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If ${|n\rangle}$ are eigenvectors of an operator $A$ then $|n\rangle\langle n| $ can be expressed in terms of a finite order polynomial

$$|n\rangle\langle n| =\prod_{m\ne n} \frac{A-a_m}{a_n-a_m}$$

if the eigenvalues $a_n$ of $A$ are distinct. I am looking for a way to do a similar thing but with degenerate eigenvalues.

My difficulty is that the derivation of this relation starts out by considering the product $\prod_{m\ne n}(A-a_m) \mathbf{I}$ and then uses the relation $\mathbf{I}=\sum_k | k\rangle\langle k|$ to proceed to the result above. Starting with the product excluding the $n=m$ term is a bit awkward since it does not allow me to generalize to a case where two eigenvalues are the same.

For the case with degenerate eigenvalues can I exclude additional terms from the product to obtain the desired result? I am just looking for a hint on how to approach this, not a worked out solution.

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  • $\begingroup$ Isn't this more of a rational function than a polynomial? $\endgroup$ Commented Sep 10, 2014 at 20:32

1 Answer 1

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Hints:

  1. Assume that $H$ is a complex Hilbert space.

  2. Assume that $A:H\to H$ is a normal operator$^1$. Then a version of the Spectral Theorem says that $A$ is orthonormally diagonalizable.

  3. Let $(\lambda_i)_{i\in I}$ denote the set of different eigenvalues of $A$ with corresponding multiplicities $(m_i)_{i\in I}$.

  4. Let $P_i$ be the orthogonal projection operator on the eigenspace $\ker(A-\lambda_i {\bf 1})\subseteq H$.

  5. Then the generalization of OP's formula reads $$ P_i ~=~ \prod_{j\in I\backslash\{i\}} \frac{A-\lambda_j}{\lambda_i-\lambda_j}. $$

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$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

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  • $\begingroup$ In short it denotes the eigenspace for the eigenvalue $\lambda_i$. Equivalently, it is the kernel for the operator $A-\lambda_i {\bf 1}$. $\endgroup$
    – Qmechanic
    Commented Sep 10, 2014 at 19:20
  • $\begingroup$ and OP's formula? $\endgroup$
    – Anode
    Commented Sep 10, 2014 at 19:22
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    $\begingroup$ OP := original poster. In this case: You. $\endgroup$
    – Qmechanic
    Commented Sep 10, 2014 at 19:24
  • $\begingroup$ If, for example, I had \lambda_2=\lambda_3 and i=1..3 then I would multiply over i=1,2 ? $\endgroup$
    – Anode
    Commented Sep 10, 2014 at 19:30
  • $\begingroup$ The map $I\ni i\mapsto \lambda_i\in \mathbb{C}$ is implicitly supposed to be injective in my terminology. $\endgroup$
    – Qmechanic
    Commented Sep 10, 2014 at 19:40

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