0
$\begingroup$

I am trying to perform a perturbation for a system but I get really confused when trying to calculate an expectation for a column vector wave function. Hamiltonian is a 2×2 diagonal matrix and I am trying to perform a perturbation.

Perturbation matrix: \begin{equation} H = \begin{bmatrix} 0 & \lambda\\ \lambda & 0 \end{bmatrix}\qquad 2 \times 2 \;\text{matrix} \end{equation}

This is the perturbation of the Hamiltonian. By diagonal system Hamiltonian, I come up with wave functions like $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ column vector wave functions (2×1).

What I am trying to perform is this: \begin{equation} \langle\Psi_0| H | \Psi_1\rangle \end{equation}

Now, isn't there a dimensional problem? $\Psi_0$ and $\Psi_1$ are 2×1 matrices and $H$ is 2×2 matrix. So, left multiplication will not work.

In fact, this is a two-state system with defined Hamiltonians for Interaction and system energies. The interacted system is a CLASSICAL oscillator. I am trying to solve the spontaneous emission possibility.

$\endgroup$
2
  • 2
    $\begingroup$ transpose $\psi_0$? $\endgroup$
    – sleepy
    Commented Jun 11, 2021 at 8:40
  • $\begingroup$ Your $H$ seems to be anti-diagonal rather than diagonal. Is it correct? $\endgroup$ Commented Jun 11, 2021 at 8:58

1 Answer 1

1
$\begingroup$

Let us represent the state vectors as $\left| \psi_0 \right> = a\left| \phi_1 \right> + b\left| \phi_2 \right> \equiv \begin{pmatrix} a \\b \end{pmatrix}$ and $\left| \psi_1 \right> = c\left| \phi_1 \right> + d\left| \phi_2 \right> \equiv \begin{pmatrix} c \\ d \end{pmatrix}$, where $\left| \phi_1 \right>$ and $\left| \phi_2 \right> $ are the basis states, the "bra"s would be the corresponding conjugate transpose, i.e. take the complex conjugate of all the elements and tranpose the matrix, $\left< \psi_0 \right| \equiv \begin{pmatrix} a^* & b^* \end{pmatrix}$. The desired expression can then be represented as

$\left< \psi_0 \right| H \left| \psi_1 \right> = \begin{pmatrix} a^* & b^* \end{pmatrix} \begin{pmatrix} 0 & \lambda \\ \lambda & 0 \end{pmatrix} \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a^* & b^* \end{pmatrix} \begin{pmatrix} \lambda d \\ \lambda c \end{pmatrix} = \lambda \left( a^* d + b^* c\right) $

$\endgroup$
1
  • $\begingroup$ Thank you! Of course the answer is the transpose of it. $\endgroup$
    – bidon
    Commented Jun 11, 2021 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.