Determinat of a continous matrix?

[wenty]

Determinat of a "continous" matrix?

a(x,y) is a matrix element ,and x,y is the row and column index.
If x,y are real numbers, how to calculate the determinat of this matrix or the inverse matrix?

An example of this kind of matrix is <k|H|k'>,a Hamiltonian in momentum representation.

 

[dextercioby]

So it's an infinite dimensional matrix...

If the matrix elements $\hat{H}_{k,k'}=:\langle k|\hat{H}|k' \rangle$ are real,then this matrix is symmetric.Then u can find an orthogonal matrix which would diagonalize the hamiltonian matrix.

Then

$$\det\left(\hat{H}_{k,k'}}\right)=exp \ \left(trace \ ln \hat{H}^{diag}_{k,k'}\right)$$



Daniel.

 

[wenty]

So it's an infinite dimensional matrix...

If the matrix elements $\hat{H}_{k,k'}=:\langle k|\hat{H}|k' \rangle$ are real,then this matrix is symmetric.Then u can find an orthogonal matrix which would diagonalize the hamiltonian matrix.

Then

$$\det\left(\hat{H}_{k,k'}}\right)=exp \ \left(trace \ ln \hat{H}^{diag}_{k,k'}\right)$$



Daniel.

The problem is how to find this orthogonal matrix.In ordinary case,to find this matrix we should solve an equation which needs to know the determinant of the given matrix.Then the question remains.

 

[dextercioby]

How about u search for a basis (a new set of kets $|k\rangle$) made up of eigen vectors (in general sense,the spectrum in continuous) of the Hamiltonian,and then everything would be tremendously simple...?

Daniel.

 

[wenty]

How about u search for a basis (a new set of kets $|k\rangle$) made up of eigen vectors (in general sense,the spectrum in continuous) of the Hamiltonian,and then everything would be tremendously simple...?

Daniel.

But if what you only know is a(i,j),which isn's the matrix elements of the Hamiltonian,how do you find the eigen vectors?