- #1

Qubix

- 82

- 1

P(a,b)=(a',b'), am I correct in assuming that I can change an operator A, from the (a,b) basis to the (a',b') basis by applying

A' = P_transposed * A * P ?

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In summary, the conversation discusses the use of transformation matrices to change operators from one basis to another. The change of basis matrix is given by a combination of scalars and the unitary transformation matrix is used to change operators between two bases. In some cases, the transpose of the transformation matrix can be used instead of the inverse, but this is a special case.

- #1

Qubix

- 82

- 1

P(a,b)=(a',b'), am I correct in assuming that I can change an operator A, from the (a,b) basis to the (a',b') basis by applying

A' = P_transposed * A * P ?

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- #2

kduna

- 52

- 7

Similarly, ##b = d_1a' + d_2b'##. The the change of basis matrix, ##P##, from ##(a,b)## to ##(a', b')## is given by:

##P = \left( \begin{array}{cc}

c_1 & d_1 \\

c_2 & d_2 \\ \end{array} \right) ##.

Then ## A' = PAP^{-1} ##.

There are times when you can get away with using the transpose instead of the inverse, but that is only when the transpose is actually equal to the inverse. This is a very special case.

- #3

Qubix

- 82

- 1

a' = 1/sqr(2) ( a + b)

b' = 1/sqr(2) (a - b)

am I correct in saying that the unitary transformation between them is

U = 1/sqr(2) ## \left( \begin{array}{cc}

1 & 1 \\

1 & -1 \\ \end{array} \right) ##. ?

and then

Then ## A' = UAU^{-1} ##.

- #4

kduna

- 52

- 7

Qubix said:

a' = 1/sqr(2) ( a + b)

b' = 1/sqr(2) (a - b)

am I correct in saying that the unitary transformation between them is

U = 1/sqr(2) ## \left( \begin{array}{cc}

1 & 1 \\

1 & -1 \\ \end{array} \right) ##. ?

and then

Then ## A' = UAU^{-1} ##.

That looks great. In this case, not only is ##U^T = U^{-1}##. But you have that both of those are ##U## itself.

The purpose of transforming operators with matrix P is to change the basis in which a linear transformation is represented. This can make calculations and analysis of the transformation easier and more efficient.

To transform an operator with matrix P, you first need to determine the matrix representation of the operator in the original basis. Then, you multiply this matrix by the matrix P, which represents the transformation between the original basis and the new basis. The result is the matrix representation of the operator in the new basis.

Transforming operators with matrix P can simplify calculations and analysis by changing the basis to one that is more convenient. It can also help identify patterns and relationships between different operators and their representations in different bases.

Yes, any linear operator can be transformed with matrix P as long as the dimensions of the original and new bases are the same. This includes operators in vector spaces such as 2D and 3D space, as well as operators in more abstract vector spaces like function spaces.

Matrix P is used to transform an operator into a new basis, and the eigenvectors and eigenvalues of the original operator will still be valid for the transformed operator. This means that the transformed operator will have the same eigenvectors and eigenvalues as the original operator, but represented in the new basis.

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