Diagonalization and change of basis

In summary, the given conversation discusses a matrix represented by a basis of vectors, with corresponding eigenvalues and eigenvectors. The solutions to further problems involve representing these vectors in a canonical basis, which is a common approach in finite dimensions. However, this convention may not be considered canonical in mathematics, and is more commonly used in quantum mechanics.
  • #1
RicardoMP
49
2
I have the following matrix given by a basis [itex]\left|1\right\rangle[/itex] and [itex]\left|2\right\rangle[/itex]:
[itex]
\begin{bmatrix}
E_0 &-A \\
-A & E_0
\end{bmatrix}
[/itex]

Eventually I found the matrix eigenvalues [itex]E_I=E_0-A[/itex] and [itex]E_{II}=E_0+A[/itex] and eigenvectors [itex]\left|I\right\rangle = \begin{bmatrix}
\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}
\end{bmatrix} and \left|II\right\rangle=\begin{bmatrix}
\frac{1}{\sqrt{2}}\\
-\frac{1}{\sqrt{2}}
\end{bmatrix}[/itex].
I found out in the solutions of further problems that I can write these vectors as [itex]\left|I\right\rangle=\frac{1}{\sqrt{2}}\left|1\right\rangle+\frac{1}{\sqrt{2}}\left|2\right\rangle[/itex] and[itex]\left|II\right\rangle=\frac{1}{\sqrt{2}}\left|1\right\rangle-\frac{1}{\sqrt{2}}\left|2\right\rangle[/itex]
But why do we assume that [itex]\left|1\right\rangle=
\begin{bmatrix}
1 \\
0
\end{bmatrix}
[/itex] and [itex]\left|2\right\rangle=
\begin{bmatrix}
0 \\
1
\end{bmatrix} ?
[/itex]
Is this canonical basis, a basis of every matrix?
 
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  • #2
The column matrix representation of a state ##\newcommand{\ket}[1]{\left| #1 \right\rangle} \ket{\psi}## in a basis ##\ket{1}##, ##\ket{2}## is given by
$$
\newcommand{\braket}[2]{\langle #1 | #2 \rangle}
\begin{pmatrix}
\braket{1}{\psi} \\ \braket{2}{\psi}
\end{pmatrix}
$$
so clearly, if you want to represent the state ##\ket{1}## in this matrix representation, you would get
$$
\begin{pmatrix}
\braket{1}{1} \\ \braket{2}{1}
\end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
$$
 
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Likes RicardoMP and stevendaryl
  • #3
to be crystal clear, you have real symmetric matrix, with eigenvectors

##\mathbf v_1 \propto
\begin{bmatrix}
1\\
1
\end{bmatrix}##

and

##\mathbf v_2 \propto
\begin{bmatrix}
1\\
-1
\end{bmatrix}##

these are orthogonal to each other. If you choose to scale each by ##\frac{1}{\sqrt{2}}## then you may call them orthonormal -- in general this is extremely desirable and hence that is why it was rescaled.

RicardoMP said:
But why do we assume that [itex]\left|1\right\rangle=
\begin{bmatrix}
1 \\
0
\end{bmatrix}
[/itex] and [itex]\left|2\right\rangle=
\begin{bmatrix}
0 \\
1
\end{bmatrix} ?
[/itex]
This appears to be a question about conventions in the use of Dirac Notation, which is not a math question but something for the QM forums.

In general the use of standard basis vectors' coordinates is a common approach when dealing in finite dimensons. Technically for a relevant canonical form you'd look to the Jordan Canonical Form. I don't think using coordinates of the standard basis is canonical per se in math, but again your questions seems to be more about QM conventions than math.
 
  • Like
Likes RicardoMP
  • #4
No, it's an abuse of notation. A ket is an abstract vector, independent of any basis. Column vectors always refer to a basis. Obviously tacitly you assumed that the basis to use should be ##\{|1 \rangle,|2 \rangle \}##. Of course the components of the basis vectors wrt. this basis itself are the "canonical basis" vectors of ##\mathbb{C}^n## (in your case ##n=2##).
 

1. What is diagonalization?

Diagonalization is a process in linear algebra where a square matrix is transformed into a diagonal matrix. This is done by finding a change of basis matrix that can be used to represent the original matrix in a simpler form.

2. Why is diagonalization important?

Diagonalization is important because it simplifies the representation of a matrix and makes it easier to perform calculations. It also allows for the identification of important properties and relationships between the matrix and its eigenvalues and eigenvectors.

3. What is a change of basis?

A change of basis is a transformation that involves changing the coordinate system used to represent a vector or a matrix. This transformation can be represented by a change of basis matrix, which is used to convert the original vector or matrix into its new representation.

4. How is diagonalization related to eigenvalues and eigenvectors?

Diagonalization is closely related to eigenvalues and eigenvectors because the diagonal matrix that results from this process contains the eigenvalues of the original matrix on its main diagonal. The eigenvectors of the original matrix are also used to construct the change of basis matrix.

5. What are the applications of diagonalization?

Diagonalization has many applications in various fields of science and engineering, including mechanics, physics, and signal processing. It is commonly used to solve systems of differential equations, analyze linear transformations, and perform data compression and dimensionality reduction.

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