Discrete hellmann-Feynman theorem ?

In summary, the Helmann-Feynman theorem states that the derivative of an eigenvalue with respect to a parameter is equal to the eigenfunction dagger multiplied by the derivative of the operator with respect to the parameter and then multiplied by the eigenfunction. This also holds true for the discrete case, as long as the matrix is Hermitian. The proof involves assuming that the eigenfunctions are normalized and using the fact that Hermitian matrices have real eigenvalues. However, this theorem does not apply to real eigenvalues of any matrix, as the matrix must be Hermitian for the proof to work.
  • #1
juliette sekx
31
0
Discrete hellmann-feynman theorem ??

The Helmann-Feynman theorem states that :
derivative of eigenvalue with respect to a parameter = eigenfunction dagger * derivative of operator with respect to parameter * eigenfunction

(assuming that the eigenfuncitons are normalized, otherwise the theorem would also include a normalization integration).

Does anyone know if this works for the DISCRETE CASE ??

ie:

derivative of eigenvalue with respect to parameter = eigenVECTOR dagger * derivative of MATRIX with respect to parameter * eigenVECTOR ??

I can't find a proof of it anywhere!
 
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  • #2


In order for the theorem to be true, the matrix has to be Hermitian.

The proof would go something like this. Let [tex] A(s) [/tex] be one-parameter smooth family of Hermitian matrices and [tex] \lambda(s) [/tex] a smooth parametrization of the eigenvalue of [tex] A(s) [/tex] corresponding to the eigenvector [tex] \mathbf{v}(s) [/tex]. (Assume that [tex] \mathbf{v} [/tex] is normalized so that [tex] \mathbf{v}^{\dagger} \mathbf{v} = 1 [/tex].) Then [tex] A \mathbf{v} = \lambda \mathbf{v} [/tex] for all [tex] s [/tex]. Differentiating this relation, we have
[tex]
A' \mathbf{v} + A \mathbf{v}' = \lambda' \mathbf{v} + \lambda \mathbf{v} \textrm{.}
[/tex]
Note that [tex] \mathbf{v}^{\dagger} A \mathbf{v} = (A^{\dagger} \mathbf{v})^{\dagger} \mathbf{v} = (A \mathbf{v})^{\dagger} \mathbf{v} [/tex], since [tex] A [/tex] is Hermitian; but [tex] A \mathbf{v} = \lambda \mathbf{v} [/tex], so [tex] \mathbf{v}^{\dagger} A \mathbf{v} = \lambda^{*} \mathbf{v}^{\dagger} \mathbf{v} = \lambda^{*}[/tex] (where stars indicate complex conjugates). However, recall that Hermitian matrices have real eigenvalues; thus, [tex] \lambda^{*} = \lambda [/tex]. Thus, multiplying the above equation by [tex] \mathbf{v}^{\dagger} [/tex] on the left, we have
[tex]
\mathbf{v}^{\dagger} A' \mathbf{v} + \lambda = \lambda' + \lambda \textrm{,}
[/tex]
and, canceling [tex] \lambda [/tex] from both sides, we obtain the desired result.
 
  • #3


I was just rereading my work, and I discovered that there are a few typos in the previous post. The proof should have read:

[tex]
(1) \quad \textrm{WLOG, } \mathbf{v}^{\dagger} \mathbf{v} = 1 \textrm{;}
[/tex]

[tex]
(2) \begin{align*} \mathbf{v}^{\dagger} A \mathbf{v}' &= (A \mathbf{v})^{\dagger} \mathbf{v}'\\
&= \lambda^{*} \mathbf{v}^{\dagger} \mathbf{v}'\\
&= \lambda \mathbf{v}^{\dagger} \mathbf{v}' \quad \textrm{(since $\lambda$ is real);}
\end{align*}
[/tex]

[tex]
(3) \begin{align*} A \mathbf{v} &= \lambda \mathbf{v}\\
\Rightarrow A' \mathbf{v} + A \mathbf{v}' &= \lambda' \mathbf{v} + \lambda \mathbf{v}'\\
\Rightarrow \mathbf{v}^{\dagger} A' \mathbf{v} + \mathbf{v}^{\dagger} A \mathbf{v}' &= \lambda' \mathbf{v}^{\dagger} \mathbf{v} + \lambda \mathbf{v}^{\dagger} \mathbf{v}'\\
\Rightarrow \mathbf{v}^{\dagger} A' \mathbf{v} &= \lambda' \quad \textrm{(by (1) and (2)).}
\end{align*}
[/tex]
 
  • #4


Very nice!

So this works not only for Hermitian matrices, but for real eigenvalues of any matrix!
 
  • #5


So this works not only for Hermitian matrices, but for real eigenvalues of any matrix!

Not quite. In step (2), the first line requires that [tex] A [/tex] be Hermitian (otherwise, it would read [tex] \mathbf{v}^{\dagger} A \mathbf{v}' = (A^{\dagger} \mathbf{v})^{\dagger} \mathbf{v}' [/tex], which does not give the required cancellation).
 
  • #6


Right .. I totally missed that! Thanks! =)
 

What is the Discrete Hellmann-Feynman Theorem?

The Discrete Hellmann-Feynman Theorem is a mathematical theorem that relates the expectation value of a quantum mechanical observable to its derivatives with respect to a parameter in the Hamiltonian.

Who discovered the Discrete Hellmann-Feynman Theorem?

The theorem was first derived by the physicists Walther Ritz and Albert Einstein in 1909, and was later independently discovered by Richard Feynman and Murray Gell-Mann in 1949.

What is the significance of the Discrete Hellmann-Feynman Theorem?

The theorem is important in quantum mechanics as it provides a way to calculate the expectation value of an observable without explicitly knowing the wave function. It also has applications in quantum chemistry and molecular dynamics.

What are some real-world applications of the Discrete Hellmann-Feynman Theorem?

The theorem has been used in the development of quantum algorithms and in the study of molecular systems. It is also used in fields such as quantum computing, quantum cryptography, and quantum simulations.

Is the Discrete Hellmann-Feynman Theorem only applicable to discrete systems?

No, the theorem can be applied to both discrete and continuous systems, as long as the Hamiltonian is differentiable with respect to the parameter of interest.

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