Change of basis of density matrix

[Bizarre123]

I have a density matrix in one basis and need to change it to another. I know the eigenvectors and eigenvalues of the basis I want to change to. How do I do this?

Any help really appreciated- thanks!

 

[andrien]

All I can say is that TRACE(ρA) is basis independent.

 

[Bizarre123]

Thanks- but in this case I need to know more than the trace?

 

[andrien]

Any other base state can be represented as a linear combination of original base states(complete).So may be by equality of traces some relation can be obtained for it.

 

[Bill_K]

If your density matrix is ρ = Σ|i>p_ij<j| in the original basis with states |i> and probabilities p_ij, and the new basis states are |α>, then expand the old basis in terms of the new: |i> = Σ|α><α|i>. This gives you ρ = ΣΣΣ|α><α|i>p_ij<j|β><β| = Σ|α>P_αβ<β| where P_αβ = ΣΣ<α|i>p_ij<j|β>.

 

[Dickfore]

By using the matrix identities:
$$\mathbf{\rho} \cdot \mathbf{U} = \mathbf{U} \cdot \mathbf{\Lambda}$$
where U is a unitary matrix ($\mathbf{U}^{\dagger} \cdot \mathbf{U} = \mathbf{U} \cdot \mathbf{U}^{\dagger} = 1$) whose columns are the normalized eigenvectors of the density matrix, and $\Lambda$ is a diagonal matrix with the corresponding eigenvalues along the main diagonal.

 

[tom.stoer]

I think we should distinguish between the density matrix ρ_mn and the density operator ρ.

Using a basis {|n>} this reads

$$\rho = \sum_{mn}\rho_{mn}|m\rangle\langle n|$$

Now let's introduce a different basis {|k'>}

$$\rho = \sum_{k^\prime} |k^\prime\rangle\langle k^\prime| \;\;\sum_{mn}\rho_{mn}|m\rangle\langle n|\;\; \sum_{l^\prime} |l^\prime\rangle\langle l^\prime| = \sum_{k^\prime l^\prime} \left(\sum_{mn}\langle k^\prime|m\rangle\,\rho_{mn}\,\langle n|l^\prime\rangle \right) |k^\prime\rangle\langle l^\prime| = \sum_{k^\prime l^\prime}\rho_{k^\prime l^\prime} |k^\prime\rangle\langle l^\prime|$$

 

[Dickfore]

The change-of-basis identity:
$$\rho_{k' l'} = \sum_{m n}{\langle k' \vert m \rangle \, \rho_{m n} \, \langle n \vert l' \rangle}$$
with the identification of the matrix:
$$U_{n l'} \equiv \langle n \vert l'\rangle$$
can be rewritten as:
$$\mathbf{\rho}' = \mathbf{U}^{\dagger} \cdot \mathbf{\rho} \cdot \mathbf{U}$$
where we used the fact that:
$$\hat{U}^{\dagger}_{k' m} = U^{\ast}_{m k'} = (\langle m \vert k' \rangle)^{\ast} = \langle k' \vert m \rangle$$

The unitarity of the similarity transformation matrix U is an expression of the orthonormality of the old and new bases:
$$\langle m \vert n \rangle = \sum_{k'} {\langle m \vert k' \rangle \, \langle k' \vert n \rangle} = \sum_{k'} {U_{m k'} \, U^{\dagger}_{k' n}} = \left[\mathbf{U} \cdot \mathbf{U}^{\dagger}\right]_{m n} = \delta_{m n} \Rightarrow \mathbf{U} \cdot \mathbf{U}^{\dagger} = \mathbf{1}$$
$$\langle k' \vert l' \rangle = \sum_{m} \langle k' \vert m \rangle \, \langle m \vert l' \rangle = \sum_{m} U^{\dagger}_{k' m} \, U_{m l'} = \left[ \mathbf{U}^{\dagger} \cdot \mathbf{U} \right]_{k' l'} = \delta_{k' l'} \Rightarrow \mathbf{U}^{\dagger} \cdot \mathbf{U} = \mathbf{1}$$

 

[tom.stoer]

exactly - I hope Bizarre123 is still interested ;-)

 

[Bizarre123]

Thanks for the replies- they've been really helpful! Yes, I was getting the density operator and the density matrix confused.