8.21 Quantum Tomography and Evolution of Open Systems

Quantum Tomography

Output electrons, we want to learn \(\rho\), then we do measurement to get \(\rho\) where \(\rho\) is a density matrix

Let's choose standard basis \(\{|x\rang\}_{x\in [d]},d:=|A|\) and \(\Lambda_x=|x\rang\lang x|\)

Then the probability for outcome \(x\) is \(p_{x}=\text{Tr}[\Lambda_{x}\rho]=\text{Tr}[|x\rang\lang x|\rho]=\lang x|\rho|x\rang\) is on the diagonal of \(\rho\)

We do many times measurements and count them, then dived the total number of measurements, we get the probability of \(x\)(e.g. \(x=2\)), that is we can estimate \(p_x\) in experiments

Then we can know the diagonal exactly.

For the rest of matrix:

Informationally Complete POVM

A POVM \(\{\Lambda_x\}_{x\in [m]}\) is called informationally complete if \(\text{Herm}(A)=\text{span}_\R\{\Lambda_x\}_{x\in[m]}\)

In the previous example: \(\Lambda_x=|x\rang\lang x|\) is not informationally complete because \(\text{span}_\R\{\Lambda_x\}_{x\in [d]}=\left\{\sum^d_{x=1}r_x|x\rang\lang x|:r_x\in \R\right\}\) is diagonal

Recall: \(\dim(\text{Herm}(A))=|A|^2\) where \(|A|=d\)

If \(\{\Lambda_x\}_{x\in [m]}\) is IC-POVM, then \(m\geq d^2\)

If \(m=d^2\), then \(\{\Lambda_x\}_{x\in[m]}\) is a basis of \(\text{Herm}(A)\)

Example

\(\{\Gamma_{0}=|0\rang\lang 0|,\Gamma_{1}=|1\rang\lang 1|,\Gamma_{2}=|+\rang\lang +|,\Gamma_{3}=|+i\rang\lang +i|\}\) where \(|+\rang=\frac{1}{\sqrt{2}}(|0\rang+|1\rang)\) and \(|+i\rang=\frac{1}{\sqrt{2}}(|0\rang+i|1\rang)\)

This list satisfies \(\Gamma_{x}\geq 0\) and a basis since eigenvalues, but the sum is not identity, thus it's not POVM

Does such \(\Lambda_x\) always exists? Yes

Definition: \(\Gamma:=\sum^{3}_{y=0}\Gamma_y>0\) (Hermitian, positive, then invertible since Hermitian ensures all eigenvalues are real and positive ensure all eigenvalues are positive, then \(\Gamma\) is injective and since it's endomorphism, then it's invertible)

Then \(\Lambda_{x}:=\Gamma^{-\frac{1}{2}}\Gamma_{x}\Gamma^{-\frac{1}{2}}\) where \(\Gamma\) is diagonal, then \(\Gamma^{-\frac{1}{2}}=\sum_j\lambda_j^{-\frac{1}{2}}|x\rang\lang x|\)

Why not \(\Lambda_x=\Gamma^{-1}\Gamma_{x}\)? Because this is not positive, then not POVM

Then \(\sum^{3}_{x=0}\Lambda_{x}=\sum_{x=0}^{3}\Gamma^{-\frac{1}{2}}\Gamma_{x}\Gamma^{-\frac{1}{2}} =\Gamma^{-\frac{1}{2}}(\sum^{3}_{x=0}\Gamma_{x})\Gamma^{-\frac{1}{2}}=\Gamma^{-\frac{1}{2}} \Gamma\Gamma^{-\frac{1}{2}}=I\)

This \(\Lambda_x\) is also positive: \(\lang \psi|\Lambda_x|\psi\rang=\lang \psi|\Gamma^{-\frac{1}{2}}\Gamma_x\Gamma^{-\frac{1}{2}}|\psi\rang=\lang \phi|\Gamma_x|\phi\rang>0\) since \(\Gamma_x\) is positive

Thus \(\Lambda_x\) is IC-POVM

Let \(H \in \text{Herm}(A)\). Let \(H' = \Gamma^{1/2} H \Gamma^{1/2}\).
We need to show that there exist \(\{r_{x}\}_{x=0}^{3}\subset\mathbb{R}\) s.t. \(H=\sum_{x=0}^{3}r_{x}\Lambda_{x}\).
Since \(\{\Gamma_x\}_{x=0}^3\) is a basis, there exists \(\{r_{x}\}_{x=0}^{3}\) s.t. \(H=\sum_{x=0}^{3}r_{x}\Gamma_{x}\Rightarrow\underbrace{\Gamma^{-1/2}H^{\prime}\Gamma^{-1/2}} _{H}=\sum_{x=0}^{3}r_{x}\underbrace{\Gamma^{-1/2}\Gamma_{x}\Gamma^{-1/2}}_{\Lambda_{x}}\)

Take \(m=d^2\), therefore \(\{\Lambda_x\}^\R_{x=1}\) is a basis of \(\text{Herm}(A)\)

Let \(\{\Gamma_y\}_{y=1}^{d^2}\) be the dual basis of \(\{\Lambda_x\}_{x=1}^{d^2}\)

Theorem: \(\text{Tr}[\Lambda_{x}\Gamma_{y}]=\delta_{xy}\) where \(\Gamma_y\) doesn't have to be positive, it is only Hermitian

Show that at least one eigenvalue of \(\Gamma_y\) is negative

Proof

Suppose \(\Gamma_{y}\ge 0\), then \(\Lambda_x = \sum_{k=1}^d a_k |u_k\rangle \langle u_k|\), \(a_k \ge 0\), \(\Gamma_{y}=\sum_{m=1}^{r}b_{m}|v_{m}\rangle\langle v_{m}|\), \(b_m > 0\), \(r \ge 1\) by spectral decomposition.

Since \(\text{Tr}[\Lambda_{x}\Gamma_{y}]=\delta_{xy}\), when \(x \ne y\), \(\text{Tr}[\Lambda_{x}\Gamma_{y}]=0\).

\(\Rightarrow \text{Tr}\left[ \sum_{k,m=1}^{d}a_{k}b_{m}|u_{k}\rangle \langle u_{k} | v_{m}\rangle \langle v_{m}| \right] = 0 \Rightarrow \sum_{k,m=1}^{d}a_{k}b_{m}\text{Tr} [|u_{k}\rangle \langle u_{k}|v_{m}\rangle \langle v_{m}|] = 0\).

Then \(\text{Tr}[|u_{k}\rangle\langle u_{k}|v_{m}\rangle\langle v_{m}|]=0\Rightarrow\text{Tr} [\langle u_{k}|v_{m}\rangle\langle v_{m}|u_{k}\rangle]=0\Rightarrow\langle u_{k}| v_{m}\rangle=0\) when \(x \ne y\).

For \(S = \text{span} \{ |v_m\rangle \}_{m=1}^r\), \(|u_k\rangle \in S^\perp \Rightarrow \{ \Lambda_x \}_{x \ne y} \subset \text{Herm}(S^\perp)\).

\(\Rightarrow \text{dim} \{ \Lambda_x \}_{x \ne y} \le \text{dim Herm}(S^\perp) \Rightarrow d^2-1 \le (d-r)^2 \le (d-1)^2\). Contradiction

\(\rho=\sum_{y=1}^{d^2}q_{y}\Gamma_{y}\) since mixed state is hermitian and we know \(p_{x}=\text{Tr}[\Lambda_{x}\rho]\) \(=\sum_{y=1}^{d^2}q_{y}\text{Tr}[\Lambda_{x}\Gamma_{y}]=\sum_{y=1}^{d^2}q_{y}\delta _{xy}=q_{x}\)

Thus \(\rho=\sum_{x=1}^{d^{2}}p_{x}\Gamma_{x}\), thus we can know the matrix

Evolution of Open Systems

Any evolution has to take density matrix to density matrix in physics, because the information is described in density matrix

The Axiomatic Approach:

Axiom 1

Evolution is described with a linear Transformation: \(\mathcal{E}: \mathcal{L}(A) \rightarrow \mathcal{L}(B)\) where \(\mathcal{L}(A): = \mathcal{L}(A,A) \cong \mathbb{C}^{d\times d}\) and \(d := |A|\)

Take density matrices to density matrices

Alice prepares the state \(\rho_x\) and \(\rho_x\) interacts with enviroment then it becomes \(\mathcal{E}(\rho_{x})\) and send to Bob

If we don't know \(x\), density matrix is an ensemble, thus we must have \(\mathcal{E}(\sum_{x}p_{x}\rho_{x})=\sum_{x}p_{x}\mathcal{E}(\rho_{x} )\)

Axiom 2

\(\mathcal{E}\) is Trace Preserving

Motivation: \(\text{Tr}[\mathcal{E}(\rho)]=1\) if \(\rho\) is a density matrix (\(\rho\geq 0,\text{Tr}[\rho]=1\))

Lemma

Let \(\mathcal{E}:\mathbb{C}^{d\times d}\to \mathbb{C}^{d'\times d'}\) be a linear transformation.
If \(\text{Tr}[\mathcal{E}(\rho)]=1,\forall\rho\geq0,\text{Tr}[\rho]=1\), then \(\text{Tr}[\mathcal{E}(M)]=\text{Tr}[M]\)

Proof

We show \(\mathrm{Tr}[\mathcal{E}(\Lambda)] = \mathrm{Tr}[\Lambda] \quad \forall \Lambda \ge 0\)
Let \(\lambda = \mathrm{Tr}[\Lambda]\), then \(\mathrm{Tr}\left[\mathcal{E}\underbrace{\left(\frac{1}{\lambda}\Lambda\right)}_{\text{density matrix}}\right] = 1 \Rightarrow \frac{1}{\lambda}\mathrm{Tr}(\mathcal{E}(\Lambda)) = 1 \Rightarrow \mathrm{Tr}(\mathcal{E} (\Lambda)) = \lambda = \mathrm{Tr}[\Lambda]\)
Hermitian Matrices: We show \(\mathrm{Tr}(\mathcal{E}(H)) = \mathrm{Tr}(H) \quad \forall H \in \mathrm{Herm}(A)\)
We know \(H=\Lambda_{+}-\Lambda_{-}\), then \(\mathrm{Tr}[\mathcal{E}(H)]=\mathrm{Tr}[\mathcal{E}(\Lambda_{+}-\Lambda_{-})]=\mathrm{Tr} [\mathcal{E}(\Lambda_{+})]-\mathrm{Tr}[\mathcal{E}(\Lambda_{-})]\)
\(= \mathrm{Tr}[\Lambda_{+}] - \mathrm{Tr}[\Lambda_{-}] = \mathrm{Tr}[H]\)
Show it for all complex matrix: \(\text{Tr}[\mathcal{E}(M)]=\text{Tr}[M]\)
We know \(M=H_{1}+iH_{2}\), then \(\text{Tr}[\mathcal{E}(H_{1}+iH_{2})]=\text{Tr}[\mathcal{E}(H_{1})]+i\text{Tr}[\mathcal{E}(H_{2})]=\text{Tr}[H_1]+i\text{Tr}[H_2]=\text{Tr}[H_1+iH_2]=\text{Tr}[M]\)

Axiom 3

\(\mathcal{E}\) is complete positive.

Definition

A linear map \(\mathcal{E} \in \mathcal{L}(A \rightarrow B)\) is called \(k\)-positive if \(\left(id_{K}^{R \rightarrow R} \otimes \mathcal{E}^{A \rightarrow B}\right)\left(\rho^{R A}\right) \geq 0 \quad \forall \rho^{R A} \geq 0\) where \(K:=|R|\), \(\forall R\)
Furthermore, \(\mathcal{E}\) is called completely positive (CP) if it is \(k\)-positive for all \(k \in \mathbb{N}\).
Every Physical operation is a CPTP, linear map(complete positive, trace preserving, linear map)

Consider a composite system consisting of two subsystems \(A\) and \(B\). Such a system is described by a bipartite density operator \(\rho^{AB} \in \mathcal{D}(A \otimes B)\).
If the subsystem \(A\) undergoes a physical evolution described by a linear map \(\mathcal{E}\in \mathcal{L}(A \to A)\), while system \(A\) does not evolve and remain intact, then the state \(\rho^{AB}\)will evolve to the state \(\sigma^{AB}:=(\mathcal{E}^{A\rightarrow A}\otimes id^{B\rightarrow B})(\rho^{AB} )\)

If \(\mathcal{E}\) represents a physical evolution, then both \(\mathcal{E}\) and \(\mathrm{id}^{A} \otimes \mathcal{E}\) must take one density matrix to another density matrix. In particular, the linear map \(\mathrm{id} \otimes \mathcal{E} \in\) \(\mathcal{L}(AB \rightarrow A B')\) must also be a positive map for any system \(A\).
Thus it is possible that \(\mathcal{E}\) are positive while \(\mathrm{id}^{A} \otimes \mathcal{E}\) is not positive. The following example shows this.

Example

\(\Phi_{+}^{AB}=|\Phi_{+}\rangle\langle\Phi_{+}|,|\Phi_{+}\rangle=\frac{1}{\sqrt{2}} (|00\rangle+|11\rangle)\)

\(\Phi_{+}^{AB}=\frac{1}{2}(|0\rangle\langle0|\otimes|0\rangle\langle0|+|0\rangle\langle 1|\otimes|0\rangle\langle1|+|1\rangle\langle0|\otimes|1\rangle\langle0|+|1\rangle \langle1|\otimes|1\rangle\langle1|)\)

Let \(\mathcal{E}(\rho) = \rho^T\), then \((\mathcal{E} \otimes id^B)(\Phi_+) = \frac{1}{2}(|0\rangle\langle0|\otimes|0\rangle\langle0| + |1\rangle\langle0|\otimes|0\rangle\langle1| + |0\rangle\langle1|\otimes|1\rangle\langle0| + |1\rangle\langle1|\otimes|1\rangle\langle1|)\)

\(\Phi_{+}=\frac{1}{2} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix}\) is positive but \(\mathcal{E}\otimes \text{id}(\Phi_{+}) = \frac{1}{2} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\) is not

Then \(\mathcal{E}\) take a density matrix to a negative matrix is not valid.

Conclusion

Such a linear CPTP map is called a quantum channel. The set of all quantum channels in \(\mathcal{L}(A \rightarrow B)\) will be denoted by \(CPTP(A \rightarrow B)\).