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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?

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

3 Answers 3

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$\vert\omega_1\rangle$ 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 $\langle \omega_i\vert\omega_j\rangle=\delta_{ij}$.

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  • $\begingroup$ @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. $\endgroup$ May 20, 2018 at 1:32
  • $\begingroup$ one thing to ask. What does "$\Omega$ diagonal in an eigenbasis" mean? $\endgroup$
    – R004
    May 20, 2018 at 1:57
  • $\begingroup$ 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. $\endgroup$ May 20, 2018 at 2:03
  • $\begingroup$ 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? $\endgroup$
    – R004
    May 20, 2018 at 3:02
  • 2
    $\begingroup$ @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. $\endgroup$ May 20, 2018 at 3:51
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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 \begin{equation} \mathbf{y}=\Omega\, \mathbf{x} \tag{01} \end{equation} 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
\begin{equation} \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} \end{equation} that is \begin{equation} (\mathbf{e}_{i})_{j}=\delta_{ij} \tag{02$^\prime$} \end{equation} Then equation (01) is expressed in matrix from \begin{equation} \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} \end{equation} that is \begin{equation} y_{i}=a_{ij}x_{j} \qquad \text{(Einstein summation convention)} \tag{03$^\prime$} \end{equation} 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 $\:\{\mathbf{e_{\sigma}}\}\:$ \begin{equation} \mathbf{u_{\rho}}=s_{\sigma\rho}\mathbf{e_{\sigma}} \tag{04} \end{equation} so that \begin{equation} \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} \end{equation} Then \begin{equation} \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} \end{equation} and inversely \begin{equation} \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$^\prime$} \end{equation} where $\:\mathrm S\:$ the invertible matrix \begin{equation} \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} \end{equation} Applying $\:\mathrm S^{\boldsymbol{-1}}\:$ on equation (01) we have \begin{equation} \mathrm S^{\boldsymbol{-1}}\, \mathbf{y}=\left(\mathrm S^{\boldsymbol{-1}}\,\Omega\,\mathrm S\right)\, \mathrm S^{\boldsymbol{-1}}\,\mathbf{x} \tag{08} \end{equation} that is \begin{equation} \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} \end{equation} and \begin{equation} \Omega'\equiv\mathrm S^{\boldsymbol{-1}}\,\Omega\,\mathrm S \tag{10} \end{equation} 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 $\:\omega_{\sigma}\in \mathbb{R}\:$ \begin{equation} \Omega\,\boldsymbol{\omega_{\sigma}}=\omega_{\sigma}\,\boldsymbol{\omega_{\sigma}} \qquad \text{(without summation for repeated index } \boldsymbol{\sigma}) \tag{11} \end{equation} 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) \begin{equation} \boldsymbol{\omega_{\rho}}=\mathrm w_{\sigma\rho} \mathbf{e_{\sigma}} \tag{12} \end{equation} where \begin{equation} \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} \end{equation} 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 \begin{equation} \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} \end{equation} 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 \begin{equation} \Omega\cdot\mathrm W = \mathrm W \cdot \mathrm{diag}\{\omega_{1},\omega_{2},\omega_{3},\cdots,\omega_{n}\} \tag{16} \end{equation} where \begin{equation} \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} \end{equation} and finally \begin{equation} \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} \end{equation} This diagonal matrix $\:\Omega'\:$ is the matrix representation of the hermitian operator $\:\Omega\:$ with respect to the complete orthonormal basis of its own eigenvectors.

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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.

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