Problem 6 By considering the commutator, show that the following Hermitian matrices may be|simultaneously diagonalized. Find the eigenvectors common to both and verify that under a unitarytransformation to this basis, both matricies arediagonalized21 0 111(4)0 0 01-121 0 1-1Since is degenerate and A is not, you must be prudent in deciding which matrix dictates the choiceof basis

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Asked Sep 26, 2019
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It's a problem in the field of quantum mechanics.

Problem 6 By considering the commutator, show that the following Hermitian matrices may be
|simultaneously diagonalized. Find the eigenvectors common to both and verify that under a unitary
transformation to this basis, both matricies are
diagonalized
2
1 0 1
1
1
(4)
0 0 0
1
-1
2
1 0 1
-1
Since is degenerate and A is not, you must be prudent in deciding which matrix dictates the choice
of basis
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Problem 6 By considering the commutator, show that the following Hermitian matrices may be |simultaneously diagonalized. Find the eigenvectors common to both and verify that under a unitary transformation to this basis, both matricies are diagonalized 2 1 0 1 1 1 (4) 0 0 0 1 -1 2 1 0 1 -1 Since is degenerate and A is not, you must be prudent in deciding which matrix dictates the choice of basis

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Expert Answer

Step 1

For simultaneous diagonalization of the matrix, the matrices specified by Ω and Λ must commute,

For the matrices to commute, their commutator brackets must evaluate to zero,

2.AJ0
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2.AJ0

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Step 2

Evaluating the commutator bracket, we have

0 12
1 1
0 1
1
1
2
1
A-A 0
0 0 1
0 0 0
-1
1 0
-1
_
1 0 1
-1
-1
1 0
1
1
2
1
2
3 0
3
3 0
3
0 0
0 0
0
10
3 0
3
3
10
3
0
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0 12 1 1 0 1 1 1 2 1 A-A 0 0 0 1 0 0 0 -1 1 0 -1 _ 1 0 1 -1 -1 1 0 1 1 2 1 2 3 0 3 3 0 3 0 0 0 0 0 10 3 0 3 3 10 3 0

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Step 3

The commutator for Ω and Λ equals zero which proves that it is simultaneously diagonalizable

 

To calculate the eigenvectors of Ω, first calculate the eigenvalues of the Ω by using the  characteristic equation

Ω-...

1 0 1
10 0
0 0 0-2 0 1 0=0
1 0
1
1
0 0
(1-20
1
0
-2
0
1-2)
1
(1-2)(2(-1)1(2)= 0
a(1-1-2)-0
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1 0 1 10 0 0 0 0-2 0 1 0=0 1 0 1 1 0 0 (1-20 1 0 -2 0 1-2) 1 (1-2)(2(-1)1(2)= 0 a(1-1-2)-0

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