3.1.Observables
In quantum mechanics observables are used to indicate quantities which
could be measured in an experiment, and observable also refers
to the self-adjoint operator associated to such a
measurement. Specifically, there is a one-to-one correspondence
between measurable quantities \(M\) and self-adjoint operators
\(\hat{M}\). One example is energy and the Hamiltonian operator
\(\hat{H}\).
Now, in quantum mechanics the possible values of a measurement of \(M\)
are the eigenvalues of \(\hat{M}\) (ignoring experimental error, we do
not commit experimental errors.) Typically for a given state
\(\ket{\psi}\) we cannot predict with certainty the result of a
measurement. instead we can give probabilities for the different
Observables correspond to Hermitian matrices. Their eigenvalues are the possible measurement outcomes.
outcomes. Note that for a Hilbert space \(\mathcal{H}\) of finite
dimension \(N\), the operator \(\hat{M}\) will have a finite number of
eigenvalues, in fact the number is precisely \(N\) (if we count the
multiplicity of any degenerate eigenvalues) since this is equivalent
to asking for the spectrum of an \(N \times N\) Hermitian matrix.
Dictionary Linear Algebra \(\leftrightarrow\) QM:
Self-adjoint operators \(\hat{H}\)\(\leftrightarrow\) Quantum mechanical observables;
Eigenvalues of a self-adjoint operator \(\hat{H}\)\(\leftrightarrow\) Possible outcomes of measuring that quantum mechanical observable; (Prove that the eigenvalues of a self-adjoint operator must be real numbers. You will never measure a complex outcome in a lab!)
Eigenstates of a self-adjoint operator \(\ket{\psi_E}\), i.e. \(\hat{H} \ket{\psi_E} = E \ket{\psi_E}\) for some real eigenvalue \(E\in \mathbb{R}\)\(\leftrightarrow\) States of definite outcome. If you measure \(\hat{H}\) on \(\ket{\psi_E}\) with probability \(p=1\) you will find outcome \(E\), the corresponding eigenvalue.
Def: The
spectrum of an operator \(\hat{H}\) is the
set
\begin{equation}
\mbox{Spec}(\hat{H}) = \{\lambda \in \mathbb{C}\,\text{s.t.}\,
\hat{H}-\lambda \hat{I} \,\text{is non invertible}\}.
\end{equation}
For a finite-dimensional Hilbert space this is identical to the set of all
finitely many eigenvalues of \(\hat{H}\).
Using basic result from linear algebra, the spectrum of a self-adjoint operator
\(\hat{M}\) is a set of real eigenvalues \(\lambda_n\), each with a corresponding
eigenstate \(\ket{n}\). Eigenstates corresponding to different eigenvalues are
automatically orthogonal. If there is degeneracy then for each eigenspace of
dimension greater than one we can always choose a basis of orthogonal
eigenstates (Apply Grahm-Schmidt procedure eigenspace by eigenspace). Of course, we can always normalise the eigenstates, then we will
have \(N\) orthonormal eigenstates giving us an orthonormal basis for \(\mathcal{H}\).
The spectral representation of an operator expresses it in terms of its eigenvectors and eigenvalues.
This also give us the
spectral representation of \(\hat{M}\) (corresponding
to diagonalisation of the matrix \(M\))
\[ \hat{M} = \sum_n \lambda_n \ket{n}\bra{n} . \]
Note that for the identity operator, the only eigenvalue is \(1\) with degeneracy
\(N\), so we can choose any orthonormal basis of \(\mathcal{H}\) and
\[ \hat{I} = \sum_n \ket{n}\bra{n} . \]
This is a very useful expression. We can often use it in calculations by
“inserting the identity as a complete sum of states.”
Now when a measurement of M is made on a state
\[\ket{\psi}=\sum_n c_n \ket{n}\,,\]
we will get the
result \(\lambda_n\) with probability \(p_n = \left\vert \ip{n}{\hat{\psi}} \right\vert^2 = \vert c_m\vert^2\)
which is just the magnitude squared of the coefficient of \(\ket{n}\) if we write
\(\psi\) in the basis \(\{\ket{n}\}\). After the measurement, if the result is
\(\lambda_n\), the state will then have definite value of M, \(\lambda_n\), so
measuring M again will give the same result. Therefore the state is no longer
\(\ket{\psi}\) but is \(\ket{n}\). Note that this “collapse of the wavefunction”
is not a unitary process, and is not reversible.
One way to describe this measurement process is in terms of the set of
projection operators \(\hat{P}_n = \ket{n}\bra{n}\) formed from the eigenstates
of \(\hat{M}\). Then the probability of result \(\lambda_n\) is
\(p_n = \ipop{\psi}{\hat{P}_n}{\psi}\) and the resulting state is
\(\frac{1}{\sqrt{p_n}} {\hat{P}_n} \ket{\psi}\) which is the state \(\ket{n}\) up
to an irrelevant overall phase.
Remember that a projector \(\hat{P}\) is a linear operator such that
\(\hat{P}^\dagger = \hat{P}\) and \(\hat{P}^2 = \hat{P}\) and check that
indeed \(\hat{P}_n = \ket{n}\bra{n}\) has all these properties.
Projection operators can be used to express the measurement process.
The above discussion of measurement assume the spectrum of \(\hat{M}\) is not
degenerate. If we have degeneracy then we can generalise the definition of the
projection operators. Consider an eigenvalue \(\lambda\). We define the
projection operator to be a sum over the eigenstates with that eigenvalue, i.e.
\[ \hat{P}_{\lambda} = \sum_{n : \lambda_n = \lambda} \ket{n}\bra{n} . \]
Then we still have the result that the probability is
\(p_{\lambda} = \ipop{\psi}{\hat{P}_{\lambda}}{\psi}\) and the resulting state is
\(\frac{1}{\sqrt{p_{\lambda}}} {\hat{P}_{\lambda}} \ket{\psi}\).
An important point to note is that a state can only have definite
values for two observables, say \(A\) and \(B\), if it is a simultaneous
eigenstate of \(\hat{A}\) and \(\hat{B}\). This is not possible for two
generic operators. However, if \(\com{A}{B} = 0\) then we can always
find simultaneous eigenstates. In this case we say that the
observables \(A\) and \(B\) are compatible. If the observables are not
compatible then measuring \(A\), then \(B\), then \(A\) again will not necessarily
give the same result for the two measurements of \(A\). That is because
the state after the first measurement of \(A\) is not an eigenstate of
\(\hat{B}\), and so a measurement of \(B\) will change the state (to some
eigenstate of \(\hat{B}\).) This will not be an eigenstate of \(A\), so the
result of the second measurement of \(A\) cannot be determined with
certainty.
Let \(\mathcal{H} = \mbox{span}\{\ket{-2},\ket{-1}, \ket{1},\ket{2}\}\) be a four dimensional Hilbert space with orthonormal basis vectors given by the eigenvectors of an hermitian operator \(\hat{A}\) as
\begin{align*}
&\hat{A} \ket{-2}= -2 \ket{-2}\,,\qquad \hat{A} \ket{-1} = -1 \ket{-1}\,,\\
&\hat{A} \ket{1} = 1 \ket{1}\,,\qquad\qquad \hat{A} \ket{2} = 2 \ket{2}\,,
\end{align*}
giving the spectral decomposition
\begin{align*}\hat{A} &= -2\ket{-2}\bra{-2}-1 \ket{-1}\bra{-1}+1 \ket{1}\bra{1}+2\ket{2}\bra{2}\\
&= ({\color{red}-2}) \hat{P}_{{\color{red}-2}} +({\color{orange}-1}) \hat{P}_{{\color{orange}-1}} + ({\color{blue}+1}) \hat{P}_{{\color{blue}+1}} +({\color{violet}+2}) \hat{P}_{{\color{violet}+2}}\\
& = \sum_{\lambda \in \mbox{Spec}(\hat{A})}{\color{red}\lambda} \hat{P}_{\color{red}\lambda} \,.
\end{align*}
in terms of the projectors \(\hat{P}_{\lambda} = \ket{\lambda}\bra{\lambda}\).
If we prepare a state \(\ket{\psi} \in \mathcal{H}\) and measure the
observable \(\hat{A}\) we can only find one of the values
\(\{-2,-1,1,2\}\).
Suppose we prepared the state
\[\ket{\psi} = 2 \ket{-2} + (1+i) \ket{-1} +3i \ket{1}\]
we know that
if we were to measure \(\hat{A}\) on \(\ket{\psi}\) we will never find the
outcome \(+2\) since the coefficient of \(\ket{2}\) in the expansion for
\(\ket{\psi}\) vanishes.
To compute the probabilities of measuring \(\{-2,-1,1,2\}\) we have to
normalise the state, i.e. we need to impose \(\ip{\psi}{\psi} = 1\). We
compute
\begin{multline}
\ip{\psi}{\psi} = \Big[ 2\bra{-2} +(1-i) \bra{-1} +(-3i) \bra{1} \Big] \\
\times \Big[ 2 \ket{-2} + (1+i) \ket{-1} +3i \ket{1}\Big]= 4+2+9=15
\end{multline}
and use this result to define the normalised state
\[\ket{\tilde{\psi}} =
\frac{\ket{\psi}}{\sqrt{15}}=\frac{2}{\sqrt{15}} \ket{-2} +
\frac{(1+i)}{\sqrt{15}} \ket{-1} + \frac{3i}{\sqrt{15}} \ket{1}\,.\]
The probability of measuring \(\hat{A}\) and finding outcome \(-2\) is
then the modulus square of the coefficient in front of \(\ket{-2}\),
i.e. \(p_{-2} = \frac{4}{15}\), similarly \(p_{-1} =
\frac{2}{15},\,p_{+1}= \frac{9}{15}\) and of course \(p_{+2} =0\). The
total probability is \(1\) as it should since
\(\ip{\tilde{\psi}}{\tilde{\psi}}=1\). Using the projector
\(\hat{P}_{-2} = \ket{-2}\bra{-2}\) we have \(p_{-2} = \bra{\tilde{\psi}}
\hat{P}_{-2}\ket{\tilde{\psi}}\).
If we prepare many copies of the same state \(\ket{\psi}\), measure
\(\hat{A}\) and then average, we find the expectation value
\begin{align*}
\langle A \rangle_\psi &= \frac{\bra{\psi} \hat{A} \ket{\psi}}{\ip{\psi}{\psi}} = \bra{\tilde{\psi}} \hat{A} \ket{\tilde{\psi}} \\
&= \frac{4}{15} \bra{-2} \hat{A} \ket{-2}+ \frac{2}{15} \bra{-1} \hat{A} \ket{-1}+ \frac{9}{15} \bra{1} \hat{A} \ket{1}\\
&= p_{{\color{red}-2}}({\color{red}-2}) + p_{{\color{orange}-1}}({\color{orange}-1})+ p_{{\color{blue}+1}}({\color{blue}+1})+ p_{{\color{violet}+2}}({\color{violet}+2}) = -\frac{1}{15}\,.
\end{align*}
3.2.Density matrices
The above sections give the standard Dirac notation description of
QM. The states previously described are what we will now call
pure states. This means that the states are definite, i.e. we
assume that (at least in principle) we know what the state of the
system is. Any uncertainties in predictions are due to the nature of
Pure states are fully determined, mixed states arise when
we do not know the state of a system. A sum of basis states is still a pure state.
QM. However, we can also
consider
mixed states which arise when we do not know with
certainty the state of a system. Here we assume that we have some
probabilistic knowledge, such as the system is in state \(\ket{\psi}\)
with probability \(p\), and in state \(\ket{\phi}\) with probability
\(1-p\). This type of uncertainty is ‘classical uncertainty’ in the
sense that it just describes our lack of knowledge about a
system. Indeed, whether the state is pure or mixed may be a matter
of perspective since one person may have more knowledge about the
system than other (we will see this later when we discuss the
reduced density matrix for a bipartite system).
For a pure state \(\ket{\psi}\) we define the
density operator or, as more
commonly called, the
density matrix to be
\[ \hat{\rho} = \ket{\psi}\bra{\psi} . \]
Note that when our Hilbert space is \(n\)-dimensional if we think of ket
vectors \(\ket{\psi}\) as \(n\) components column vectors \(\textbf{z}\) and
bra vectors \(\bra{\phi}\) as \(n\) components row vectors
\(\textbf{w}^\dagger\), then an operator of the form
\(\ket{\psi}\bra{\phi}\) can be thought of as \(\textbf{z}
\textbf{w}^\dagger\), hence a \(n\times 1\) matrix times a \(1\times n\)
matrix, i.e. a \(n\times n\) matrix, while the inner product \(
\textbf{w}^\dagger \textbf{z}\) as a \(1\times n\) matrix times a
\(n\times 1\) matrix resulting in a \(1\times 1\) matrix, i.e. a complex
number.
For pure states there is a one-to-one mapping between the density
matrix and the state, so we can work with one or the other. For
example we have the following correspondence:
\[ \begin{array}{lcl}
\hat{M}\ket{\psi} = \lambda\ket{\psi} & \longleftrightarrow & \hat{M}\hat{\rho} = \lambda\hat{\rho} \\
\ket{\psi} \rightarrow \hat{U}\ket{\psi} & \longleftrightarrow & \hat{\rho} \rightarrow \hat{U}\hat{\rho}\hat{U}^{\dagger}
\end{array} \]
Inner products of states arise when multiplying operators or when taking traces.
In particular, if we label the orthonormal basis states \(\ket{n}\) for some range
of integers \(n\), we define the trace of \(\hat{A}\) to be:
\[ \Tr(\hat{A}) = \sum_n \ipop{n}{\hat{A}}{n} \,, \]
you can think of
\(\ipop{m}{\hat{A}}{n}\) as the \(m^{th}\) row, \(n^{th}\) column entry of
the matrix representation of \(\hat{A}\) operator in the standard basis,
hence the trace just defined corresponds indeed to the sum of the
diagonal entries.
The trace of the density matrix \(\hat{\rho}\)
is equal to one, \(\Tr(\hat{\rho})=1\).
Note that
\begin{equation}
\begin{aligned}
\Tr(\hat{\rho}) &= \sum_n \ipop{n}{\hat{\rho}}{n} = \sum_n \ip{n}{\psi} \ip{\psi}{n}\\
& = \sum_n \ip{\psi}{n} \ip{n}{\psi} = \ipop{\psi}{\hat{I}}{\psi} = 1\,,
\end{aligned}
\end{equation}
where in the last two steps we used the spectral representation of the identity operator and the fact that \(\ket{\psi}\) is normalised.
Similar manipulations show that in general
\(\Tr (\ket{\phi}\bra{\psi}) = \ip{\psi}{\phi} \). Also note that for a pure state
\(\Tr(\hat{\rho}^2) = 1\) since \(\hat{\rho}\) is a projector and we know that for projectors we have \(\hat{\rho}^2 = \hat{\rho}\).
Mixed states describe situations where there is uncertainty
about the state of the system due to lack of knowledge, i.e. this is
the usual ‘classical' uncertainty we have if we don’t know everything
The density operator (or density matrix) can be used to describe the state of a system, for both pure and mixed states.
about the system. We can describe mixed states in terms of an ensemble
of pure states, each with a given probability of being the state of
the system, e.g. \(\{ ( p_i, \ket{i})\}\) with \(\ket{i}\) not necessarily
orthogonal but chosen with unit norm (if not just normalised them one
by one). The density matrix is just the linear combination of the
density matrices for each of the pure states, weighted by the
probability, i.e.
\[ \hat{\rho} = \sum_i p_i \ket{i}\bra{i} . \]
Note that there is no
requirement for the state \(\ket{i}\) to be orthogonal (although we
assume they are normalised) and also such a mixed state density matrix
does not correspond to a unique ensemble. There will be in general
more than one ensemble \(\{ ( p_i, \ket{i})\}\) giving rise to the same
density matrix for the same mixed state.
Of course, the probabilities \(p_i\) cannot be negative and must sum to
\(1\). We can also generalise the definition of the mixed state density
matrix to allow ensembles including mixed states.I.e. we can have
\(\hat{\rho} = \sum_i p_i \hat{\rho}_i\) where the \(\hat{\rho}_i\) are
mixed and/or pure state density matrices.
Such ensembles can only give a pure state in the trivial case where
there is only one pure state, which must then have probability
\(1\). However, given a density matrix it is often not immediately
obvious whether it describes a pure or a mixed state. A test for this
(see later discussion) is to calculate \(\Tr(\hat{\rho}^2)\) which will
be \(1\) for a pure state and less than 1 for a mixed state.
By construction, density matrices are
normalised such that \(\mbox{Tr}\, \hat{\rho} = 1\);
Hermitian \(\hat{\rho}^\dagger = \hat{\rho}\);
positive operators, meaning that for any state
\(\ket{\psi}\), \(\ipop{\psi}{\hat{\rho}}{\psi} \ge 0\) (Note this can
be equal to zero even for \(\ket{\psi}\neq0\). In matrix language this
is called
semi-positive-definite).
If we measure then the results for pure states in Dirac notation
generalise to all pure or mixed density matrices as:
The value of the result is \(\lambda\) with probability
\(p_{\lambda} = \Tr(\hat{P}_{\lambda}\, \hat{\rho}) = \Tr(\hat{P}_{\lambda}\,\hat{\rho}\,\hat{P}_{\lambda})\).
The density matrix after measuring M to be \(\lambda\) is
\[\hat{\rho}\to \frac{1}{p_{\lambda}}\, \hat{P}_{\lambda}\, \hat{\rho}\, \hat{P}_{\lambda} = \frac{1}{\Tr(\hat{P}_{\lambda}\,\hat{\rho}\,\hat{P}_{\lambda})} \, \hat{P}_{\lambda}\, \hat{\rho}\, \hat{P}_{\lambda}\,.\]
Given a two dimensional Hilbert space \(\mathcal{H} = \mbox{span}\{\ket{0},\ket{1}\}\) decide whether the matrix \(\rho = \frac{1}{9}\left(\begin{matrix} 5 & 2-4i \\ 2+4i & 4\end{matrix}\right)\) is
a density matrix for a pure state,
a density matrix for a mixed state,
not a density matrix,
when we represented the basis vector using the standard basis.
First of all to be a density matrix the matrix \(\rho\) needs to be with
\(\mbox{Tr}\rho =1\), hermitian \(\rho^\dagger= \rho\) and semi-positive
definite, i.e. \(\textbf{z}^\dagger \rho\, \textbf{z} \geq 0\) for all
\(\textbf{z}\in\mathbb{C}^2\).
It is simple to check that \(\rho\) is indeed
hermitian and with trace equal to \(1\). Instead of checking that \(\rho\)
is semi-positive definite let us see how a density matrix for a pure
state \(\ket{\psi} = a\ket{0}+b\ket{1}\) looks like. Let us assume the
state is normalised so \(|a|^2+|b|^2=1\) and pass to vector
representation \(\ket{\psi}\to \left(\begin{matrix}
a\\b\end{matrix}\right)\) then the matrix associate to its density
operator is
\[
\hat{\rho_\psi} = \ket{\psi}\bra{\psi} \to \rho_\psi = \left(\begin{matrix} a\\b\end{matrix}\right) \left(\begin{matrix} a\\b\end{matrix}\right)^\dagger = \left(\begin{matrix} |a|^2 & a b^*\\
b a^* & |b|^2 \end{matrix}\right)\,.
\]
It is simple to see that if we chose \(b=\frac{2}{3}\) and \(a=\frac{1-2i}{3}\) we obtain precisely the matrix \(\rho\) under question. Note that this is not the only possibility! We can multiply \(\ket{\psi} = \frac{1-2i}{3}\ket{0}+\frac{2}{3}\ket{1}\) by any phase \(e^{i \alpha}\) without changing its density matrix.
3.3.Pure versus mixed states
In the example above we were able to find explicitly the pure state
whose density operator was the matrix provided, however we would like
to know whether a certain density operator given comes from a pure
state or a mixed one.To this end we have the following theorem.
Let \(\hat{\rho}\) be a density operator on a Hilbert space \(\mathcal{H}\), i.e. \(\mbox{Tr}\,\hat{\rho}=1\), \(\hat{\rho}^\dagger =\hat{\rho}\) and \(\hat{\rho}\) positive operator.
The density operator \(\hat{\rho}\) corresponds to a pure state if and only if \(\mbox{Tr} \,\hat{\rho}^2 =1\).
Proof:\((\Rightarrow)\) Let us assume that \(\hat{\rho} = \ket{\psi}\bra{\psi}\) is the density matrix associated to a pure state \(\ket{\psi}\in \mathcal{H}\).
It is simple to compute \(\hat{\rho}^2 = \ket{\psi}\ip{\psi}{\psi} \bra{\psi} =\hat{\rho}\) since the state is normalised, hence \(\mbox{Tr}\,\hat{\rho}^2 = \mbox{Tr} \hat{\rho} = 1\).
\((\Leftarrow)\) Conversely let us suppose that \(\hat{\rho}\) is the density operator corresponding to the ensemble \(\{p_i,\ket{\psi_i}\}\), i.e. \(\hat{\rho} = \sum_i p_i \hat{\rho}_i = \sum_i p_i \ket{\psi_i}\bra{\psi_i}\).
We want to compute \(\mbox{Tr} \hat{\rho}^2\):
\begin{align*}
\mbox{Tr} \hat{\rho}^2 &= \sum_n \bra{n} \hat{\rho}^2 \ket{n} = \sum_{n,i,j} p_i p_j \ip{n}{\psi_i} \ip{\psi_i}{\psi_j} \ip{\psi_j}{n}\\
& = \sum_{ij} p_i p_j \ip{\psi_i}{\psi_j} \left(\sum_n \ip{\psi_j}{n}\ip{n}{\psi_i}\right)
\\
& = \sum_{ij} p_i p_j \ip{\psi_i}{\psi_j} \bra{\psi_j}\hat{I} \ket{\psi_i} \\
&= \sum_{ij} p_i p_j \ip{\psi_i}{\psi_j}\ip{\psi_j}{\psi_i } \\
&= \sum_{ij} p_i p_j \vert \ip{\psi_i}{\psi_j}\vert^2\leq \sum_{ij} p_i p_j \leq 1\,.
\end{align*}
For a pure state, the density matrix has \(\Tr(\hat{\rho}^2)=1\). For a mixed state, we have instead \(\Tr(\hat{\rho}^2) \lt{} 1\).
In the second line we used the spectral decomposition of the identity operator \(\hat{I} = \sum_n \ket{n}\bra{n}\), while in the third line we made use of the complex Cauchy-Schwarz inequality
\[
\vert \ip{\psi_i}{\psi_j}\vert^2 \leq \ip{\psi_i}{\psi_i} \ip{\psi_j}{\psi_j} \leq 1
\]
since the states \(\ket{\psi_i}\) are normalised. Finally in the
last step we used the fact that the \(p_i\) are probabilities and
\(\sum_i p_i =1\).
We also know that the equality in the Cauchy-Schwarz inequality holds
if and only if the vectors \(\ket{\psi_i}\) and \(\ket{\psi_j}\) are
collinear, i.e. \(\ket{\psi_i} = a \ket{\psi_j}\) for some complex
number \(a\in\mathbb{C}\) that can only be a phase \(a=e^{i \alpha}\)
since all the vectors must have length one.
Hence we have that \(\mbox{Tr} \,\hat{\rho}^2 \leq 1\) with equality if
and only if all vectors are collinear with one another, i.e. they are
all a multiple of say the first one \(\ket{\psi_i} = e^{i \alpha_i}
\ket{\psi_1}\) but this means that the density matrix
\[
\hat{\rho} = \sum_i p_i \ket{\psi_i}\bra{\psi_i} = \sum_i p_i \ket{\psi_1}\bra{\psi_1} = \ket{\psi_1}\bra{\psi_1}\,,
\]
hence \(\mbox{Tr}\, \hat{\rho}^2 =1 \) and \(\hat{\rho}\) is a pure state.
We then have a complete characterization of pure vs mixed states! We just need to compute \(\mbox{Tr}\,\hat{\rho}^2\) if this number is less than one we know that we have a mixed state, if we find one we know the state is pure.
We will shortly give a geometric characterization for pure and mixed state in the simplest case of a two dimension Hilbert space, i.e. what we call a qubit.
Suppose we have a three dimensional Hilbert space with orthonormal basis \(\mathcal{H} =\mbox{span}\{\ket{1},\ket{3},\ket{5}\}\).
Compute the density matrix associated to the ensemble \(\{ (2/3, \ket{\psi_1}),(1/3,\ket{\psi_2})\}\) where \(\ket{\psi_1} = \frac{1}{\sqrt{2}} (\ket{1}-\ket{3})\) and \(\ket{\psi_2} = \frac{1}{\sqrt{2}}(\ket{3} + i \ket{5}) \).
First of all we notice that the state \(\ket{\psi_1},\ket{\psi_2}\) are normalised, had they not we would have had to normalise them before proceeding.
The density operator associated with this mixed state is then
\begin{align*}
\hat{\rho} &= \frac{2}{3} \ket{\psi_1}\bra{\psi_1} + \frac{1}{3} \ket{\psi_2} \bra{\psi_2}\\
&=\frac{2}{6}( \ket{1}-\ket{3})(\bra{1}-\bra{3}) + \frac{1}{6} (\ket{3} + i \ket{5})(\bra{3} -i \bra{5})\\
&= \frac{2}{6} \ket{1}\bra{1} -\frac{2}{6} \ket{1}\bra{3}-\frac{2}{6}\ket{3}\bra{1} \\
&\qquad +\frac{1}{2}\ket{3}\bra{3}+\frac{i}{6}\ket{3}\bra{5} -\frac{i}{6} \ket{5}\bra{3} +\frac{1}{6}\ket{5}\bra{5}\,.
\end{align*}
If we represent the three basis vectors using the standard basis we can write the density operator as the \(3\times3 \) matrix
\begin{align*}
\rho &= \frac{2}{3}\left(\begin{matrix}1/\sqrt{2}\\-1/\sqrt{2}\\0\end{matrix}\right) \left(\begin{matrix}1/\sqrt{2}\\-1/\sqrt{2}\\0\end{matrix}\right) ^\dagger+ \frac{1}{3}\left(\begin{matrix}0\\1/\sqrt{2}\\i/\sqrt{2} \end{matrix}\right) \left(\begin{matrix}0\\1/\sqrt{2}\\i/\sqrt{2} \end{matrix}\right) ^\dagger = \\
& =\left(\begin{matrix}\frac{1}{3} & -\frac{1}{3} & 0\\
-\frac{1}{3} & \frac{1}{2} & \frac{i}{6} \\
0& -\frac{i}{6} & \frac{1}{6}\end{matrix}\right)\,.
\end{align*}
It is simple to check now that the trace of this matrix is of course one, while \(\mbox{Tr}\rho^2 = \frac{2}{3}\lt{}1\) since the state is a mixed state.
Finally if we have the observable \(\hat{A}\) with spectrum \(\{1,\ket{1}; 3,\ket{3}; 5,\ket{5} \}\), we can easily compute the expectation value on this state
\[
\mbox{Tr} \left[ \hat{\rho} \hat{A} \right]= \mbox{Tr} \left[ \rho \left(\begin{matrix} 1 & 0 & 0\\ 0 & 3 & 0\\ 0& 0& 5\end{matrix}\right) \right]= \frac{8}{3}\,,
\]
which you can also check using the abstract operator formalism.
Let \(\mathcal{H} =\mbox{span}\{\ket{1},...,\ket{6}\}\) be a six dimensional Hilbert space with orthonormal basis \(\ket{i}\) given by the eigenvectors with eigenvalues \(\{1,2,3,4,5,6\}\) for the hermitian operator \(\hat{A}\).
Consider the normalised mixed state given by the ensemble \(\{(\frac{1}{6},\ket{1}),(\frac{1}{6},\ket{2}),(\frac{1}{6},\ket{3}),(\frac{1}{6},\ket{4}),(\frac{1}{6},\ket{5}),(\frac{1}{6},\ket{6})\}\)
\[
\hat{\rho} = \sum_{i=1}^6 \frac{1}{6} \ket{i}\bra{i} = \frac{1}{6}\hat{I}\,,\qquad \rho =\frac{\mathbb{I}_6}{6}\,,
\]
where we used the standard basis to represent the basis vectors and obtain the matrix representation for \(\hat{\rho}\) given by \(\rho\).
This state is in a certain sense (that we will quantify later on) the most mixed, it is an equally probable ensemble of the six basis vectors. Its trace is clearly one while \(\mbox{Tr}(\hat{\rho}^2) =\mbox{Tr}(\rho^2) = \frac{1}{6}\lt{}1\).
The expectation value of the observable \(\hat{A}\), with spectrum precisely \(\{1,\ket{1};\,...\,; 6,\ket{6}\}\), on this state is given by
\[
\langle A\rangle = \mbox{Tr}( \hat{\rho} \hat{A} ) = \sum_{i=1}^6 \frac{1}{6} \times i =\frac{7}{2}\,,
\]
if you want the state \(\hat{\rho}\) is the most “classically” uncertain of all the states, it is exactly the same ensemble of a six-faced die for which the average outcome is precisely \(7/2\).
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