Eigenvalues with Dirac notation

Question

I wonder if there is a way to find the eigenvalues and eigenstates of an operator in Dirac notation without writing it in matrix form.

For example, say $A=\left|\phi_1 \right> \left< \phi_2 \right|+\left|\phi_2 \right> \left< \phi_1 \right|$. Writing it in the matrix form in basis $\left|\phi_1 \right>,\left|\phi_2 \right>$ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} the diagonalization is easy. But is there a way to do the same thing using the Dirac notation directly?

Answer 1

In general no, there is no easier way to find eigenvalues (and eigenstates) of an operator in Dirac notation, when compared to matrix notation.

Besides some obvious cases, as with a diagonal operator $\Lambda = \sum_{i=0}^n \lambda_i |\phi_i\rangle\!\langle\phi_i|$, with $\{|\phi_i\rangle\}$ being a normal basis, where eigendecomposition is already given.

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Two further notes:

1. Consider the case $A = |\phi_1\rangle\!\langle\phi_1| + |\phi_2\rangle\!\langle\phi_2|$, with $\langle\phi_2|\phi_1\rangle \neq 0$. To better visualize, let's go to matrix form. Say:

$\begin{align} |\phi_1\rangle \rightarrow \begin{bmatrix}0 \\ 1 \end{bmatrix}\ \text{ and }\ |\phi_2\rangle \rightarrow \begin{bmatrix}1 \\ 1 \end{bmatrix} \\[1.5ex] A = |\phi_1\rangle\!\langle\phi_1| + |\phi_2\rangle\!\langle\phi_2| \rightarrow \begin{bmatrix}1 & 1 \\ 1 & 2 \end{bmatrix} \end{align}$

$A$ has a conjugate pair of eigenvalues, instead of eigenvalue $1$ as a distracted reader might be lead to think.

2. Also consider $A = |\phi_1\rangle\!\langle\phi_2| + |\phi_2\rangle\!\langle\phi_1|$ with $\langle\phi_2|\phi_1\rangle \notin \mathbb{R}$, I leave it to you to check that the eigenvectors of $\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}$ in the basis $\{|\phi_1\rangle, |\phi_2\rangle\}$ do not correspond to eigenstates of $A$.

Answer 2

When you wrote your operator in matrix form, you evidently assumed your kets are orthonormal, $$ \langle \phi_i|\phi_j\rangle= \delta_{ij}, $$
so that $$ \langle \phi_i|A|\phi_j\rangle=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. $$

Consequently, you find eigenvalues and eigenvectors identically in both cases, since they are equivalent. In bra-ket notation your characteristic polynomial is again $\lambda^2-1$, with the usual roots 1 and -1.

The linear algebra is the same, so their eigenvectors are $$ A(a|\phi_1\rangle+ b|\phi_2) = \pm (a|\phi_1\rangle+ b|\phi_2) , $$ so $a=b=1/\sqrt 2$ or $a=-b=1/\sqrt 2$, respectively. It is the same linear algebraic diagonalization! Neither more nor less difficult!