Hermitian operator
- Quantum Mechanics 1.3
from 「Modern Quantum Mechanics」: Sakurai, J. J.
What we should consider are the Hermitians.
Hermitian operator
A Hermitian operator, or self-adjoint operator is an operator that is equal to its own hermitian conjugate.
- $ \hat{A}=\hat{A}^\dag $
Since operators with such properties are treated very importantly in quantum mechanics, we’re going to summarize their properties from now on.
Real expectation values
\[ \expct{\hat{A}} = \brkt{\psi}{\hat{A}\psi} \nt \begin{align*} \expct{\hat{A}}^\ast &= \brkt{\psi}{\hat{A}\psi}^\ast \nl &= \brkt{\hat{A}\psi}{\psi} \nl &= \brkt{\hat{A}^\dag \psi}{\psi} \nl &= \brkt{\psi}{\hat{A}\psi} \end{align*} \nt \therefore\; \expct{\hat{A}} = \expct{\hat{A}}^\ast \]
This implies that the physical quantities of Hermitian operators are observable (since it’s real); in this respect, Hermitian operators are called observables.
Orthogonality of eigenfunctions
\[ \begin{cases} \hat{A}\ket{\psi_n} = a_n\ket{\psi_n} \nl \hat{A}\ket{\psi_m} = a_m\ket{\psi_m} \end{cases} \nt \begin{cases} \brkt{\psi_n}{\hat{A}\psi_m} = a_m\brkt{\psi_n}{\psi_m} \nl \brkt{\hat{A}\psi_n}{\psi_m} = a_n^\ast\brkt{\psi_n}{\psi_m} = a_n\brkt{\psi_n}{\psi_m} \end{cases} \nt \Rightarrow (a_n-a_m)\brkt{\psi_n}{\psi_m}=0 \nt \therefore \brkt{\psi_n}{\psi_m}=\delta_{nm} \]
Measurements and probabilities
The result above implies that any wavefunction can be represented as the sum of eigenfunctions of an observable, which are elements of an orthogonal basis.
\[ \ket{\psi}=\sum_n c_n\ket{\psi_n} \; \left( \sum_n |c_n|^2=\brkt{\psi}{\psi} \right) \]
If then, the probability for observing the state $\ket{\psi_n}$ is proportional to \(\abs{\brkt{\psi_n}{\psi}}^2 = \abs{c_n}^2\); suppose it’s $k|c_n|^2$. The overall probability $ k\brkt{\psi}{\psi} $ should be $1$, and this results:
- $ P(\psi_n) = \dfrac{\abs{c_n}^2}{\brkt{\psi}{\psi}} $
And if \(\ket{\psi}\) is normalized,
- $ P(\psi_n)=\abs{c_n}^2 $.
Keep in mind that in quantum mechanics, a measure of the physical quantity becomes one of the eigenvalues, not the expectation value.
Normalization
And this also implies that if state vector $\ket{f}$ and $\ket{g}$ are in following relation, they’re representing identical physical system.
- $ \ket{f}=c\ket{g} \;(c\in\Complex) $
Then we can normalize any state vectors using phase factors $e^{iz}$, of which norm is $1$.
- $ \ket{\psi} \to \dfrac{e^{iz}\ket{\psi}}{\sqrt{\brkt{\psi}{\psi}}} $
Wavefunction collapse
The measurement transforms the system to become an eigenvector for the measured value, which is called a wavefunction collapse. (It can be understood that it collapses the probability of an eigenvalue different from the measured value.)
\[ \ket{\psi} \xrightarrow[\text{eigenvalue}:\; l \;/\; \text{eigenvector}:\; \ket{l}]{\text{observable}:\; L} \ket{l} \]
Extra
For any operator $\hat{A}$, all of the following operators are hermitian.
- $ \hat{A}+\hat{A}^\dag $
- $ i(\hat{A}-\hat{A}^\dag) $
- $ \hat{A}\hat{A}^\dag $
Also, there’s a property about a commutator:
- $ [\hat{A},\hat{B}]=0 \Rightarrow \hat{A}\hat{B}=\left(\hat{A}\hat{B}\right)^\dag $
We’ll talk about commutators later.
