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According to my skript:

A pure state is a ray: $\quad$ $\{λψ\}$, where $ψ ∈ \mathcal H$, $||ψ|| =1$ fixed and $λ ∈ \mathbb C$, $|λ| = 1$.
Pure states are uniquely given by 1-dimensional orthogonal projectors $Π$ on $\mathcal H$: $\quad$ $Πφ = |ψ\rangle\langleψ|φ\rangle$.
Let $Π(\mathcal H)$ be the set of all such projectors.
Definition. A symmetry is a map $S: Π(\mathcal H) \rightarrow Π(\mathcal H'), Π \rightarrow Π'$, such that $$tr(Π_1Π_2) = tr(Π_1'Π_2')$$ or equivalently in rays $$|\langleψ_1|ψ_2\rangle|^2 = |\langleψ_1'|ψ_2'\rangle|^2,$$
that means, that the probabilites are invariant. Furthermore one can express the expectation value of an observable $A$ through $Π$: $$\langle A\rangle_Π = tr(AΠ) = \langleψ|A|ψ\rangle.$$

Questions:
1) How does one prove $$tr(Π_1Π_2) = |\langleψ_1|ψ_2\rangle|^2?$$ 2) How are the $|ψ_1\rangle, |ψ_2\rangle$ related?

Thanks you in advance. I am confused due to reading too many things from different sources. I mean if I follow another book I can understand the relation with the trace and the expectation value even in the more general case of mixed states but the specific $|ψ_1\rangle, |ψ_2\rangle$ confuse me. Aren't they belonging to a complete orthonormal base $\{ |φ_n\rangle \}_{n=1,2,…}$ and therefore one could write $|ψ_1\rangle = \sum_n ψ_{1_n}|φ_n\rangle$ and then using this base to write out $tr(Π_1Π_2)$?

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  • $\begingroup$ A pure state $|\psi>$ is one of the states in $\mathcal H$, with the property that any power of its density matrix $M$, is equal to the density matrix, i.e. $M^n = M$. Let's show that for the state $|\psi>$. We have $M = |\psi><\psi|$ and $M^2 = |\psi><\psi|\psi><\psi| = |\psi><\psi|$. That, on condition that $|psi>$ is normalized, i.e. $||\psi||^2 = <\psi|\psi> = 1$ . The story with the 1-dimensional orthogonal projectors you don't need it. $\endgroup$
    – Sofia
    Commented Dec 28, 2014 at 20:21
  • $\begingroup$ (continuation) In a Hilbert space $\mathcal H$ you can choose a base of states as you say $ \{|\phi>_n ⟩ \}_{n=1,2,...D}$, where D is the dimension of the Hilbert space. E.g. for a particle of spin $\hbar/2$, $D = 2$. However, the choice of a base is not unique. In your Hilbert space, you can choose a base s.t. one of the base-vectors is your function $\psi>$. The other $D-1$ base vectors have to be orthogonal on $\psi>$, and orthogonal on one another. When we pass from one base to another, we say that we rotate the base. $\endgroup$
    – Sofia
    Commented Dec 28, 2014 at 20:34
  • $\begingroup$ (continuation) The language of the projectors is equivalent. If we chose a base $ \{ |\phi>_n⟩ \}_{ n=1,2,...D}$, then with each $|\phi>_n⟩$ we can construct a projector $\Pi _n = |\phi>_n><\phi|_n$. Now, the expectation value of an operator $A$ is not what you say. In some state $|\psi>$, the expectation value is $<\psi|A|\psi>$. $\endgroup$
    – Sofia
    Commented Dec 28, 2014 at 20:39

1 Answer 1

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1) The trace of the rank-one operator $|\psi_1\rangle\langle\psi_2|$ is $\operatorname{Tr}(|\psi_1\rangle\langle\psi_2|) = \langle\psi_2|\psi_1\rangle$ (think of the trace of the matrix $[v_iw_j]$ where $v_i$ and $w_j$ are the components of vectors $v$ and $w$ respectively in an orthonormal basis), so for the product of $\Pi_1\Pi_2$ you get $$\begin{align}\operatorname{Tr}(|\psi_1\rangle\langle\psi_1|\psi_2\rangle\langle\psi_2|) &= \langle\psi_1|\psi_2\rangle\operatorname{Tr}(|\psi_1\rangle\langle\psi_2|)\\&=\langle\psi_1|\psi_2\rangle\langle\psi_2|\psi_1\rangle\\&=|\langle\psi_1|\psi_2\rangle|^2\end{align}$$ 2) $|\psi_1\rangle$ and $|\psi_2\rangle$ share no relation in general, they are arbitrary unitary vectors from your Hilbert space that define independent pure states. But as they define the same GNS representation, up to unitary equivalence, they can always be linked by a unitary operator.

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  • $\begingroup$ Thank you! Is this justified by: $tr(|ψ_1\rangle \langleψ_2|) = \sum_i \langle φ_i|ψ_1\rangle \langle ψ_2|φ_i\rangle = \sum_i \langle ψ_2|φ_i\rangle \langle φ_i|ψ_1\rangle = \langle ψ_2|ψ_1\rangle$? $\endgroup$
    – Red Pencil
    Commented Dec 28, 2014 at 21:31
  • $\begingroup$ yes, provided $|\phi_i\rangle$ form an orthonormal basis of the Hilbert space and you allow the sum to be an integral when needed $\endgroup$
    – Phoenix87
    Commented Dec 28, 2014 at 21:35
  • $\begingroup$ Dunno why I was struggling with this thing. I think the main problem was that I did not think of pulling out $\langle ψ_1|ψ_2\rangle$ out of the trace... :x Cheers! $\endgroup$
    – Red Pencil
    Commented Dec 28, 2014 at 21:38

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