# Hermitian operator in an orthonormal eigenbasis

In page 36 of Shankar's Principles of Quantum Mechanics is given a theorem:

Theorem 10. To every Hermitian Operator $\Omega$, there exists (at least) a basis consisting of its orthonormal eigenvectors. It is diagonal in this eigenbasis and has its eigenvalues as its diagonal entries.

There is a part of the proof that I do not understand. Turning to page 36, one reads that, corresponding to the eigenvalue $\omega_{1}$ is a normalized eigenvector $|\omega_{1}\rangle$. Considering $|\omega_{1}\rangle$ to be a basis, $\Omega$ has a matrix form $$\begin{bmatrix} \omega_{1} & 0 &. &. &. & 0\\ 0& \omega_{2}& . & . & . &. \\ . & . & \omega_{3}&. & . &. \\ . &. &. &. &. &. \\ . & .& .& .& . &. \\ 0& . &. & .& .&\omega_{n} \end{bmatrix}$$
Where the dots indicate a series of zeroes. This leads me to question. Why does $\Omega$ take that form? Also, why is the first column the image of $|\omega_{1}\rangle$ after $\Omega$ has acted on it?

• We need more context here. Do you know what a hermitian operator is? A basis? An eigenvector? What part of the statement is confusing? Commented May 19, 2018 at 15:43
• Related post by OP: physics.stackexchange.com/q/406803/2451 Commented May 19, 2018 at 16:08
• @Javier I have edited the post.
– R004
Commented May 19, 2018 at 16:12
• That's quite better, but it would also help if you posted the relevant equations. Not everyone has that book. Commented May 19, 2018 at 16:14
• @Javier Sure. I will do that now.
– R004
Commented May 19, 2018 at 16:16

## 3 Answers

$\vert\omega_1\rangle$ would not be a basis, but the set $\{\vert\omega_1\rangle,\vert\omega_2\rangle, \ldots,\vert\omega_n\rangle\}$ is a basis. The eigenvector $\vert\omega_i\rangle$ is such that $\Omega\vert\omega_i\rangle=\omega_i\vert\omega_i\rangle$ so...

With the identification $$\vert\omega_1\rangle \to \left(\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0\end{array}\right)\, ,\quad \vert\omega_2\rangle \to \left(\begin{array}{c} 0 \\ 1 \\ \vdots \\ 0\end{array}\right)\, ,\ldots \, , \vert\omega_n\rangle \to \left(\begin{array}{c} 0 \\ \vdots \\ 0 \\ 1 \end{array}\right)$$ and the matrix representation of $\Omega$ as you suggest you find by simple matrix multiplication that $$\Omega \vert \omega_n\rangle = \omega_n\vert\omega_n\rangle$$ as per the properties of eigenstates of $\Omega$. Note that under this identification the vectors $\vert\omega_i\rangle$ are an orthonormal basis since $\langle \omega_i\vert\omega_j\rangle=\delta_{ij}$.

• @R004 Hermiticity does not depend on the basis if this basis is orthonormal. $\Omega$ would no longer by diagonal in the new basis but it would still be hermitian. Commented May 20, 2018 at 1:32
• one thing to ask. What does "$\Omega$ diagonal in an eigenbasis" mean?
– R004
Commented May 20, 2018 at 1:57
• It means the matrix only has entries on the diagonal, as the matrix in your question. If your previous comment you allow for the eigenvalues $\omega_1$ to occur twice, so that a linear combination of eigenvectors with this eigenvalue is also an eigenvector. Commented May 20, 2018 at 2:03
• please bare with me for a moment. The theorem has in it "To every Hermitian operator" that confuses me. Does "every" speak for all Hermitian operators?
– R004
Commented May 20, 2018 at 3:02
• @R004 Yes every hermitian operator can be brought to diagonal form. The change of basis that results in the transformation of $\Omega$ to diagonal form is the change of basis that takes the initial basis states to the new basis having the eigenvectors of $\Omega$ as basis states. Commented May 20, 2018 at 3:51

Let $\:\mathbb{H}\:$ a n-dimensional Hilbert space and $\:\Omega\:$ any hermitian operator in it. If $\:\mathbf{x} \in \mathbb{H}\:$ then its image $\:\mathbf{y} \:$ under $\:\Omega\:$ is $$\mathbf{y}=\Omega\, \mathbf{x} \tag{01}$$ This is an equation containing vectors and operators and is valid independently of the coordinates. Let choose as orthonormal basis the following complete set of mutually orthogonal normalized vectors
$$\mathbf{e}_1\!=\! \begin{bmatrix} 1\\ 0\\ 0\\ \vdots\\ 0\\ 0 \end{bmatrix}, \:\mathbf{e}_2\!=\! \begin{bmatrix} 0\\ 1\\ 0\\ \vdots\\ 0\\ 0 \end{bmatrix}, \cdots\cdots, \:\mathbf{e}_{n-1}\!=\! \begin{bmatrix} 0\\ 0\\ 0\\ \vdots\\ 1\\ 0 \end{bmatrix}, \:\mathbf{e}_{n}\!=\! \begin{bmatrix} 0\\ 0\\ 0\\ \vdots\\ 0\\ 1 \end{bmatrix} \tag{02}$$ that is $$(\mathbf{e}_{i})_{j}=\delta_{ij} \tag{02^\prime}$$ Then equation (01) is expressed in matrix from $$\mathbf{y}= \begin{bmatrix} y_{1}\\ y_{2}\\ y_{3}\\ \vdots\\ \vdots\\ y_{n} \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\ a_{21} & a_{22} & a_{23} & \cdots & a_{2n}\\ a_{31} & a_{32} & a_{33} & \cdots & a_{3n}\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ a_{n1} & a_{n2} & a_{n3} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \vdots\\ \vdots\\ x_{n} \end{bmatrix} = \Omega\, \mathbf{x} \tag{03}$$ that is $$y_{i}=a_{ij}x_{j} \qquad \text{(Einstein summation convention)} \tag{03^\prime}$$ Now, suppose we want to express our equations on a different basis say $\:\{\mathbf{u}_{\sigma}\}$, not necessarily orthonormal. To do so , we express the members of the new basis $\:\{\mathbf{u}_{\rho}\}$ in the old basis $\:\{\mathbf{e_{\sigma}}\}\:$ $$\mathbf{u_{\rho}}=s_{\sigma\rho}\mathbf{e_{\sigma}} \tag{04}$$ so that $$\mathbf{x'}=x'_{\rho}\mathbf{u_{\rho}}=x'_{\rho}s_{\sigma\rho} \mathbf{e_{\sigma}}=x_{\sigma}\mathbf{e_{\sigma}} \quad \Longrightarrow \quad x_{\sigma}=s_{\sigma\rho}x'_{\rho} \tag{05}$$ Then $$\mathbf{x}= \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \vdots\\ \vdots\\ x_{n} \end{bmatrix} = \begin{bmatrix} s_{11} & s_{12} & s_{13} & \cdots & s_{1n}\\ s_{21} & s_{22} & s_{23} & \cdots & s_{2n}\\ s_{31} & s_{32} & s_{33} & \cdots & s_{3n}\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ s_{n1} & s_{n2} & s_{n3} & \cdots & s_{nn} \end{bmatrix} \begin{bmatrix} x'_{1}\\ x'_{2}\\ x'_{3}\\ \vdots\\ \vdots\\ x'_{n} \end{bmatrix} =\mathrm S\, \mathbf{x'} \tag{06}$$ and inversely $$\mathbf{x'}= \begin{bmatrix} x'_{1}\\ x'_{2}\\ x'_{3}\\ \vdots\\ \vdots\\ x'_{n} \end{bmatrix} = \begin{bmatrix} s_{11} & s_{12} & s_{13} & \cdots & s_{1n}\\ s_{21} & s_{22} & s_{23} & \cdots & s_{2n}\\ s_{31} & s_{32} & s_{33} & \cdots & s_{3n}\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ s_{n1} & s_{n2} & s_{n3} & \cdots & s_{nn} \end{bmatrix}^{\boldsymbol{-1}} \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \vdots\\ \vdots\\ x_{n} \end{bmatrix} =\mathrm S^{\boldsymbol{-1}}\, \mathbf{x} \tag{06^\prime}$$ where $\:\mathrm S\:$ the invertible matrix $$\mathrm S\equiv \begin{bmatrix} s_{11} & s_{12} & s_{13} & \cdots & s_{1n}\\ s_{21} & s_{22} & s_{23} & \cdots & s_{2n}\\ s_{31} & s_{32} & s_{33} & \cdots & s_{3n}\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ s_{n1} & s_{n2} & s_{n3} & \cdots & s_{nn} \end{bmatrix} \tag{07}$$ Applying $\:\mathrm S^{\boldsymbol{-1}}\:$ on equation (01) we have $$\mathrm S^{\boldsymbol{-1}}\, \mathbf{y}=\left(\mathrm S^{\boldsymbol{-1}}\,\Omega\,\mathrm S\right)\, \mathrm S^{\boldsymbol{-1}}\,\mathbf{x} \tag{08}$$ that is $$\mathbf{y'}=\Omega'\,\mathbf{x'} \quad \text{where} \quad \mathbf{y'}=\mathrm S^{\boldsymbol{-1}}\, \mathbf{y} \quad, \quad \mathbf{x'}=\mathrm S^{\boldsymbol{-1}}\, \mathbf{x} \tag{09}$$ and $$\Omega'\equiv\mathrm S^{\boldsymbol{-1}}\,\Omega\,\mathrm S \tag{10}$$ This is the matrix representation of $\:\Omega\:$ in the new basis.

Now, suppose that $\:\Omega\:$ is hermitian. Then it has a complete set of eigenvectors $\:\{\boldsymbol{\omega}_{\sigma}\}\:$ with real eigenvalues $\:\omega_{\sigma}\in \mathbb{R}\:$ $$\Omega\,\boldsymbol{\omega_{\sigma}}=\omega_{\sigma}\,\boldsymbol{\omega_{\sigma}} \qquad \text{(without summation for repeated index } \boldsymbol{\sigma}) \tag{11}$$ To express equation (01) with respect to the basis of eigenvectors $\:\{\boldsymbol{\omega}_{\sigma}\}\:$ given that its representation with respect to $\:\mathbf{e_{\rho}}\:$ is (03) let in analogy to (04) $$\boldsymbol{\omega_{\rho}}=\mathrm w_{\sigma\rho} \mathbf{e_{\sigma}} \tag{12}$$ where $$\mathrm W\equiv \begin{bmatrix} \mathrm w_{11} & \mathrm w_{12} & \mathrm w_{13} & \cdots & \mathrm w_{1n}\\ \mathrm w_{21} & \mathrm w_{22} & \mathrm w_{23} & \cdots & \mathrm w_{2n}\\ \mathrm w_{31} & \mathrm w_{32} & \mathrm w_{33} & \cdots & \mathrm w_{3n}\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \mathrm w_{n1} & \mathrm w_{n2} & \mathrm w_{n3} & \cdots & \mathrm w_{nn} \end{bmatrix} = \begin{bmatrix} | & | & | & \cdots & |\\ | & | & | & \cdots & |\\ \boldsymbol{\omega_{1}} & \boldsymbol{\omega_{2}} & \boldsymbol{\omega_{3}} & \cdots & \boldsymbol{\omega_{n}} \\ | & | & | & \cdots & |\\ | & | & | & \cdots & |\\ \downarrow & \downarrow & \downarrow & \cdots & \downarrow \end{bmatrix} \tag{13}$$ This means that the components of the eigenvector $\:\boldsymbol{\omega_{\sigma}}\:$ with respect to $\:\mathbf{e_{\rho}}\:$ is the $\sigma$-column of the matrix $\:\mathrm W\:$. So $$\Omega\,\boldsymbol{\omega_{\sigma}}=\omega_{\sigma}\,\boldsymbol{\omega_{\sigma}} \Longrightarrow \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\ a_{21} & a_{22} & a_{23} & \cdots & a_{2n}\\ a_{31} & a_{32} & a_{33} & \cdots & a_{3n}\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ a_{n1} & a_{n2} & a_{n3} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} \mathrm w_{1\sigma}\\ \mathrm w_{2\sigma}\\ \mathrm w_{3\sigma}\\ \vdots\\ \vdots\\ \mathrm w_{n\sigma} \end{bmatrix} =\omega_{\sigma} \begin{bmatrix} \mathrm w_{1\sigma}\\ \mathrm w_{2\sigma}\\ \mathrm w_{3\sigma}\\ \vdots\\ \vdots\\ \mathrm w_{n\sigma} \end{bmatrix} = \begin{bmatrix} \omega_{\sigma}\mathrm w_{1\sigma}\\ \omega_{\sigma}\mathrm w_{2\sigma}\\ \omega_{\sigma}\mathrm w_{3\sigma}\\ \vdots\\ \vdots\\ \omega_{\sigma}\mathrm w_{n\sigma} \end{bmatrix} \tag{14}$$ and \begin{align} \Omega\,\mathrm W & = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\ a_{21} & a_{22} & a_{23} & \cdots & a_{2n}\\ a_{31} & a_{32} & a_{33} & \cdots & a_{3n}\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ a_{n1} & a_{n2} & a_{n3} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} \mathrm w_{11} & \mathrm w_{12} & \mathrm w_{13} & \cdots & \mathrm w_{1n}\\ \mathrm w_{21} & \mathrm w_{22} & \mathrm w_{23} & \cdots & \mathrm w_{2n}\\ \mathrm w_{31} & \mathrm w_{32} & \mathrm w_{33} & \cdots & \mathrm w_{3n}\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \mathrm w_{n1} & \mathrm w_{n2} & \mathrm w_{n3} & \cdots & \mathrm w_{nn} \end{bmatrix} \nonumber\\ & = \begin{bmatrix} \omega_{1}\mathrm w_{11} & \omega_{2}\mathrm w_{12} & \omega_{3}\mathrm w_{13} & \cdots & \omega_{1}\mathrm w_{1n}\\ \omega_{1}\mathrm w_{21} & \omega_{2}\mathrm w_{22} & \omega_{3}\mathrm w_{23} & \cdots & \omega_{2}\mathrm w_{n2}\\ \omega_{1}\mathrm w_{31} & \omega_{2}\mathrm w_{32} & \omega_{3}\mathrm w_{33} & \cdots & \omega_{3}\mathrm w_{n3}\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \omega_{1}\mathrm w_{n1} & \omega_{2}\mathrm w_{n2} & \omega_{3}\mathrm w_{n3} & \cdots & \omega_{n}\mathrm w_{nn} \end{bmatrix} \nonumber\\ &= \begin{bmatrix} \mathrm w_{11} & \mathrm w_{12} & \mathrm w_{13} & \cdots & \mathrm w_{1n}\\ \mathrm w_{21} & \mathrm w_{22} & \mathrm w_{23} & \cdots & \mathrm w_{2n}\\ \mathrm w_{31} & \mathrm w_{32} & \mathrm w_{33} & \cdots & \mathrm w_{3n}\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \mathrm w_{n1} & \mathrm w_{n2} & \mathrm w_{n3} & \cdots & \mathrm w_{nn} \end{bmatrix} \begin{bmatrix} \omega_{1} & 0 & 0 & \cdots & 0\\ 0 & \omega_{2} & 0 & \cdots & 0\\ 0 & 0 & \omega_{3} & \cdots & 0\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ 0 & 0 & 0 & \cdots & \omega_{n} \end{bmatrix} \tag{15} \end{align} so $$\Omega\cdot\mathrm W = \mathrm W \cdot \mathrm{diag}\{\omega_{1},\omega_{2},\omega_{3},\cdots,\omega_{n}\} \tag{16}$$ where $$\mathrm{diag}\{\omega_{1},\omega_{2},\omega_{3},\cdots,\omega_{n}\}\equiv \begin{bmatrix} \omega_{1} & 0 & 0 & \cdots & 0\\ 0 & \omega_{2} & 0 & \cdots & 0\\ 0 & 0 & \omega_{3} & \cdots & 0\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ 0 & 0 & 0 & \cdots & \omega_{n} \end{bmatrix} \tag{17}$$ and finally $$\Omega'=\mathrm W ^{\boldsymbol{-1}}\Omega\,\mathrm W= \mathrm{diag}\{\omega_{1},\omega_{2},\omega_{3},\cdots,\omega_{n}\}= \begin{bmatrix} \omega_{1} & 0 & 0 & \cdots & 0\\ 0 & \omega_{2} & 0 & \cdots & 0\\ 0 & 0 & \omega_{3} & \cdots & 0\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ 0 & 0 & 0 & \cdots & \omega_{n} \end{bmatrix} \tag{18}$$ This diagonal matrix $\:\Omega'\:$ is the matrix representation of the hermitian operator $\:\Omega\:$ with respect to the complete orthonormal basis of its own eigenvectors.

Hermitian operators are important because their eigenvectors corresponding to different eigenvalues are orthogonal to each other (and can be normalized if required), and they form a basis for the Hilbert space on which the operators act.

Take, for instance, the $\sigma_z$ operator. Its eigenvalues are $\pm 1$ and its eigenvectors are $(1,0)^T, (0,1)^T$. You can express any vector of the 2-dimensional Hilbert space as a linear combination of these eigenvectors.