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.

  • 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


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) $

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

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