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### 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.

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

The density matrix

\begin{equation}\label{eqn:qmLecture3:20}

\hat{\rho} = \ket{\Psi}\bra{\Psi}

\end{equation}

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

\begin{equation}\label{eqn:qmLecture3:40}

\hat{\rho} \ket{\phi} = \ket{\Psi} \lr{ \braket{ \Psi }{\phi} }.

\end{equation}

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

\begin{equation}\label{eqn:qmLecture3:60}

\ket{\Psi}

= \sum_{m,n} C_{m, n} \ket{m}_{\textrm{L}} \ket{n}_{\textrm{R}},

\end{equation}

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

\begin{equation}\label{eqn:qmLecture3:460}

\ket{\Psi}

= \sum_{m,n} C_{m, n} \ket{m} \ket{n}.

\end{equation}

The density operator is

\begin{equation}\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’}.

\end{equation}

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

\begin{equation}\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}

\end{equation}

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

\begin{equation}\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}

\end{equation}

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

\begin{equation}\label{eqn:qmLecture3:140}

S_{\textrm{Total}} =

S_1^z

+

S_1^z,

\end{equation}

as sketched in fig. 3.

The average of some observable is

\begin{equation}\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’}.

\end{equation}

Consider the trace of the density operator observable product

\begin{equation}\label{eqn:qmLecture3:180}

\textrm{Tr}( \hat{\rho} \hatA )

= \sum_{m, n} \braket{m n}{\Psi} \bra{\Psi} \hatA \ket{m, n}.

\end{equation}

Let

\begin{equation}\label{eqn:qmLecture3:200}

\ket{\Psi} = \sum_{m, n} C_{m n} \ket{m, n},

\end{equation}

so that

\begin{equation}\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}

\end{equation}

This is just

\begin{equation}\label{eqn:qmLecture3:240}

\boxed{

\bra{\Psi} \hatA \ket{\Psi} = \textrm{Tr}( \hat{\rho} \hatA ).

}

\end{equation}

### Left observables

Consider

\begin{equation}\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}

\end{equation}

We see

\begin{equation}\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}} }.

\end{equation}

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

\begin{equation}\label{eqn:qmLecture3:300}

\ket{\psi_1} = \inv{\sqrt{2}} \lr{ \ket{ \uparrow \downarrow } – \ket{ \downarrow \uparrow } }

\end{equation}

\begin{equation}\label{eqn:qmLecture3:320}

\ket{\psi_2} = \inv{\sqrt{2}} \lr{ \ket{ \uparrow \downarrow } + \ket{ \uparrow \uparrow } }.

\end{equation}

For the first pure state the density operator is

\begin{equation}\label{eqn:qmLecture3:360}

\hat{\rho} = \inv{2}

\lr{ \ket{ \uparrow \downarrow } – \ket{ \downarrow \uparrow } }

\lr{ \bra{ \uparrow \downarrow } – \bra{ \downarrow \uparrow } }

\end{equation}

What are the reduced density matrices?

\begin{equation}\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}

\end{equation}

so the matrix representation of this reduced density operator is

\begin{equation}\label{eqn:qmLecture3:380}

\hat{\rho}_{\textrm{L}}

=

\inv{2}

\begin{bmatrix}

1 & 0 \\

0 & 1

\end{bmatrix}.

\end{equation}

For the second pure state the density operator is

\begin{equation}\label{eqn:qmLecture3:400}

\hat{\rho} = \inv{2}

\lr{ \ket{ \uparrow \downarrow } + \ket{ \uparrow \uparrow } }

\lr{ \bra{ \uparrow \downarrow } + \bra{ \uparrow \uparrow } }.

\end{equation}

This has a reduced density matrice

\begin{equation}\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}

\end{equation}

This has a matrix representation

\begin{equation}\label{eqn:qmLecture3:440}

\hat{\rho}_{\textrm{L}}

=

\begin{bmatrix}

1 & 0 \\

0 & 0

\end{bmatrix}.

\end{equation}

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.