# What is a Hermitian

1. Jul 24, 2014

### Greg Bernhardt

Definition/Summary

The Hermitian transpose or Hermitian conjugate (or conjugate transpose) $M^{\dagger}$ of a matrix $M$ is the complex conjugate of its transpose $M^T$.

A matrix is Hermitian if it is its own Hermitian transpose: $M^{\dagger}\ =\ M$.

An operator $A$ is Hermitian (or self-adjoint) if it is its own adjoint: $\langle Ax|y\rangle\ =\ \langle x|Ay\rangle$ (in the finite-dimensional case, that means that its matrix is Hermitian).

In quantum theory, an observable must be represented by a Hermitian operator (on a Hilbert space).

For other uses of the adjective "Hermitian", see http://en.wikipedia.org/wiki/Hermitian.

Equations

$$\int \psi _1 ^* (\hat{O} \psi _2 ) \, dx = \int (\hat{O}\psi _1 )^* \psi _2 \, dx$$

$$\langle b |\hat{O} |c \rangle = \langle c |\hat{O} |b \rangle ^*$$

Extended explanation

A matrix $M$ is hermitian if:
$$M^{\dagger} = (M^T)^* = M ,$$
where $\dagger$ is called the hermitian conjugate, and is thus a combination of matrix transpose and complex conjugation of each entry in the matrix.

In quantum mechanics, observable quantities are assigned by hermitian operators. Examples of those are:

(with continuous spectrum)
position operator
$$\hat{x},$$

momentum operator
$$-i\hbar \dfrac{\partial}{\partial x},$$

(with discrete spectrum)
z-component of angular momentum operator
$$\hat{L}_z .$$

In terms of wave functions, an operator $\hat{O}$ is hermitian if:
$$\int \psi _1 ^* (\hat{O} \psi _2 ) \, dx = \int (\hat{O}\psi _1 )^* \psi _2 \, dx$$

In terms of bra-ket:
$$\langle b |\hat{O} |c \rangle = \langle c |\hat{O} |b \rangle$$

Now, using the wave function formalism, some valuable identities will be presented:

Let us consider two hermitian operators $\hat{A}$ and $\hat{B}$.

The expectation value:
$$<\hat{A}> = \int \psi ^* (\hat{A} \psi ) dx = \int (\hat{A}\psi )^* \psi dx,$$
is real, proof:
$$<\hat{A}>^* = \int ((\hat{A}\psi)^*\psi)^* dx = \int (\hat{A}\psi)\psi^* dx = \int \psi^* (\hat{A}\psi) dx = <\hat{A}>$$
Since $\hat{A}$ was said to be hermitian, and $\psi _1 = \psi _2$ when we do expectation values.

Expectation value of $\hat{A}^2$:
$$<\hat{A}^2> = \int \psi ^* (\hat{A}(\hat{A}\psi)) dx = \int \psi^* (\hat{A}\tilde{\psi})dx =$$
$$(\tilde{\psi} = \hat{A}\psi \: \text{ is a new wavefunction} )$$
$$\int (\hat{A}\psi)^*\tilde{\psi}dx = \int (\hat{A}\psi)^*(\hat{A}\psi) dx$$

Now we can show another useful result:
$$\int \psi^* (\hat{A}(\hat{B}\psi))dx = \int(\psi^*(\hat{B}(\hat{A}\psi)))^*dx ,$$
prove this as an exercise.

Two more useful things:
$$I = \int \psi^*(\hat{A}\hat{B}+\hat{B}\hat{A})\psi = I^*$$
is real, show this as an exercise.

The operators always to the right if not indicated otherwise. Thus:
$$I = \int \psi^*(\hat{A}(\hat{B}\psi))dx + \int \psi^* (\hat{B}(\hat{A}\psi)) dx$$

$$J = \int \psi^*(\hat{A}\hat{B}-\hat{B}\hat{A})\psi = -J^*$$
is imaginary, show this as an exercise.

These identities are needed to prove the uncertainty relations of quantum mechanics.

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