Ensembles for spin one half

Mixed ensemble averages

In [1], Sakurai leaves it to the reader to verify that knowledge of the three ensemble averages [S_x], [S_y],[S_z] is sufficient to reconstruct the density operator for a spin one half system.

I’ll do this in two parts, the first using a spin-up/down ensemble to see what form this has, then the general case. The general case is a bit messy algebraically. After first attempting it the hard way, I did the grunt work portion of that calculation in Mathematica, but then realized it’s not so bad to do it manually.

Consider first an ensemble with density operator

\label{eqn:ensemblesForSpinOneHalf:20}
\rho =
w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-},

where these are the $$\BS \cdot (\pm \zcap)$$ eigenstates. The traces are

\label{eqn:ensemblesForSpinOneHalf:40}
\begin{aligned}
\textrm{Tr}( \rho \sigma_x )
&=
\bra{+} \rho \sigma_x \ket{+}
+
\bra{-} \rho \sigma_x \ket{-} \\
&=
\bra{+} \rho \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \ket{+}
+
\bra{-} \rho \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \ket{-} \\
&=
\bra{+} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{-}
+
\bra{-} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{+} \\
&=
\bra{+} w_{-} \ket{-}
+
\bra{-} w_{+} \ket{+} \\
&=
0,
\end{aligned}

\label{eqn:ensemblesForSpinOneHalf:60}
\begin{aligned}
\textrm{Tr}( \rho \sigma_y )
&=
\bra{+} \rho \sigma_y \ket{+}
+
\bra{-} \rho \sigma_y \ket{-} \\
&=
\bra{+} \rho \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \ket{+}
+
\bra{-} \rho \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \ket{-} \\
&=
i \bra{+} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{-}

i \bra{-} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{+} \\
&=
i \bra{+} w_{-} \ket{-}

i \bra{-} w_{+} \ket{+} \\
&=
0,
\end{aligned}

and
\label{eqn:ensemblesForSpinOneHalf:100}
\begin{aligned}
\textrm{Tr}( \rho \sigma_z )
&=
\bra{+} \rho \sigma_z \ket{+}
+
\bra{-} \rho \sigma_z \ket{-} \\
&=
\bra{+} \rho \ket{+}

\bra{-} \rho \ket{-} \\
&=
\bra{+} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{+}

\bra{-} \lr{ w_{+} \ket{+}\bra{+} + w_{-} \ket{-}\bra{-} } \ket{-} \\
&=
\bra{+} w_{+} \ket{+}

\bra{-} w_{-} \ket{-} \\
&=
w_{+} – w_{-}.
\end{aligned}

Since $$w_{+} + w_{-} = 1$$, this gives

\label{eqn:ensemblesForSpinOneHalf:80}
\boxed{
\begin{aligned}
w_{+} &= \frac{1 + \textrm{Tr}( \rho \sigma_z )}{2} \\
w_{-} &= \frac{1 – \textrm{Tr}( \rho \sigma_z )}{2}
\end{aligned}
}

Attempting to do a similar set of trace expansions this way for a more general spin basis turns out to be a really bad idea and horribly messy. So much so that I resorted to \href{https://raw.githubusercontent.com/peeterjoot/mathematica/master/phy1520/spinOneHalfSymbolicManipulation.nb}{Mathematica to do this symbolic work}. However, it’s not so bad if the trace is done completely in matrix form.

Using the basis

\label{eqn:ensemblesForSpinOneHalf:120}
\begin{aligned}
\ket{\BS \cdot \ncap ; + } &=
\begin{bmatrix}
\cos(\theta/2) \\
\sin(\theta/2) e^{i \phi}
\end{bmatrix} \\
\ket{\BS \cdot \ncap ; – } &=
\begin{bmatrix}
\sin(\theta/2) e^{-i \phi} \\
-\cos(\theta/2) \\
\end{bmatrix},
\end{aligned}

the projector matrices are

\label{eqn:ensemblesForSpinOneHalf:140}
\begin{aligned}
\ket{\BS \cdot \ncap ; + } \bra{\BS \cdot \ncap ; + }
&=
\begin{bmatrix}
\cos(\theta/2) \\
\sin(\theta/2) e^{i \phi}
\end{bmatrix}
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos^2(\theta/2) & \cos(\theta/2) \sin(\theta/2) e^{-i \phi} \\
\sin(\theta/2) \cos(\theta/2) e^{i \phi} & \sin^2(\theta/2)
\end{bmatrix},
\end{aligned}

\label{eqn:ensemblesForSpinOneHalf:160}
\begin{aligned}
\ket{\BS \cdot \ncap ; – } \bra{\BS \cdot \ncap ; – }
&=
\begin{bmatrix}
\sin(\theta/2) e^{-i \phi} \\
-\cos(\theta/2) \\
\end{bmatrix}
\begin{bmatrix}
\sin(\theta/2) e^{i \phi} & -\cos(\theta/2) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\sin^2(\theta/2) & -\cos(\theta/2) \sin(\theta/2) e^{-i \phi} \\
-\cos(\theta/2) \sin(\theta/2) e^{i \phi} & \cos^2(\theta/2)
\end{bmatrix}
\end{aligned}

With $$C = \cos(\theta/2), S = \sin(\theta/2)$$, a general density operator in this basis has the form

\label{eqn:ensemblesForSpinOneHalf:180}
\begin{aligned}
\rho
&=
w_{+}
\begin{bmatrix}
C^2 & C S e^{-i \phi} \\
S C e^{i \phi} & S^2
\end{bmatrix}
+
w_{-}
\begin{bmatrix}
S^2 & -C S e^{-i \phi} \\
-C S e^{i \phi} & C^2
\end{bmatrix} \\
&=
\begin{bmatrix}
w_{+} C^2 + w_{-} S^2 & (w_{+} – w_{-})C S e^{-i \phi} \\
(w_{+} -w_{-} ) S C e^{i \phi} & w_{+} S^2 + w_{-} C^2
\end{bmatrix}.
\end{aligned}

The products with the Pauli matrices are

\label{eqn:ensemblesForSpinOneHalf:200}
\begin{aligned}
\rho \sigma_x
&=
\begin{bmatrix}
w_{+} C^2 + w_{-} S^2 & (w_{+} – w_{-})C S e^{-i \phi} \\
(w_{+} -w_{-} ) S C e^{i \phi} & w_{+} S^2 + w_{-} C^2
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \\
&=
\begin{bmatrix}
(w_{+} – w_{-})C S e^{-i \phi} & w_{+} C^2 + w_{-} S^2 \\
w_{+} S^2 + w_{-} C^2 & (w_{+} -w_{-} ) S C e^{i \phi} \\
\end{bmatrix}
\end{aligned}

\label{eqn:ensemblesForSpinOneHalf:220}
\begin{aligned}
\rho \sigma_y
&=
\begin{bmatrix}
w_{+} C^2 + w_{-} S^2 & (w_{+} – w_{-})C S e^{-i \phi} \\
(w_{+} -w_{-} ) S C e^{i \phi} & w_{+} S^2 + w_{-} C^2
\end{bmatrix}
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \\
&=
i
\begin{bmatrix}
(w_{+} – w_{-})C S e^{-i \phi} & -w_{+} C^2 – w_{-} S^2 \\
w_{+} S^2 + w_{-} C^2 & -(w_{+} -w_{-} ) S C e^{i \phi} \\
\end{bmatrix}
\end{aligned}

\label{eqn:ensemblesForSpinOneHalf:240}
\begin{aligned}
\rho \sigma_z
&=
\begin{bmatrix}
w_{+} C^2 + w_{-} S^2 & (w_{+} – w_{-})C S e^{-i \phi} \\
(w_{+} -w_{-} ) S C e^{i \phi} & w_{+} S^2 + w_{-} C^2
\end{bmatrix}
\begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\
&=
\begin{bmatrix}
w_{+} C^2 + w_{-} S^2 & -(w_{+} – w_{-})C S e^{-i \phi} \\
(w_{+} -w_{-} ) S C e^{i \phi} & – (w_{+} S^2 + w_{-} C^2)
\end{bmatrix}
\end{aligned}

The respective traces can be read right off the matrices
\label{eqn:ensemblesForSpinOneHalf:260}
\begin{aligned}
\textrm{Tr}( \rho \sigma_x ) &= (w_{+} – w_{-}) \sin\theta \cos\phi \\
\textrm{Tr}( \rho \sigma_y ) &= (w_{+} – w_{-}) \sin\theta \sin\phi \\
\textrm{Tr}( \rho \sigma_z ) &= (w_{+} – w_{-}) \cos\theta \\
\end{aligned}.

This gives

\label{eqn:ensemblesForSpinOneHalf:280}
(w_{+} – w_{-}) \ncap = \lr{ \textrm{Tr}( \rho \sigma_x ), \textrm{Tr}( \rho \sigma_y ), \textrm{Tr}( \rho \sigma_z ) },

or

\label{eqn:ensemblesForSpinOneHalf:281}
\boxed{
w_{\pm} = \frac{1 \pm \sqrt{ \textrm{Tr}^2( \rho \sigma_x ) + \textrm{Tr}^2( \rho \sigma_y ) + \textrm{Tr}^2( \rho \sigma_z )} }{2} .
}

So, as claimed, it’s possible to completely describe the ensemble weight factors using the ensemble averages of $$[S_x], [S_y], [S_z]$$. I used the Pauli matrices instead, but the difference is just an $$\Hbar/2$$ scaling adjustment.

Pure ensemble

It turns out that doing the above is also pr. 3.10(b). Part (a) of that problem is to show how the expectation values $$\expectation{S_x}, \expectation{S_y},\expectation{S_x}$$ fully determine the spin orientation for a pure ensemble.

Suppose that the system is in the state $$\ket{\BS \cdot \ncap ; + }$$ as defined above, then the expectation values of $$\sigma_x, \sigma_y, \sigma_z$$ with respect to this state are

\label{eqn:ensemblesForSpinOneHalf:300}
\begin{aligned}
\expectation{\sigma_x}
&=
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
\cos(\theta/2) \\
\sin(\theta/2) e^{i \phi}
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix}
\sin(\theta/2) e^{i \phi} \\
\cos(\theta/2) \\
\end{bmatrix} \\
&=
\sin\theta \cos\phi,
\end{aligned}

\label{eqn:ensemblesForSpinOneHalf:340}
\begin{aligned}
\expectation{\sigma_y}
&=
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
\cos(\theta/2) \\
\sin(\theta/2) e^{i \phi}
\end{bmatrix} \\
&=
i
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix}
-\sin(\theta/2) e^{i \phi} \\
\cos(\theta/2) \\
\end{bmatrix} \\
&=
\sin\theta \sin\phi,
\end{aligned}

\label{eqn:ensemblesForSpinOneHalf:360}
\begin{aligned}
\expectation{\sigma_z}
&=
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}
\begin{bmatrix}
\cos(\theta/2) \\
\sin(\theta/2) e^{i \phi}
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\theta/2) &
\sin(\theta/2) e^{-i \phi}
\end{bmatrix}
\begin{bmatrix}
\cos(\theta/2) \\
-\sin(\theta/2) e^{i \phi}
\end{bmatrix} \\
&=
\cos\theta.
\end{aligned}

So we have
\label{eqn:ensemblesForSpinOneHalf:380}
\boxed{
\ncap = \lr{ \expectation{\sigma_x}, \expectation{\sigma_y}, \expectation{\sigma_z} }.
}

The spin direction is completely determined by this vector of expectation values (or equivalently, the expectation values of $$S_x, S_y, S_z$$).

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

PHY1520H Graduate Quantum Mechanics. Lecture 3: Density matrix (cont.). Taught by Prof. Arun Paramekanti

September 24, 2015 phy1520 No comments , , , , , , , , ,

Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 3 content.

Density matrix (cont.)

An example of a partitioned system with four total states (two spin 1/2 particles) is sketched in fig. 1.

fig. 1. Two spins

An example of a partitioned system with eight total states (three spin 1/2 particles) is sketched in fig. 2.

fig. 2. Three spins

The density matrix

\label{eqn:qmLecture3:20}
\hat{\rho} = \ket{\Psi}\bra{\Psi}

is clearly an operator as can be seen by applying it to a state

\label{eqn:qmLecture3:40}
\hat{\rho} \ket{\phi} = \ket{\Psi} \lr{ \braket{ \Psi }{\phi} }.

The quantity in braces is just a complex number.

After expanding the pure state $$\ket{\Psi}$$ in terms of basis states for each of the two partitions

\label{eqn:qmLecture3:60}
\ket{\Psi}
= \sum_{m,n} C_{m, n} \ket{m}_{\textrm{L}} \ket{n}_{\textrm{R}},

With $$\textrm{L}$$ and $$\textrm{R}$$ implied for $$\ket{m}, \ket{n}$$ indexed states respectively, this can be written

\label{eqn:qmLecture3:460}
\ket{\Psi}
= \sum_{m,n} C_{m, n} \ket{m} \ket{n}.

The density operator is

\label{eqn:qmLecture3:80}
\hat{\rho} =
\sum_{m,n}
C_{m, n}
C_{m’, n’}^\conj
\ket{m} \ket{n}
\sum_{m’,n’}
\bra{m’} \bra{n’}.

Suppose we trace over the right partition of the state space, defining such a trace as the reduced density operator $$\hat{\rho}_{\textrm{red}}$$

\label{eqn:qmLecture3:100}
\begin{aligned}
\hat{\rho}_{\textrm{red}}
&\equiv
\textrm{Tr}_{\textrm{R}}(\hat{\rho}) \\
&= \sum_{\tilde{n}} \bra{\tilde{n}} \hat{\rho} \ket{ \tilde{n}} \\
&= \sum_{\tilde{n}}
\bra{\tilde{n} }
\lr{
\sum_{m,n}
C_{m, n}
\ket{m} \ket{n}
}
\lr{
\sum_{m’,n’}
C_{m’, n’}^\conj
\bra{m’} \bra{n’}
}
\ket{ \tilde{n} } \\
&=
\sum_{\tilde{n}}
\sum_{m,n}
\sum_{m’,n’}
C_{m, n}
C_{m’, n’}^\conj
\ket{m} \delta_{\tilde{n} n}
\bra{m’ }
\delta_{ \tilde{n} n’ } \\
&=
\sum_{\tilde{n}, m, m’}
C_{m, \tilde{n}}
C_{m’, \tilde{n}}^\conj
\ket{m} \bra{m’ }
\end{aligned}

Computing the matrix element of $$\hat{\rho}_{\textrm{red}}$$, we have

\label{eqn:qmLecture3:120}
\begin{aligned}
\bra{\tilde{m}} \hat{\rho}_{\textrm{red}} \ket{\tilde{m}}
&=
\sum_{m, m’, \tilde{n}} C_{m, \tilde{n}} C_{m’, \tilde{n}}^\conj \braket{ \tilde{m}}{m} \braket{m’}{\tilde{m}} \\
&=
\sum_{\tilde{n}} \Abs{C_{\tilde{m}, \tilde{n}} }^2.
\end{aligned}

This is the probability that the left partition is in state $$\tilde{m}$$.

Average of an observable

Suppose we have two spin half particles. For such a system the total magnetization is

\label{eqn:qmLecture3:140}
S_{\textrm{Total}} =
S_1^z
+
S_1^z,

as sketched in fig. 3.

fig. 3. Magnetic moments from two spins.

The average of some observable is

\label{eqn:qmLecture3:160}
\expectation{\hatA}
= \sum_{m, n, m’, n’} C_{m, n}^\conj C_{m’, n’}
\bra{m}\bra{n} \hatA \ket{n’} \ket{m’}.

Consider the trace of the density operator observable product

\label{eqn:qmLecture3:180}
\textrm{Tr}( \hat{\rho} \hatA )
= \sum_{m, n} \braket{m n}{\Psi} \bra{\Psi} \hatA \ket{m, n}.

Let

\label{eqn:qmLecture3:200}
\ket{\Psi} = \sum_{m, n} C_{m n} \ket{m, n},

so that

\label{eqn:qmLecture3:220}
\begin{aligned}
\textrm{Tr}( \hat{\rho} \hatA )
&= \sum_{m, n, m’, n’, m”, n”} C_{m’, n’} C_{m”, n”}^\conj
\braket{m n}{m’, n’} \bra{m”, n”} \hatA \ket{m, n} \\
&= \sum_{m, n, m”, n”} C_{m, n} C_{m”, n”}^\conj
\bra{m”, n”} \hatA \ket{m, n}.
\end{aligned}

This is just

\label{eqn:qmLecture3:240}
\boxed{
\bra{\Psi} \hatA \ket{\Psi} = \textrm{Tr}( \hat{\rho} \hatA ).
}

Left observables

Consider

\label{eqn:qmLecture3:260}
\begin{aligned}
\bra{\Psi} \hatA_{\textrm{L}} \ket{\Psi}
&= \textrm{Tr}(\hat{\rho} \hatA_{\textrm{L}}) \\
&=
\textrm{Tr}_{\textrm{L}}
\textrm{Tr}_{\textrm{R}}
(\hat{\rho} \hatA_{\textrm{L}}) \\
&=
\textrm{Tr}_{\textrm{L}}
\lr{
\lr{
\textrm{Tr}_{\textrm{R}} \hat{\rho}
}
\hatA_{\textrm{L}})
} \\
&=
\textrm{Tr}_{\textrm{L}}
\lr{
\hat{\rho}_{\textrm{red}}
\hatA_{\textrm{L}})
}.
\end{aligned}

We see

\label{eqn:qmLecture3:280}
\bra{\Psi} \hatA_{\textrm{L}} \ket{\Psi}
=
\textrm{Tr}_{\textrm{L}} \lr{ \hat{\rho}_{\textrm{red}, \textrm{L}} \hatA_{\textrm{L}} }.

We find that we don’t need to know the state of the complete system to answer questions about portions of the system, but instead just need $$\hat{\rho}$$, a “probability operator” that provides all the required information about the partitioning of the system.

Pure states vs. mixed states

For pure states we can assign a state vector and talk about reduced scenarios. For mixed states we must work with reduced density matrix.

Example: Two particle spin half pure states

Consider

\label{eqn:qmLecture3:300}
\ket{\psi_1} = \inv{\sqrt{2}} \lr{ \ket{ \uparrow \downarrow } – \ket{ \downarrow \uparrow } }

\label{eqn:qmLecture3:320}
\ket{\psi_2} = \inv{\sqrt{2}} \lr{ \ket{ \uparrow \downarrow } + \ket{ \uparrow \uparrow } }.

For the first pure state the density operator is
\label{eqn:qmLecture3:360}
\hat{\rho} = \inv{2}
\lr{ \ket{ \uparrow \downarrow } – \ket{ \downarrow \uparrow } }
\lr{ \bra{ \uparrow \downarrow } – \bra{ \downarrow \uparrow } }

What are the reduced density matrices?

\label{eqn:qmLecture3:340}
\begin{aligned}
\hat{\rho}_{\textrm{L}}
&= \textrm{Tr}_{\textrm{R}} \lr{ \hat{\rho} } \\
&=
\inv{2} (-1)(-1) \ket{\downarrow}\bra{\downarrow}
+\inv{2} (+1)(+1) \ket{\uparrow}\bra{\uparrow},
\end{aligned}

so the matrix representation of this reduced density operator is

\label{eqn:qmLecture3:380}
\hat{\rho}_{\textrm{L}}
=
\inv{2}
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}.

For the second pure state the density operator is
\label{eqn:qmLecture3:400}
\hat{\rho} = \inv{2}
\lr{ \ket{ \uparrow \downarrow } + \ket{ \uparrow \uparrow } }
\lr{ \bra{ \uparrow \downarrow } + \bra{ \uparrow \uparrow } }.

This has a reduced density matrice

\label{eqn:qmLecture3:420}
\begin{aligned}
\hat{\rho}_{\textrm{L}}
&= \textrm{Tr}_{\textrm{R}} \lr{ \hat{\rho} } \\
&=
\inv{2} \ket{\uparrow}\bra{\uparrow}
+\inv{2} \ket{\uparrow}\bra{\uparrow} \\
&=
\ket{\uparrow}\bra{\uparrow} .
\end{aligned}

This has a matrix representation

\label{eqn:qmLecture3:440}
\hat{\rho}_{\textrm{L}}
=
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}.

In this second example, we have more information about the left partition. That will be seen as a zero entanglement entropy in the problem set. In contrast we have less information about the first state, and will find a non-zero positive entanglement entropy in that case.

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

PHY1520H Graduate Quantum Mechanics. Lecture 2: Basic concepts, time evolution, and density operators. Taught by Prof. Arun Paramekanti

Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering chap 1 (basic concepts),3 (density operator). content from [1].

Basic concepts

We’ve reviewed the basic concepts that we will encounter in Quantum Mechanics.

1. Abstract state vector. $$\ket{ \psi}$$
2. Basis states. $$\ket{ x }$$
3. Observables, special Hermitian operators. We’ll only deal with linear observables.
4. Measurement.

We can either express the wave functions $$\psi(x) = \braket{x}{\psi}$$ in terms of a basis for the observable, or can express the observable in terms of the basis of the wave function (position or momentum for example).

We saw that the position space representation of a momentum operator (also an observable) was

\label{eqn:lecture2:20}
\hat{p} \rightarrow -i \Hbar \PD{x}{}.

In general we can find the matrix element representation of any operator by considering its representation in a given basis. For example, in a position basis, that would be

\label{eqn:lecture2:40}
\bra{x’} \hat{A} \ket{x} \leftrightarrow A_{x x’}

The Hermitian property of the observable means that $$A_{x x’} = A_{x’ x}^\conj$$

\label{eqn:lecture2:60}
\int dx \bra{x’} \hat{A} \ket{x} \braket{x }{\psi} = \braket{x’}{\phi}
\leftrightarrow
A_{x’ x} \psi_x = \phi_{x’}.

Example: Measurement example

fig. 1. Polarizer apparatus

Consider a polarization apparatus as sketched in fig. 1, where the output is of the form $$I_{\textrm{out}} = I_{\textrm{in}} \cos^2 \theta$$.

A general input state can be written in terms of each of the possible polarizations

\label{eqn:lecture2:80}
\alpha \ket{ \updownarrow } + \beta \ket{ \leftrightarrow } \sim
\cos\theta \ket{ \updownarrow } + \sin\theta \ket{ \leftrightarrow }

Here $$\abs{\alpha}^2$$ is the probability that the input state is in the upwards polarization state, and $$\abs{\beta}^2$$ is the probability that the input state is in the downwards polarization state.

The measurement of the polarization results in an output state that has a specific polarization. That measurement is said to collapse the wavefunction.

When attempting a measurement, looking for a specific value, effects the state of the system, and is call a strong or projective measurement. Such a measurement is

• (i) Probabilistic.
• (ii) Requires many measurements.

This measurement process results a determination of the eigenvalue of the operator. The eigenvalue production of measurement is why we demand that operators be Hermitian.

It is also possible to try to do a weaker (perturbative) measurement, where some information is extracted from the input state without completely altering it.

Time evolution

1. Schrodinger picture.
The time evolution process is governed by a Schrodinger equation of the following form\label{eqn:lecture2:100}
i \Hbar \PD{t}{} \ket{\Psi(t)} = \hat{H} \ket{\Psi(t)}.

This Hamiltonian could be, for example,

\label{eqn:lecture2:120}
\hat{H} = \frac{\hat{p}^2}{2m} + V(x),

Such a representation of time evolution is expressed in terms of operators $$\hat{x}, \hat{p}, \hat{H}, \cdots$$ that are independent of time.

2. Heisenberg picture.Suppose we have a state $$\ket{\Psi(t)}$$ and operate on this with an operator

\label{eqn:lecture2:140}
\hat{A} \ket{\Psi(t)}.

This will have time evolution of the form

\label{eqn:lecture2:160}
\hat{A} e^{-i \hat{H} t/\Hbar} \ket{\Psi(0)},

or in matrix element form

\label{eqn:lecture2:180}
\bra{\phi(t)} \hat{A} \ket{\Psi(t)}
=
\bra{\phi(0)}
e^{i \hat{H} t/\Hbar}
\hat{A} e^{-i \hat{H} t/\Hbar} \ket{\Psi(0)}.

We work with states that do not evolve in time $$\ket{\phi(0)}, \ket{\Psi(0)}, \cdots$$, but operators do evolve in time according to

\label{eqn:lecture2:200}
\hat{A}(t) =
e^{i \hat{H} t/\Hbar}
\hat{A} e^{-i \hat{H} t/\Hbar}.

Density operator

We can have situations where it is impossible to determine a single state that describes the system. For example, given the gas in the room that you are sitting in, there are things that we can measure, but it is impossible to describe the state that describes all the particles and also impossible to construct a Hamiltonian that governs all the interactions of those many many particles.

We need a probabilistic description to even describe such a complex system.

Suppose we have a complex system that can be partitioned into two subsets, left and right, as sketched in fig. 2.

fig. 2. System partitioned into separate set of states

If the states in each partition can be enumerated separately, we can write the state of the system as sums over the probability amplitudes that for the combined states.

\label{eqn:lecture2:220}
\ket{\Psi}
=
\sum_{m, n} C_{m,n} \ket{m} \ket{n}

Here $$C_{m, n}$$ is the probability amplitude to find the state in the combined state $$\ket{m} \ket{n}$$.

As an example of such a system, we could investigate a two particle configuration where spin up or spin down can be separately measured for each particle.

\label{eqn:lecture2:240}
\ket{\psi} = \inv{\sqrt{2}} \lr{
\ket{\uparrow}\ket{\downarrow}
+
\ket{\downarrow}\ket{\rightarrow}
}

Considering such a system we could ask questions such as

• What is the probability that the left half is in state $$m$$? This would be\label{eqn:lecture2:260}
\sum_n \Abs{C_{m, n}}^2
• Probability that the left half is in state $$m$$, and the
probability that the right half is in state $$n$$? That is\label{eqn:lecture2:280}
\Abs{C_{m, n}}^2

We define the density operator

\label{eqn:lecture2:300}
\hat{\rho} = \ket{\Psi} \bra{\Psi}.

This is idempotent

\label{eqn:lecture2:320}
\hat{\rho}^2 =
\lr{ \ket{\Psi} \bra{\Psi} }
\lr{ \ket{\Psi} \bra{\Psi} }
=
\ket{\Psi} \bra{\Psi}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Entropy when density operator has zero eigenvalues

In the class notes and the text [1] the Von Neumann entropy is defined as

\label{eqn:densityMatrixEntropy:20}
S = -\textrm{Tr} \rho \ln \rho.

In one of our problems I had trouble evaluating this, having calculated a density operator matrix representation

\label{eqn:densityMatrixEntropy:40}
\rho = E \wedge E^{-1},

where

\label{eqn:densityMatrixEntropy:60}
E = \inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix},

and
\label{eqn:densityMatrixEntropy:100}
\wedge =
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}.

The usual method of evaluating a function of a matrix is to assume the function has a power series representation, and that a similarity transformation of the form $$A = E \wedge E^{-1}$$ is possible, so that

\label{eqn:densityMatrixEntropy:80}
f(A) = E f(\wedge) E^{-1},

however, when attempting to do this with the matrix of \ref{eqn:densityMatrixEntropy:40} leads to an undesirable result

\label{eqn:densityMatrixEntropy:120}
\ln \rho =
\inv{2}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}
\begin{bmatrix}
\ln 1 & 0 \\
0 & \ln 0
\end{bmatrix}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}.

The $$\ln 0$$ makes the evaluation of this matrix logarithm rather unpleasant. To give meaning to the entropy expression, we have to do two things, the first is treating the trace operation as a higher precedence than the logarithms that it contains. That is

\label{eqn:densityMatrixEntropy:140}
\begin{aligned}
-\textrm{Tr} ( \rho \ln \rho )
&=
-\textrm{Tr} ( E \wedge E^{-1} E \ln \wedge E^{-1} ) \\
&=
-\textrm{Tr} ( E \wedge \ln \wedge E^{-1} ) \\
&=
-\textrm{Tr} ( E^{-1} E \wedge \ln \wedge ) \\
&=
-\textrm{Tr} ( \wedge \ln \wedge ) \\
&=
– \sum_k \wedge_{kk} \ln \wedge_{kk}.
\end{aligned}

Now the matrix of the logarithm need not be evaluated, but we still need to give meaning to $$\wedge_{kk} \ln \wedge_{kk}$$ for zero diagonal entries. This can be done by considering a limiting scenerio

\label{eqn:densityMatrixEntropy:160}
\begin{aligned}
-\lim_{a \rightarrow 0} a \ln a
&=
-\lim_{x \rightarrow \infty} e^{-x} \ln e^{-x} \\
&=
\lim_{x \rightarrow \infty} x e^{-x} \\
&=
0.
\end{aligned}

The entropy can now be expressed in the unambiguous form, summing over all the non-zero eigenvalues of the density operator

\label{eqn:densityMatrixEntropy:180}
\boxed{
S = – \sum_{ \wedge_{kk} \ne 0} \wedge_{kk} \ln \wedge_{kk}.
}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.