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:
<br /> \mathbf{\rho} \cdot \mathbf{U} = \mathbf{U} \cdot \mathbf{\Lambda}<br />
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:
<br /> \rho_{k&#039; l&#039;} = \sum_{m n}{\langle k&#039; \vert m \rangle \, \rho_{m n} \, \langle n \vert l&#039; \rangle}<br />
with the identification of the matrix:
<br /> U_{n l&#039;} \equiv \langle n \vert l&#039;\rangle<br />
can be rewritten as:
<br /> \mathbf{\rho}&#039; = \mathbf{U}^{\dagger} \cdot \mathbf{\rho} \cdot \mathbf{U}<br />
where we used the fact that:
<br /> \hat{U}^{\dagger}_{k&#039; m} = U^{\ast}_{m k&#039;} = (\langle m \vert k&#039; \rangle)^{\ast} = \langle k&#039; \vert m \rangle<br />

The unitarity of the similarity transformation matrix U is an expression of the orthonormality of the old and new bases:
<br /> \langle m \vert n \rangle = \sum_{k&#039;} {\langle m \vert k&#039; \rangle \, \langle k&#039; \vert n \rangle} = \sum_{k&#039;} {U_{m k&#039;} \, U^{\dagger}_{k&#039; n}} = \left[\mathbf{U} \cdot \mathbf{U}^{\dagger}\right]_{m n} = \delta_{m n} \Rightarrow \mathbf{U} \cdot \mathbf{U}^{\dagger} = \mathbf{1}<br />
<br /> \langle k&#039; \vert l&#039; \rangle = \sum_{m} \langle k&#039; \vert m \rangle \, \langle m \vert l&#039; \rangle = \sum_{m} U^{\dagger}_{k&#039; m} \, U_{m l&#039;} = \left[ \mathbf{U}^{\dagger} \cdot \mathbf{U} \right]_{k&#039; l&#039;} = \delta_{k&#039; l&#039;} \Rightarrow \mathbf{U}^{\dagger} \cdot \mathbf{U} = \mathbf{1}<br />

 

[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.