen atom potential, where the bound states are eigenvectors and the scattering states are part of the continuous spectrum since they aren't normalizable. It seems to me that this operator is similar, with the e^(kn)-like sequences acting like scattering states for all k (including k=0, I don't know what I was saying before). Can anyone clarify this?
I think that analogy is correct. But also I think since the problem states that your sequence is in [itex]\ell_2[/itex] that the 'continuum' states don't count as part of the spectrum.
 
  • #7
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Hmm... I don't think I understand exactly what you mean. Why can't an operator on [itex]\ell[/itex]2 have a continuous spectrum? Maybe I'm just misunderstanding the concept of a continuous spectrum, but it seems to me that the sequences [itex]\ell[/itex]n = ekn (everything I said before about wanting to expand this into polynomials was rubbish) behave exactly like the scattering states of a potential. (A[itex]\ell[/itex])n = (ek + e-k - 2)[itex]\ell[/itex]n, but the sequence isn't square summable. Why does the fact that it is a sequence space rather than a function space change things?
 
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  • #8
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You can write [itex]A=S+S^*-2I[/itex] with S the shift operator.

Now, do you know the spectral mapping theorem?? This states that for certain f

[tex]\sigma( f(L))=f(\sigma(L))[/tex]

Use this.
 
  • #9
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Hmm... I don't know that theorem and I don't remember it being mentioned in class (we don't have any clear cut textbook or syllabus) but I might have missed it. I'll take a look once I get the chance.
 
  • #10
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Okay, I've been trying to understand the spectral mapping theorem. So x is an operator, σ(x) is its spectrum, and f(x) is some function. Suppose σ(x) = [0, 1], and f(x) = x2 - 3I. Then the spectral mapping theorem says that the spectrum of the operator A = x2 - 3I is given by the image of f(y) = y2 - 3 on y[itex]\in[/itex][0, 1], which in this case would be [-3, -2]. Is this right?
 
  • #11
Dick
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Okay, I've been trying to understand the spectral mapping theorem. So x is an operator, σ(x) is its spectrum, and f(x) is some function. Suppose σ(x) = [0, 1], and f(x) = x2 - 3I. Then the spectral mapping theorem says that the spectrum of the operator A = x2 - 3I is given by the image of f(y) = y2 - 3 on y[itex]\in[/itex][0, 1], which in this case would be [-3, -2]. Is this right?
I've been brushing up on the subject a little since I last realized I didn't know what I was talking about. And I think I know enough to confirm that that sounds right.
 
  • #12
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I've been brushing up on the subject a little since I last realized I didn't know what I was talking about. And I think I know enough to confirm that that sounds right.
In that case, since A = S + S* - 2I, and we know that the spectrum of S is |λ| = 1, the spectrum of A should be given by ei[itex]\theta[/itex] + e-i[itex]\theta[/itex] - 2 = 2cos[itex]\theta[/itex] - 2, 0 ≤ [itex]\theta[/itex] < 2∏, which is [-4, 0]. Is this correct? This seems like it might be equivalent to constructing the sequences ekn, but restricting k to be imaginary... I'm not sure if makes sense to think of it that way though.
 
  • #13
Dick
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In that case, since A = S + S* - 2I, and we know that the spectrum of S is |λ| = 1, the spectrum of A should be given by ei[itex]\theta[/itex] + e-i[itex]\theta[/itex] - 2 = 2cos[itex]\theta[/itex] - 2, 0 ≤ [itex]\theta[/itex] < 2∏, which is [-4, 0]. Is this correct? This seems like it might be equivalent to constructing the sequences ekn, but restricting k to be imaginary... I'm not sure if makes sense to think of it that way though.
Sure, A=f(S) where f(z)=z+1/z-2. So if you know the spectrum of S, then you know the spectrum of f(S). I think. I'm a little vague on some of this as well, so I hope micromass will check in.
 

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