Density matrix representation of a density operator

[cianfa72]

TL;DR

About the representation of a density operator by means of a density matrix

A doubt related to the use of density matrix associated to a density operator $\rho$.

A quantum pure state $\ket {\psi}$ can be written as density operator as $\ket{\psi} \bra{\psi}$.

In this link the density matrix $\rho$ for the pure state $\ket {\psi} =(\ket{\psi_1} + \ket{\psi_2})/\sqrt{2}$ is given by $$\rho = \frac 1 2 \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$ I believe this density matrix representation makes sense only w.r.t. a basis that includes the vectors $\{ \ket{\psi_1}, \ket{\psi_2} \}$ assumed to be orthonormal (the dimension/size of the density square matrix should be the dimension of the underlying system's Hilbert space).

 

[Haborix]

Certainly once you write down the matrix components of an operator you have chosen a basis. Not sure if you had a specific question in mind.

 

[cianfa72]

Certainly once you write down the matrix components of an operator you have chosen a basis. Not sure if you had a specific question in mind.

Yes. The point I was making is that any density operator $\rho$ has a density matrix representation associated with a given basis chosen for the underlying Hilbert state space.

What if the Hilbert space is infinite dimensional (like the position/momentum state space) ? I don't think we get a matrix with an infinite number of rows and columns

 

[Cthugha]

Why not? This is typically the case for bosonic systems. Consider, e.g., a standard quantum mechanical harmonic oscillator in the Fock (occupation number) basis. In principle, the Hilbert space is infinite-dimensional. In practice, of course there is only a finite amount of energy available and you can truncate the Hilbert space at some point.

If you really want to look at continuous variables, it is often easier to change to a phase space formulation (e.g. the Wigner function) which contains the same information. The Wigner function is the Weyl transform of the density matrix.

 

[cianfa72]

Why not? This is typically the case for bosonic systems. Consider, e.g., a standard quantum mechanical harmonic oscillator in the Fock (occupation number) basis.

Do you mean the square density matrix representation may have actually an infinite (countable) number of rows and columns ?

 

[Cthugha]

Yes, but you of course still expect the typical properties a density operator needs to have, for example that they are trace class.

Of course one must be careful here from the mathematical and formal point of view as some things are different for infinite-dimensional stuff. For example, there is no really meaningful definition of a maximally mixed state in infinite dimensions.

 

[Nugatory]

Do you mean the square density matrix representation may have actually an infinite (countable) number of rows and columns ?

It's not even restricted to countable infinities.

 

[WWGD]

Do you mean the square density matrix representation may have actually an infinite (countable) number of rows and columns ?

And you may even have a determinant, though not always, as you will have convergence issues.

 

[WWGD]

It's not even restricted to countable infinities.

Isn't the support, i.e., the set of nonzero entries, countable?