Schrödinger equation: eigen value or differential equation

[Tanja]

I have a problem on the basis of quantum mechanics and it's so simple that I'm almost too afraid to ask. Anyway:

1.) Schrödingers differential equation is used for time indepent and time depent problems. The solution is a wave function or the linear combination of the resulting wave functions, but not a vector.

2.) The Eigen Value equation. Here the solution are states which are vectors and the Hamiltonian is a matrix.

My problems are:
1.)When becomes the Hamiltonian a matrix? If the spin is taken into account and the Hamiltonian can be expressed in terms of the Pauli matrices? Are there any other potentials that can be written in matrix form?

2.) The density of states is the sum over all eigen states. If these states are vectors, the density becomes a matrix. Here again: Is the density a matrix if the spin is regarded or are there any other examples, spin states excluded, where the density can written in matrix form?

Thanks for attending to my embarrising problem

 

[Dr Transport]

Tha Hamiltonian becomes a matrix when you are describing a system of interacting states, for example calculating the band structure of a solid. Each wavefunction descibes a band, at certain points the solutions may be sompletely separable, but away from each of these points the solutions are combinations.

You also have to solve a matrix formulation when considering perturbation of degenerate states.

As for the density of states being a matrix, if it is, it is diagonal, all the elements may not be equal, but it is definitely diagonal. ( I have done density of states calculations for anisotropic crystal systems.)

 

[blechman]

The Hamiltonian is a LINEAR OPERATOR (as in from Linear Algebra). There is a theorem that states that any linear operator can be expressed as a matrix. So the Hamiltonian is ALWAYS a matrix! As is the density of states.

Similarly, the "wavefunctions" ARE vectors in the Hilbert space - a linear space where functions are the "vectors" (sometimes called a "function space").

Remember that in linear algebra, "vectors" do not have to be those pointy things that you think about in high-school math! The subject is much more abstract than that.

 

[birulami]

There is a theorem that states that any linear operator can be expressed as a matrix.

Ooops, I would expect that this very much depends on the space on which the operator is defined. If, for example, the space is \{f| f: R\to R\} and the operator is d/dx, I wonder how you write this operator as a matrix?

Harald.

 

[jostpuur]

Ooops, I would expect that this very much depends on the space on which the operator is defined. If, for example, the space is \{f| f: R\to R\} and the operator is d/dx, I wonder how you write this operator as a matrix?

Harald.

Just what the word "matrix" means is getting a little relative. There can be infinite matrices too. Frankly, I have not seen matrices that would have uncountably elements, in strictly mathematical texts, but physicists seemingly have no problems with this either.

The derivative operator gets diagonalized in the Fourier space, so there the matrix at least is trivial to write. Similarly as a n times n matrix is a mapping

\{1,2,\ldots,n\}\times\{1,2,\ldots,n\}\to\mathbb{C},

the derivative operator matrix is a mapping (okey this is not a mapping but a distribution, but who cares...)

<br /> \mathbb{R}\times\mathbb{R}\to\mathbb{C},<br /> \quad (k_1,k_2)\mapsto ik_1 \delta(k_1-k_2)<br />

And similarly as the action of the n times n matrix on a vector is given by

<br /> (A x)_i = \sum_{j=1}^n A_{ij} x_j<br />

when

<br /> x=\sum_{i=1}^n x_i e_i<br />

the action of the derivative operator is given by

<br /> (D f)(k) = \int dk&#039;\; ik\delta(k-k&#039;) f(k&#039;) = ik f(k)<br />

when

<br /> |f\rangle = \int dk\; f(k)|k\rangle.<br />

For position representation, write the functions e^ikx as the basis

<br /> \int dk\; f(k) e^{ikx} \mapsto \int dk\; f(k) ik e^{ikx} = \frac{d}{dx} \big(\int dk\; f(k) e^{ikx}\Big)<br />

 

[Tanja]

Thanks all, for your help and the huge mathematical background. Nevertheless, my problem is based on the physics behind. It originally arised as I thought about the density matrix and the time evolution operators: \rho(t) = U' \rho (t=o) U, with U = exp(\frac{i}{\hbar}\int H(t)dt. I know that this is extremely useful in calculations with spins.

But, what if I regard the charge density of electrons in a peturbation field, neglecting spin. The Hamiltonian would be a matrix, but a matrix with coordinates in real space. The resulting wave functions are vectors, but vectors with respect to all space directions. Anyway you can't compare the resulting density matrix with the one you get by spin calculation and I wonder wether the basis transformation of \rho (t=0) can be used in this case, cause the density matrix is not an evolution of Pauli matrices in the Bloch spere any more.

So, is it possible to use the time evolution with the unitary matrices in the case of a density matrix consisting of eigen vectors in real space??

And thanks again for the discussion on operators and I agree, the wave functions can definatively be seen a vector. But again, the density matrix of these states: does the time evolution still hold? I suspect the answer is no, but I would be sooo happy for any other point of views.