- #1

D_Tr

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Hi PF people!

I am not sure my question can elegantly fit in the template, but I 'll try.

I am self-studying the 8th chapter of "Mathematical Methods for Physics and Engineering", 3rd edition by Riley, Hobson, Bence. In the section about unitary matrices, it is stated that:

"A unitary matrix represents, in a particular basis, a linear operator that leaves the norms (lengths) of complex vectors unchanged. If y = Ax is represented in some coordinate system by the matrix equation y = Ax then <y|y> is given in this coordinate system by: y†y = x†A†Ax = x†x = <x|x>.

The above proof used the fact that A†A = I since A is unitary. BUT: Isn't the definition of the inner product used in this proof only valid for vectors defined in an orthonormal basis? In the same chapter, the inner product of two vectors in an arbitrary basis is straightforwardly proved to be:

<a|b> = a†Gb, where G is a square matrix and Gij = <ei|ej>, where ei, ej are basis vectors. When the components are with respect to an orthonormal basis we can calculate the inner product by just multiplying the corresponding components and adding the products (G becomes the identity matrix), but in an non orthonormal basis this is not the case.

Question: Do unitary matrices preserve length in a non-orthonormal basis? If yes, how do we prove it? Why does the book use the formula for the orthonormal basis? Some sources I found on google just do what the book does whitout reference to the type of basis used.

The relevant equations are the ones in 1, I do not think that anything else is needed.

In a non orthonormal basis, we have:

<y|y> = y†Gy = (Ax)†G(Ax) = x†A†GAx

<x|x> is x†Gx in an arbitrary basis. Additionally, G is hermitian, since <ei|ej> = <ej|ei>*, and A is unitary.

Can we prove that x†A†GAx = x†Gx or does this simply not hold? I am really baffled... What I was thinking was the following example: If we construct a matrix that transforms vector [a, b, c], into [b, c, a], this matrix is unitary (the matrix actually is [0 1 0, 0 0 1, 1 0 0]), since times its hermitian conjugate (just its transpose, since it is real) gives the identity matrix. The permuted vector definitely has the same norm in an orthonormal basis, but in an orthogonal basis where we chose one basis vector to be much longer than the others, wouldn't this permutation generally change the vector's length? So if what I am thinking is correct, this unitary matrix changes the vector's norm in a non orthonormal basis...

Thanks for reading :)

I am not sure my question can elegantly fit in the template, but I 'll try.

## Homework Statement

I am self-studying the 8th chapter of "Mathematical Methods for Physics and Engineering", 3rd edition by Riley, Hobson, Bence. In the section about unitary matrices, it is stated that:

"A unitary matrix represents, in a particular basis, a linear operator that leaves the norms (lengths) of complex vectors unchanged. If y = Ax is represented in some coordinate system by the matrix equation y = Ax then <y|y> is given in this coordinate system by: y†y = x†A†Ax = x†x = <x|x>.

The above proof used the fact that A†A = I since A is unitary. BUT: Isn't the definition of the inner product used in this proof only valid for vectors defined in an orthonormal basis? In the same chapter, the inner product of two vectors in an arbitrary basis is straightforwardly proved to be:

<a|b> = a†Gb, where G is a square matrix and Gij = <ei|ej>, where ei, ej are basis vectors. When the components are with respect to an orthonormal basis we can calculate the inner product by just multiplying the corresponding components and adding the products (G becomes the identity matrix), but in an non orthonormal basis this is not the case.

Question: Do unitary matrices preserve length in a non-orthonormal basis? If yes, how do we prove it? Why does the book use the formula for the orthonormal basis? Some sources I found on google just do what the book does whitout reference to the type of basis used.

## Homework Equations

The relevant equations are the ones in 1, I do not think that anything else is needed.

## The Attempt at a Solution

In a non orthonormal basis, we have:

<y|y> = y†Gy = (Ax)†G(Ax) = x†A†GAx

<x|x> is x†Gx in an arbitrary basis. Additionally, G is hermitian, since <ei|ej> = <ej|ei>*, and A is unitary.

Can we prove that x†A†GAx = x†Gx or does this simply not hold? I am really baffled... What I was thinking was the following example: If we construct a matrix that transforms vector [a, b, c], into [b, c, a], this matrix is unitary (the matrix actually is [0 1 0, 0 0 1, 1 0 0]), since times its hermitian conjugate (just its transpose, since it is real) gives the identity matrix. The permuted vector definitely has the same norm in an orthonormal basis, but in an orthogonal basis where we chose one basis vector to be much longer than the others, wouldn't this permutation generally change the vector's length? So if what I am thinking is correct, this unitary matrix changes the vector's norm in a non orthonormal basis...

Thanks for reading :)

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