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# Orthonormal Basis Vectors

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Given two sets of complete orthonormal basis vectors: {|u1>, |u2>, |u3>,...} and {|v1>, |v2>, |v3>,...}. Use Dirac notation to prove:

1. That for any linear operator Â, the trace is independent of choice of basis i.e.
Tr(Â) = Σ<ui| Â|ui> = Σ<vi| Â|vi>.

2. Given that Tr(Â) = Σ<ui| Â|ui>. Prove that Tr(ÂĈ) = Tr(ĈÂ) for any two linear operators Â and Ĉ.

https://brainmass.com/physics/scalar-and-vector-operations/orthonormal-basis-vectors-216531

## SOLUTION This solution is FREE courtesy of BrainMass!

1. The trace of an operator A is the sum of the diagonal elements in its matrix representation, relative to some (complete orthonormal) basis. That is, if

Tr (A) = Σ;_{i=1}^n < u_i | A | u_i > = < u_1 | A | u_1 > + .... + < u_n | A | u_n >

where {|u_i>} is a (complete orthonormal) basis for the (Hilbert) space on which T is defined.

To show that this definition is independent of the choice of basis, let {|v_i>} be another (orthonormal) basis; then there is a unitary matrix P such that P |u_i> = |v_i> for all i (in particular, <v_i | = <u_i | P*, where P is the conjugate transpose of P). Recall that the identity operator can be written as I = Σ_{i=1}^n |u_i><u_i|. Hence,

Tr (A)

= Σ_{i=1}^n < v_i | A | v_i >

= Σ_{i = 1}^n <u_i | P* (I) A (I) P | u_i>

= Σ_{i = 1}^n Σ_{j = 1}^n Σ_{k = 1}^n <u_i | P* | u_ j > <u_ j | A | u_k > <u_k | P | u_ i>

= Σ_{j, k} (Σ_i <u_i | P* | u_ j >< u_k | P | u_i >) <u_ j | A | u_k>

= Σ_{j, k} δ_{jk} < u_ j | A | u_k > = Σ_k <u_k | A | u_k>

since PP* = P*P = I. (The notation Σ_{j, k} is shorthand for Σ_{j = 1}^n Σ_{k = 1}^n, etc.)

2. We have

Tr(AB) = Σ_{i = 1}^n <u_i | AB|u_i > = Σ_{i=1}^n <u_i | A I B | u_i>

= Σ_{i=1}^n < u_i | (Σ_{j = 1}^n | u_ j >< u_ j| ) B |u_i>

= Σ_{i = 1}^n Σ_{j = 1}^n <u_i | A | u_ j > <u_ j |B| u_i>

= Σ_{i = 1}^n Σ_{j = 1}^n <u_ j |B| u_i><u_i |A| u_ j>

= Σ_{j = 1}^n < u_ j |B| ( Σ_{i=1}^n |u_i ><u_i | ) A |u_ j>

= Σ_{j=1}^n < u_ j |B I A| u_ j> = Tr(BA)

as claimed.

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