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There is a theorem in my notes on spectral decomposition that goes as follows

Spectral decomposition of an operator Assume that the eigenvectors of $\hat{A}$ define a basis $B=\{|\phi_j>\},$then $A_{kj}=<\phi_k|\hat{ A}|\phi_j>= α_jδ_{kj}$. Operator in this basis is a diagonal matrix with eigenvalues on the diagonal

$ \hat{A} = \sum_{kj} A_{kj} |\phi_k><\phi_j| $

So take the example $\hat{A}= \begin {pmatrix} 3 &2\\2 &3 \end{pmatrix}$

It has eigenvalues $\lambda_1=1, \lambda_2=5$ which have eigenvetors $|\phi_1>= \begin{pmatrix} 1 \\-1 \end{pmatrix}$ and $|\phi_2>= \begin{pmatrix} 1 \\1 \end{pmatrix}$ respectively.

Edit: following up on bungos comment

$|\phi_1>, |\phi_2>$ are not normalised the normalised versions are $|\phi_1'>=\begin {pmatrix} 1/\sqrt{2} \\-1/\sqrt{2} \end {pmatrix}, |\phi_2'>=\begin {pmatrix} 1/\sqrt{2} \\1/\sqrt{2} \end {pmatrix}$ respectively. So...

$<\phi_1'|\hat{A}|\phi_1'>=1=A_{11}$

$<\phi_2'|\hat{A}|\phi_2'>=5=A_{22}$

$|\phi_1'><\phi_1'|=\begin {pmatrix} 1/2 & -1/2 \\-1/2 &1/2 \end{pmatrix}$

$|\phi_2'><\phi_2'|=\begin {pmatrix} 1/2 &1/2\\1/2 &1/2 \end{pmatrix}$

So by my calculations : $\hat{A} = A_{11} |\phi_1'><\phi_1'| +A_{22} |\phi_2'><\phi_2'|=1$$\begin {pmatrix} 1/2 & -1/2\\-1/2 &1/2 \end{pmatrix}$$+5\begin {pmatrix} 1/2 & 1/2\\ 1/2 &1/2 \end{pmatrix}$=$\begin {pmatrix} 3 &2\\2 &3 \end{pmatrix}$

I think that the theorem is written incorrectly , above I have followed all the steps but of course I have not gotten a diagonal matrix at the end because that would imply 2=0 which is nonsense.

So when we do the calculation by this methos we don't see any diagonal matrices

Am I correct in my understanding of the above ?

(Note:There is another equivalent method where $A=CSC^T$, where C is a matrix with unit e-vectors as columns , and S is a diagonal matrix with e-values as entries.)

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  • $\begingroup$ Don't you need to normalize the eigenvectors? Also, I think you mixed up which eigenvalue goes with which eigenvector. $\endgroup$
    – user169852
    Jul 11, 2019 at 20:41
  • $\begingroup$ @Bungo I edited it to fix the mistakes you mentioned but I'm still doing something wrong , would you mind having a look please ? $\endgroup$
    – excalibirr
    Jul 11, 2019 at 21:20
  • $\begingroup$ @bhapi $(1,1)$ normalizes to $(1/\sqrt{2},1/\sqrt{2})$. Its the sum of squares that must add to 1. $\endgroup$ Jul 11, 2019 at 22:07
  • $\begingroup$ @eyeballfrog oops , I meant to write that, the rest of the calculations use that normalisation though $\endgroup$
    – excalibirr
    Jul 11, 2019 at 22:15
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    $\begingroup$ So this gives you a decomposition $A = EDE^T$. To go the other way (so you get the diagonal matrix as the output), multiply both sides on the left by $E^T$ and on the right by $E$ to get $E^T A E = D$. $\endgroup$
    – user169852
    Jul 11, 2019 at 22:34

1 Answer 1

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It would’ve be good if you had quoted the theorem verbatim from your source because I think that you might be misreading it and seem to be missing some important details from its statement. The theorem makes two separate but related claims. For a diagonalizable operator:

  1. Expressed in a basis of eigenvectors, the matrix of the operator is diagonal.
  2. Given an orthonormal basis of eigenvectors, the operator can be decomposed into a linear combination of projectors onto the spans of the individual eigenvectors. This is the content of the identity $\hat A = \sum_{k,j} A_{kj}\,\lvert\varphi_k\rangle \langle\varphi_j\rvert$.

In order to understand these claims correctly it’s important to distinguish between a vector $\lvert v\rangle$ and its coordinate representation relative to some basis $\mathcal B$, which I’ll denote $[v]_{\mathcal B}\in\mathbb C^n$. This distinction can be easy to forget about when the vector itself is an element of $\mathbb C^n$. Similarly, one must distinguish between the operator $\hat A$ and its matrix representation $[\hat A]_{\mathcal B'}^{\mathcal B}$.

Claim #2 above is a statement about an operator and its eigenvectors. It is true independent of the basis in which you choose to represent these vectors and operator. Note that this representation basis isn’t necessarily the orthonormal eigenbasis that’s the subject of the theorem—it can be any orthonormal basis whatsoever. Claim #1, on the other hand, is about the representation of the operator $\hat A$ in a particular basis. Specifically, it says that if $\mathcal B = \{\lvert\phi_i\rangle\}$ is an orthonormal basis that consists of eigenvectors of $\hat A$, then the matrix $[\hat A]_{\mathcal B}^{\mathcal B}$ is diagonal.

When you tried to verify the first claim by applying the decomposition in the second, you did so relative to the standard basis $\mathcal E$. That is, you computed $[\hat A]_{\mathcal E}^{\mathcal E} = \sum_i \lambda_i ([\phi_i]_{\mathcal E})^T[\phi_i]_{\mathcal E}$, but this is just your original matrix. Instead, you need to express everything relative to the eigenbasis $\mathcal B=\{\lvert\phi_i\rangle\}$: $[\hat A]_{\mathcal B}^{\mathcal B} = \sum_i \lambda_i([\phi_i]_{\mathcal B})^T[\phi_i]_{\mathcal B}$.

In order to do this, you first need an orthonormal eigenbasis. The two eigenvectors that you found are orthogonal, but you do need to normalize them as you’ve done in a later edit. We then have $$[\varphi_1]_{\mathcal B} = \begin{bmatrix}1\\0\end{bmatrix} \text{ and } [\varphi_2]_{\mathcal B} = \begin{bmatrix}0\\1\end{bmatrix},$$ so $$5 ([\varphi_1]_{\mathcal B})^T[\varphi_1]_{\mathcal B} + 1([\varphi_2]_{\mathcal B})^T[\varphi_2]_{\mathcal B} = 5\begin{bmatrix}1&0\\0&0\end{bmatrix}+\begin{bmatrix}0&0\\0&1\end{bmatrix} = \begin{bmatrix}5&0\\0&1\end{bmatrix},$$ which is diagonal as claimed.

It’s worth noting that this diagonal decomposition is just a special case of a more general outer product decomposition of an operator. For any orthonormal basis, $\hat I = \sum_i \lvert i\rangle\langle i\rvert$, and so $$\hat A = \hat I\hat A\hat I = \sum_{i,j} \langle i \rvert \hat A \lvert j \rangle \lvert i\rangle\langle j\rvert.$$ Now, if $\lvert i\rangle$ also happens to be an eigenvector of the diagonalizable operator $\hat A$ with eigenvalue $\lambda_i$, then $\hat A\lvert i\rangle = \lambda_i\lvert i\rangle$ and $\langle i\rvert \hat A\lvert j\rangle = \lambda_i\delta_{ij}$, which leads to the diagonal decomposition $$\hat A = \hat I\hat A\hat I = \sum_{i,j} \langle i \rvert \hat A \lvert j \rangle \lvert i\rangle\langle j\rvert = \sum_{i,j} \lambda_i\delta_{ij} \lvert i\rangle\langle j\rvert = \sum_i \lambda_i \lvert i\rangle\langle i\rvert.$$ Relative to this basis, the corresponding matrix is diagonal.

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  • $\begingroup$ I'm not sure I understand how we're using the change of basis here , So when we take the $A_{kj}=<\phi_k|\hat{ A}|\phi_j>= α_jδ_{kj}$, we use the $|\phi_1'>=\begin {pmatrix} 1/\sqrt{2} \\-1/\sqrt{2} \end {pmatrix}, |\phi_2'>=\begin {pmatrix} 1/\sqrt{2} \\1/\sqrt{2} \end {pmatrix}$ ? (seems as how you got the same coeeficients in your answer as mine ) but then when take the $|\phi_k><\phi_j|$ we're using the changed basis $\lvert\varphi_1\rangle = \begin{bmatrix}1\\0\end{bmatrix} \text{ and } \lvert\varphi_2\rangle = \begin{bmatrix}0\\1\end{bmatrix},$ ? Also how did you obtain the changed basis $\endgroup$
    – excalibirr
    Jul 18, 2019 at 21:03
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    $\begingroup$ @excalibirr You need to be very careful to distinguish between a vector and its representation as a coordinate tuple. The basis relative to which we’re doing the decomposition hasn’t changed: that’s always $\{\lvert\phi_i\rangle\}$. What’s changed in order to get the diagonal matrix is the basis used to represent these eigenvectors and the matrix of the operator. For any basis $\{v_1,\dots,v_n\}$ whatsoever, the coordinate vector of $v_1$ relative to this basis is always $[1,0,\dots,0]^T$, of $v_2$ is $[0,1,\dots,0]^T$ and so on. $\endgroup$
    – amd
    Jul 18, 2019 at 23:55

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