Raising e to a Non-Diagonal Matrix | Calculating eiaB Components

[Vitani11]

Homework Statement

Find all components of the matrix e^iaB. a is a constant and B is a 3x3 matrix whose first row is 0,0,-i second row is 0,0,0 and third row is i,0,0. The taylor expansion of e^iaB gives 1+iaB-a²B²/2! - ...

Homework Equations

The taylor expansion of e^iaB gives 1+iaB-a²B²/2! - ...

The Attempt at a Solution

I don't know what to do from here. If this is a diagonal matrix I would be able to multiply each element in B by ia and then raise e to the power of whatever the result is for each element in the matrix, but this doesn't qualify as a diagonal matrix. I have looked online and can't find any resources that speak about raising e to a non-diagonal matrix.

 

[Vitani11]

Right now I have that the element in place of -i is e^B and the element in place of i is e^-B, the 0's stayed 0

 

[Mark44]

Homework Statement

Find all components of the matrix e^iaB. a is a constant and B is a 3x3 matrix whose first row is 0,0,-i second row is 0,0,0 and third row is i,0,0. The taylor expansion of e^iaB gives 1+iaB-a²B²/2! - ...

Continue what you're doing here with the Maclaurin expansion. The powers of B are cyclical, oscillating between two values.

Homework Equations

The taylor expansion of e^iaB gives 1+iaB-a²B²/2! - ...

The Attempt at a Solution

I don't know what to do from here. If this is a diagonal matrix I would be able to multiply each element in B by ia and then raise e to the power of whatever the result is for each element in the matrix, but this doesn't qualify as a diagonal matrix. I have looked online and can't find any resources that speak about raising e to a non-diagonal matrix.

 

[Vitani11]

I see that it is also cos(aB)+isin(aB), but how does this help? What do you mean by oscillating between two values?

 

[StoneTemplePython]

I don't know what to do from here. If this is a diagonal matrix I would be able to multiply each element in B by ia and then raise e to the power of whatever the result is for each element in the matrix, but this doesn't qualify as a diagonal matrix.

With respect to all square matrices, the way to think about it is: act as if they were diagonal (and failing that settle for just upper triangular).

Do you know how to diagonalize a matrix? It can definitely be done in this case.

 

[vela]

I see that it is also cos(aB)+isin(aB), but how does this help? What do you mean by oscillating between two values?

The Cayley-Hamilton theorem says that B satisfies its characteristic polynomial. In this case, that means $B^3-B = 0$. It follows then that all odd powers of B are equal to B and all even powers are equal to $B^2$. If you calculate the first few powers of B, you can verify this is the case.

 

[Vitani11]

Okay so I can diagonalise this thing and find its matrix of eigenvalues and then I can just raise the elements in this diagonal matrix to the power e like I would for a matrix which was originally diagonal since it is equivalent to B?

 

[Dick]

Okay so I can diagonalise this thing and find its matrix of eigenvalues and then I can just raise the elements in this diagonal matrix to the power e like I would for a matrix which was originally diagonal since it is equivalent to B?

You aren't quite done then. You still have to take the resulting diagonal matrix and transform back to the original basis. Ie. undo the diagonalization.

 

[Vitani11]

Okay. After I do this then I can use the formula P^-1DP=B to transform this back to B.

 

[Dick]

Okay. After I do this then I can use the formula P^-1DP=B to transform this back to B.

Right.

 

[Vitani11]

Wow, that is a beautiful result. I learned how to do this two different ways now :) thank you for the help!