- #1

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e.g exp(-iaL), where L is a 4*4 matrix (like a group generator )

- Thread starter princeton118
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- #1

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e.g exp(-iaL), where L is a 4*4 matrix (like a group generator )

- #2

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By its power series:

[tex]\exp(-iaL)=\sum_{n=0}^{\infty}\frac{(-ia)^nL^n}{n!} [/tex]

[tex]\exp(-iaL)=\sum_{n=0}^{\infty}\frac{(-ia)^nL^n}{n!} [/tex]

- #3

Ben Niehoff

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[tex]L = P^{-1}DP[/tex]

where

[tex]D = \left[ \begin{array}{cccc}\lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_n \end{array} \right][/tex]

for the eigenvalues [itex]\lambda_k[/itex]. Then

[tex]e^L = e^{P^{-1}DP} = P^{-1}e^DP[/tex]

(sorry, I forget the proof of this). Then [itex]e^D[/itex] is easy to evaluate:

[tex]e^D = \left[ \begin{array}{cccc}e^{\lambda_1} & & & \\ & e^{\lambda_2} & & \\ & & \ddots & \\ & & & e^{\lambda_n} \end{array} \right][/tex]

- #4

HallsofIvy

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Also to prove Ben Niehoff's forumla, you can use the Taylors series for e

Then

[tex]e^A= I+ A+ \frac{1}{2}A^2+ \cdot\cdot\cdot+ \frac{1}{n!}A^n+ \cdot\cdot\cdot[

[tex]= I+ PDP^{-1}+ \frac{1}{2}(PDP^{-1})^2+ \cdot\cdot\cdot+ \frac{1}{n!}(PDP^{-1})^n+ \cdot\cdot\cdot[/tex]

[tex]= (PP^{-1})+ PDP^{-1}+ /frac{1}{2}(PD^2P^{-1})+ \cdot\cdot\cdot+ \frac{1}{n!}+ PD^nP^{-1}+ \cdot\cdot\cdot[/tex]

[tex]= P(I+ D+ \frac{1}{2}D^2+ \cdot\cdot\cdot+ \frac{1}{n!}D^n+ \cdot\cdot\cdot)P^{-1}[/tex]

[tex]= Pe^DP^{-1}[/tex]

and e

With that "i" you might find it better to use [itex]e^{iA}= cos(A)+ i sin(A)[/itex]. You can find cos(A) and sin(A) by using their Taylor series in exactly the same way: if A is diagonalizable- [itex]A= PDP^{-1}[/itex], then cos(A)= Pcos(D)P^{-1}, sin(A)= Psin(D)P^{-1}. Of course, cos(D) is the diagonal matrix with diagonal elements cos(a) for every a on the diagonal of D and sin(D) is the diagonal matrix with diagonal elements sin(a) for every a on the diagonal of D.

- #5

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Let [itex]A[/itex] be an operator, [itex]a_k[/itex] its eigenvalues and [itex]|\Psi_k \rangle[/itex] the eigenvectors,

i.e. you have the eigenequation [itex]A |\Psi_k \rangle = a_k |\Psi_k \rangle[/itex].

Functions of an operator [itex]A[/itex] are calculated as

[tex]f(A) = \sum_{k=1}^N f(a_k) |\Psi_k \rangle \langle \Psi_k|[/tex]

Let's call this

(see Plenio, lecture notes on quantum mechanics 2002, page 51 eq. (1.86) http://www.lsr.ph.ic.ac.uk/~plenio/teaching.html [Broken]).

Equation (1) can also be used to calculate functions of matrices.

Let A be a matrix with the eigenequation

[tex]A \vec{\Psi}_k = a_k \vec{\Psi}_k[/tex]

Then equation (1) becomes

[tex]f(A) = \sum_{k=1}^N f(a_k) \vec{\Psi}_k (\vec{\Psi}_k)^T[/tex]

[itex]a_k[/itex] is the eigenvalue, [itex]\vec{\Psi}_k[/itex] is the eigenvector (column vector) and [itex]\vec{\Psi}_k^T[/itex] is the transposed eigenvector (row vector).

In your case we have a matrix L.

You first have to calculate the eigenvalues [itex]l_k[/itex] and eigenvectors [itex]\vec{\Psi}_k[/itex] of L, i.e. you have the eigenequation

[itex]L \vec{\Psi}_k = l_k \vec{\Psi}_k[/itex]

And for your case we consider the function [itex]f(x)=\mbox{exp}(x)[/itex] such that

[itex]f(L)=\mbox{exp}(L)[/itex] and [itex]f(l_i) = \mbox{exp}(l_i)[/itex]

Plugging this into equation (1) we get

[tex]f(L) = \sum_{k=1}^N f(l_k) \vec{\Psi}_k \vec{\Psi}_k^T[/tex]

[tex]\mbox{exp}(L) = \sum_{k=1}^N \mbox{exp}(l_k) \vec{\Psi}_k \vec{\Psi}_k^T[/tex]

In order to calculate [itex]\mbox{exp}(-iaL)[/itex] you just multiply your

eigenequation [itex]L \vec{\Psi}_k = l_k \vec{\Psi}_k[/itex] by [itex](-ia)[/itex]

and get the new eigenequation

[itex](-ai)L \vec{\Psi}_k = (-ai)l_k \vec{\Psi}_k[/itex]

You can interpret this as eigenequation with the matrix

[itex](-ai)L[/itex] whose eigenvalues are [itex](-ai)l_k[/itex]

Thus, you can calculate [itex]\mbox{exp}(-aiL)[/itex] as

[tex]\mbox{exp}(-aiL) = \sum_{k=1}^N \mbox{exp}(-ail_k) \vec{\Psi}_k \vec{\Psi}_k^T[/tex]

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- #7

morphism

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Look up Jordan normal form.

- #8

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Matlab uses a scaling and squaring operation.

exp(A)~exp(A/t)^t

First the matrix is divided by some number so that the matrix can be approximated with a Pade approximation. A pade approximation provides a better fit then a taylor series. Once the matrix exponential fox A/t is approximated then matrix approximation of A is approximated via the above relation.

The method used by MATLAB is superior for most matrices. The diagonalization method is only superior when the eignevalues are nearly orthogonal. There are many cases where numeric problems may arise. For instance if the eigen values are too close together or too far apart. In such cases the numeric methods may not yield accurate computations of the Matrix expontial.

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