About Schrödinger-Heisenberg connection

[snoopies622]

I stumbled over something reading Green's Matrix Mechanics (1965) this afternoon. There was an equation very similar to one I saw in Dirac's Lectures on Quantum Field Theory (1966), where he talks about the equivalence (or near equivalence) of the Schrödinger and Heisenberg formulations of ordinary quantum mechanics:

<br /> <br /> U_{S} = e ^ {-iHt/ \hbar} U_{H} e^ {iHt/ \hbar }<br /> <br />

I take it that the U's are matrices, but what are the exponential terms? Vectors? Other matrices? If H is a matrix, what does it mean to raise a real number (e) to the power of a matrix? If instead they are real numbers, wouldn't the two exponential terms then simply cancel each other out?

 

[Jorriss]

They are infact matrices.

http://en.wikipedia.org/wiki/Matrix_exponential

Functions of matrices are generally defined by their power series.

 

[dextercioby]

They are operators. H is an operator, the exponential of an operator is defined through the series expansion as an infinite sum. Of course, technicalities such as convergence, domain mustn't be overlooked.

 

[snoopies622]

Hey thanks, I had no idea!

 

[tom.stoer]

Usually a function of an (hermitean) operator is defined using the spectral decomposition. Let A be an hermitean operator, let a be its eigenvalues and let |a> define the eigenvectors = an orthonormal basis.

Then

f(A) = f(A) \sum_a|a\rangle\langle a| = \sum_a f(a)\,|a\rangle\langle a|

 

[snoopies622]

Interesting . . Will try this on a few very simple examples and see what happens.