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Hermitian Operators (Matrices) in Quantum Mechanics

Table of contents

Hermitian operators have real eigenvalues Here we prove that Hermitian operators have real eigenvalues.
Eigenvectors of Hermitian operators are orthogonal Here we prove that eigenstates of a Hermitian operator are pairwise orthogonal.
Spectral theorem of Hermitian Operators Here we take a look at the most important property of a Hermitian operator.
Hermitian matrices Here we represent Hermitian operators as matrices and look at some examples.

This is how the mean value of an operator $H$ in the state $|\mathit{\Psi}\rangle$ is calculated in quantum mechanics.

$$ \begin{align} \langle \mathit{\Psi} | ~H \, \mathit{\Psi}\rangle \end{align} $$

The mean was represented here in Bra-Ket notation. Under the hood of the Bra-Ket notation hides a postulate of quantum mechanics in the form of an integral. This postulate tells us how we can determine mean values in quantum mechanics.

In the following example, the mean value of the operator $H$ is calculated using the one-dimensional spatial wave function $ \Psi(x) $:

$$ \begin{align} \langle \mathit{\Psi} | ~H \, \mathit{\Psi}\rangle ~=~ \int_a^b \mathit{\Psi}^{~*} \, (H \, \mathit{\Psi}) ~ \text{d}x \end{align} $$

A mean value, as we know it from everyday life, is a real number. In quantum mechanics, however, we can also obtain a complex mean, with a real and imaginary part.

For example, if we want to calculate the position mean or energy mean, we expect quantum mechanics to provide us with a real value.

A complex position or energy do not make any sense!

In such cases, we require that the mean must be equal to the complex-conjugate mean:

$$ \begin{align} \langle \mathit{\Psi} | ~H \, \mathit{\Psi}\rangle ~=~ \langle \mathit{\Psi} | ~H \, \mathit{\Psi}\rangle^* \end{align} $$

Specifically, for our example, this means that the integral 2 must be equal to the following complex-conjugate integral:

$$ \begin{align} \langle \mathit{\Psi} | ~H \, \mathit{\Psi}\rangle^* ~=~ \left( \int_a^b \mathit{\Psi}^{~*} \, (H \, \mathit{\Psi}) ~ \text{d}x \right)^* \end{align} $$

This is a requirement which is fulfilled only by a real number. As you know from mathematics, if we complex conjugate a real number, for example $5^* = 5$, then it remains unchanged. But for complex numbers the sign changes, for example $(4+2\,\mathrm{i})^* = 4-2\,\mathrm{i}$. So by this requirement we exclude complex mean values!

Our requirement has far-reaching consequences for the operator $H$ and its properties. Let's look at this by manipulating the complex conjugate integral 4 a bit. First, we can interchange the two scalars $(\hat H \, \mathit{\Psi})$ and $\mathit{\Psi}^{~*}$:

$$ \begin{align} \langle \mathit{\Psi} | ~H \, \mathit{\Psi}\rangle^* ~=~ \left( \int_a^b ( H \, \mathit{\Psi}) \, \mathit{\Psi}^{~*} ~ \text{d}x \right)^* \end{align} $$

Then apply the complex conjugation (the star) to the individual integrands:

$$ \begin{align} \langle \mathit{\Psi} | ~H \, \mathit{\Psi}\rangle^* ~=~ \int_a^b (H \, \mathit{\Psi})^* \, (\mathit{\Psi}^{~*})^* ~ \text{d}x \end{align} $$

From the complex numbers we know: $ (\mathit{\Psi}^{~*})^* ~=~ \mathit{\Psi} $. Let us then rewrite the integral into bra-ket notation:

$$ \begin{align} \langle \mathit{\Psi} | ~H \, \mathit{\Psi}\rangle^* & ~=~ \int_a^b (H \, \mathit{\Psi})^* \, \mathit{\Psi}^ ~ \text{d}x \\\\
& ~=~ \langle H \, \mathit{\Psi} ~|~ \mathit{\Psi} \rangle \end{align} $$

The rewritten complex-conjugate mean 7 in Bra-Ket notation must be equal to the original mean 2, according to our requirement:

$$ \begin{align} \langle \mathit{\Psi} | ~ H \, \mathit{\Psi}\rangle ~=~ \langle H \, \mathit{\Psi} | ~ \mathit{\Psi}\rangle \end{align} $$

In order for our requirement 3 for a real mean to be satisfied, it must be possible to interchange the operator $H$ in the scalar product. So it must not matter whether we apply $H$ to the ket or to the bra vector. The mean value remains the same. Such an operator, which can be shifted back and forth in the scalar product without changing the mean value, is called a Hermitian operator.

Example: Momentum as Hermitian operator

Let's check if the momentum operator $ \hat p ~=~ - \mathit{i}\,\hbar \, \frac{\text{d}}{\text{d}x} $ has a real mean. To do this, we must check that $ \hat p $ is a Hermitian operator, that is, Eq. 8 is satisfied.

$$ \begin{align} \langle \mathit{\Psi} ~|~ \hat p \, \mathit{\Psi} \rangle
& ~=~ \int_a^b \mathit{\Psi}^{~*} \, (\hat p \, \mathit{\Psi)} ~ \text{d}x \\\\
& ~=~ \int_a^b \mathit{\Psi}^{~*} \, \left(- \mathit{i}\,\hbar \, \frac{\text{d} \mathit{\Psi}}{\text{d}x} \right) ~ \text{d}x \end{align} $$

The $\hbar$ and $\mathit{i}$ are ordinary numbers and may be shifted back and forth. To shift the derivative operator $ \frac{\text{d}}{\text{d}x} $ to the $\mathit{\Psi}^{~*}$, we use partial integration:

$$ \begin{align} \langle \mathit{\Psi} ~|~ \hat p \, \mathit{\Psi} \rangle
& ~=~\bigl[ -\mathit{i}\,\hbar\, \mathit{\Psi}^{~*} \, \mathit{\Psi}~ \bigr]_a^b ~-~ \int_a^b \left(- \mathit{i}\,\hbar \, \frac{\text{d} \mathit{\Psi}^{~*}}{\text{d}x} \right) \, \mathit{\Psi}~ \text{d}x \\\\
& ~=~ \bigl[ -\mathit{i}\,\hbar\, \mathit{\Psi}^{~*} \, \mathit{\Psi}~ \bigr]_a^b ~+~ \int_a^b \left(\mathit{i}\,\hbar \, \frac{\text{d} \mathit{\Psi}^{~*}}{\text{d}x} \right) \, \mathit{\Psi}~ \text{d}x \end{align} $$

We can also write the term $ \left(\mathit{i}\,\hbar \, \frac{\text{d} \mathit{\Psi}^{~*}}{\text{d}x} \right) $ like this $ \left(- \mathit{i}\,\hbar \, \frac{\text{d} \mathit{\Psi}}{\text{d}x} \right)^{*} $:

$$ \begin{align} \langle \mathit{\Psi} ~|~ \hat p \, \mathit{\Psi} \rangle
& ~=~ \bigl[ -\mathit{i}\,\hbar\, \mathit{\Psi}^{~*} \, \mathit{\Psi}~ \bigr]_a^b ~+~ \int_a^b \left(-\mathit{i}\,\hbar \, \frac{\text{d} \mathit{\Psi}}{\text{d}x} \right)^{*} \, \mathit{\Psi}~ \text{d}x \\\\
& ~=~ \bigl[ -\mathit{i}\,\hbar\, \mathit{\Psi}^{~*} \, \mathit{\Psi}~ \bigr]_a^b ~+~ \int_a^b (\hat{p} \, \mathit{\Psi})^{*} \, \mathit{\Psi}~ \text{d}x \\\\
& ~=~ \bigl[ -\mathit{i}\,\hbar\, \mathit{\Psi}^{~*} \, \mathit{\Psi}~ \bigr]_a^b ~+~ \langle \hat p \, \mathit{\Psi} ~|~ \mathit{\Psi} \rangle \end{align} $$

Since the first summand is nonzero depending on $a$, $b$, and $\mathit{\Psi}$, the Hermitian property 8 is not satisfied. So the momentum $p$ in general has not a real mean. What if we consider the whole space?

$$ \begin{align} \langle \mathit{\Psi} ~|~ \hat p \, \mathit{\Psi} \rangle
~=~ \bigl[ -\mathit{i}\,\hbar\, \mathit{\Psi}^{~*} \, \mathit{\Psi}~ \bigr]_{-\infty}^{\infty} ~+~ \langle \hat p \, \mathit{\Psi} ~|~ \mathit{\Psi} \rangle \end{align} $$

For all possible wave functions $\mathit{\Psi}$ the boundary term still does not cancel. But this is not so bad, because in physics we are not interested in all mathematically conceivable wave functions $\mathit{\Psi}$. Physically 'meaningful' wave functions must be normalizable. And normalizable wave functions must approach zero for $x = \pm \infty $. So the boundary term at the chosen integration limits is dropped and we get the Hermitian property of $ \hat p $:

$$ \begin{align} \langle \mathit{\Psi} ~|~ \hat p \, \mathit{\Psi} \rangle
~=~ \langle \hat p \, \mathit{\Psi}~|~ \mathit{\Psi} \rangle \end{align} $$

Let us briefly look at the definition of an adjoint operator $ H^{\dagger} $:

$$ \begin{align} \langle \mathit{\Psi} | ~ H \, \mathit{\Psi}\rangle ~=~ \langle H^{\dagger} \, \mathit{\Psi} | ~ \mathit{\Psi}\rangle \end{align} $$

$ H^{\dagger} $ is thus an operator applied to the bra vector, yielding the same scalar product as the operator $H$ acting on the ket vector. Compare the definition of the adjoint operator 14 with the definition 8 of the Hermitian operator. From this you can see:

$$ \begin{align} H ~=~ H^{\dagger} \end{align} $$

An operator $H$ that is equal to its adjoint $H^{\dagger} $ is called a self-adjoint operator.

Since it does not matter whether the Hermitian operator is applied to the Bra or to the Ket vector first, $H$ can be placed between the Bra and Ket:

$$ \begin{align} \langle \mathit{\Psi} | ~ H \, \mathit{\Psi}\rangle ~=~ \langle H \, \mathit{\Psi} | ~ \mathit{\Psi}\rangle ~:=~ \langle \mathit{\Psi} | \, H \, | \mathit{\Psi}\rangle \end{align} $$

The operator then acts on either the left or right successor. Analogously, the Hermitian adjoint operator can be written between the Bra and Ket vectors.

A Hermitian operator has three other incredibly important properties, which we'll look at next.

Hermitian operators have real eigenvalues

Property #1

A Hermitian operator has real eigenvalues.

This statement is also valid for infinite-dimensional Hilbert spaces. We prove them here for finite dimensional spaces. For this we take the eigenvalue equation for the Hermitian operator $H$:

$$ \begin{align} H \, |\varphi\rangle ~=~ \lambda \, |\varphi\rangle \end{align} $$

Here $ |\varphi\rangle $ is an eigenvector of $H$ and $ \lambda $ is the associated eigenvalue. Then we apply the Bra vector $ \langle \varphi | $ from the left to the eigenvalue equation:

$$ \begin{align} \langle \varphi | \, H \, |\varphi\rangle ~=~ \langle \varphi | \, \lambda \, |\varphi\rangle \end{align} $$

We may pull the eigenvalue $ \lambda $ out of the scalar product. Because it is just a number:

$$ \begin{align} \langle \varphi | \, H \, |\varphi\rangle ~=~ \lambda \, \langle \varphi | \varphi\rangle \end{align} $$

Now we apply the property in Eq. 16 that defines the Hermitian operator. We can apply $H$ to the Bra vector. $H$ applied to the Bra vector gives the complex conjugate eigenvalue $\lambda^*$:

$$ \begin{align} \langle \varphi | \, H \, |\varphi\rangle ~=~ \langle \varphi | \, \lambda^* \, | \varphi\rangle \end{align} $$

This is a number that we can pull out of the scalar product:

$$ \begin{align} \langle \varphi | \, H \, |\varphi\rangle ~=~ \lambda^* \langle \varphi | \varphi\rangle \end{align} $$

Let's set Eq. 19 and 21 equal, bring everything to one side of the equation, and factor out the scalar product $ \langle \varphi | \varphi \rangle $:

$$ \begin{align} \left( \lambda ~-~ \lambda^* \right) \, \langle \varphi | \varphi\rangle ~=~ 0 \end{align} $$

For this equation to be zero, $ \lambda $ must be equal to the complex conjugate $ \lambda^* $. So it must be real:

$$ \begin{align} \lambda ~=~ \lambda^* \end{align} $$

Eigenvectors of Hermitian operators are orthogonal

Property #2

Eigenvectors of a Hermitian operator are orthogonal if they have different eigenvalues.

Let us take two different eigenvalue equations for the Hermitian operator $H$:

$$ \begin{align} H \, | \varphi_1 \rangle &~=~ \lambda \, | \varphi_1 \rangle \\\\
H \, | \varphi_2 \rangle &~=~ \mu \, | \varphi_2 \rangle \end{align} $$

Here $|\varphi_1\rangle$ and $|\varphi_2\rangle$ are two different eigenvectors of $H$. And $\lambda$ and $\mu$ are the corresponding eigenvalues. Let's apply the bra vector $ \langle \varphi_2 | $ from the left to the first eigenvalue equation. The eigenvalue $\lambda$ is a real number which we can place in front of the scalar product:

$$ \begin{align} \langle \varphi_2 | \, H \, | \varphi_1 \rangle ~=~ \lambda \, \langle \varphi_2 | \varphi_1 \rangle \end{align} $$

We express the second eigenvalue equation with the Bra vector:

$$ \begin{align} \langle \varphi_2 | \, H ~=~ \langle \varphi_2 | \, \mu \end{align} $$

Let's multiply the equation from the right by the ket vector $ | \varphi_1 \rangle $ and place the real eigenvalue $\mu$ in front of the scalar product:

$$ \begin{align} \langle \varphi_2 | \, H \, | \varphi_1 \rangle ~=~ \mu \, \langle \varphi_2 | \varphi_1 \rangle \end{align} $$

Let's set our two results Eq. 25 and 27 equal, put everything on one side of the equation, and factor out the scalar product. Then we get:

$$ \begin{align} \left( \lambda ~-~ \mu \right) \, \langle \varphi_2 | \varphi_1 \rangle ~=~ 0 \end{align} $$

If we assume that the eigenvectors are not degenerate, that is, they have different eigenvalues $\lambda$ and $\mu$, then the scalar product between the two eigenvectors of $H$ must be zero:

$$ \begin{align} \langle \varphi_2 | \varphi_1 \rangle ~=~ 0 \end{align} $$

Thus, the non-degenerate distinct eigenvectors of a Hermitian operator are always orthogonal to each other!

Spectral theorem of Hermitian Operators

Property #3

The set of eigenvectors of a Hermitian operator can be used as a basis.

This property is so important that it has a name, namely the spectral theorem. A linear, finite-dimensional operator $H$ is, as you know, a linear mapping from one Hilbert space to another Hilbert space:

$$ \begin{align} H: \mathcal{H} ~\rightarrow~ \mathcal{H} \end{align} $$

$\mathcal{H} $ can be for example a two-dimensional Hilbert space. In this case the basis of this Hilbert space consists of two 2d vectors. So, by the spectral theorem, we can take two different eigenvectors of $H$ as basis vectors and thus represent every possible ket vector from $\mathcal{H} $.

Hermitian matrices

As you know from linear algebra: you can represent a linear map, like 30, as a matrix. If the Hilbert space is two-dimensional, then the operator $H$ corresponds to a 2x2 matrix. A Hermitian operator represented as a matrix is called a Hermitian matrix.

In the matrix representation, the adjoint matrix ("$\dagger$") stands for a transposed and complex-conjugate matrix. By Eq. 15, a Hermitian matrix is equal to its transposed and complex-conjugate matrix.

Examples for Hermite matrices

$$ \begin{align} \begin{bmatrix}
0 & -\mathrm{i} \\
\mathrm{i} & 0 \\
\end{bmatrix} \end{align} $$

This is the Pauli $\sigma_y$ matrix and it is Hermitian. If you transpose and complex-conjugate it, you get the same matrix back. Also the $\sigma_x$-matrix is a Hermitian matrix:

$$ \begin{align} \begin{bmatrix}
0 & 1 \\
1 & 0 \\
\end{bmatrix} \end{align} $$

Example of a non-Hermitian matrix

$$ \begin{align} \begin{bmatrix}
1 & 2 \\
-3\mathrm{i} & 0 \\
\end{bmatrix} \end{align} $$

This is a non-Hermitian matrix because if you transpose it and complex conjugate it, you get a completely different matrix that is not the same as the original matrix.