11 KiB
Math 401, Topic 6: Postulates of quantum theory and measurement operations
Section 1: Postulates of quantum theory
This part is a review of the quantum theory, I will keep the content brief.
If you are familiar with the linear algebra defined before, you can jump right into this section to keep your time as viewing those compact notations.
Pure states
Pure state and mixed state
A pure state is a state that is represented by a unit vector in \mathscr{H}^{\otimes N}.
A mixed state is a state that is represented by a density operator in \mathscr{H}^{\otimes N}. (convex combination of pure states)
if \rho_j=|\psi_j\rangle\langle\psi_j|, then \rho=\sum_{j=1}^N p_j\rho_j is a mixed state, where p_j\geq 0 and \sum_{j=1}^N p_j=1.
Coset space
Two non-zero vectors u,v\in \mathscr{H} are said to represent the same state if u=cv for some complex number c with |c|=1.
The set of states of a quantum system is called the coset space of \mathscr{H}, u\sim v if u=cv for some complex number c with |c|=1.
The coset space is called the projective space of \mathscr{H}, denoted by P(\mathscr{H})\colon=(\mathscr{H}\setminus\{0\})/\sim.
Any vector in the form e^{i\theta}|u\rangle for some u\in \mathscr{H} and \theta\in \mathbb{R} represents the same state as |u\rangle.
Example: the system of a qubit has a Hilbert space \mathbb{C}^2, the coset space is P(\mathbb{C}^2)\cong S^2 is the Bloch sphere.
Composite systems
Tensor product
The tensor product of two Hilbert spaces \mathscr{H}_1 and \mathscr{H}_2 is the Hilbert space \mathscr{H}_1\otimes\mathscr{H}_2 with the inner product \langle u_1\otimes u_2,v_1\otimes v_2\rangle=\langle u_1,v_1\rangle\langle u_2,v_2\rangle.
The tensor product of two vectors u_1\in \mathscr{H}_1 and u_2\in \mathscr{H}_2 is the vector u_1\otimes u_2\in \mathscr{H}_1\otimes\mathscr{H}_2.
Multipartite systems
For each part in a multipartite quantum system, each part is associated a Hilbert space \mathscr{H}_i. The total system is associated a Hilbert space \mathscr{H}=\mathscr{H}_1\otimes\mathscr{H}_2\otimes\cdots\otimes\mathscr{H}_n.
The state of the total system has the form u_1\otimes u_2\otimes\cdots\otimes u_n for some u_i\in \mathscr{H}_i.
Entanglement (talk later)
A state |\psi\rangle is entangled if it cannot be expressed as a product state v_1\otimes v_2 for any single-qubit states |v_1\rangle and |v_2\rangle. In other words, an entangled state is non-separable.
Example: the Bell state |\psi^+\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) is entangled.
Assume it can be written as |\psi\rangle=|\psi_1\rangle\otimes|\psi_2\rangle where |\psi_1\rangle=a|0\rangle+b|1\rangle and |\psi_2\rangle=c|0\rangle+d|1\rangle. Then:
|\psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle
Setting this equal to |\psi^+\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) gives:
ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)
This requires:
ac=bd=\frac{1}{2}
ad=bc=0
This is a contradiction, so |\psi^+\rangle is entangled.
Mixed states and density operators
Density operator
A density operator is a Hermitian, positive semi-definite operator with trace 1.
The density operator of a pure state |\psi\rangle is \rho=|\psi\rangle\langle\psi|.
The density operator of a mixed state is given by the unit vector u_1,u_2,\cdots,u_n in \mathscr{H} with the probability p_1,p_2,\cdots,p_n, p_i\geq 0 such that \sum_{i=1}^n p_i=1.
The density operator is \rho=\sum_{i=1}^n p_i|u_i\rangle\langle u_i|.
Trace 1 proposition
Density operator on the finite dimensional Hilbert space \mathscr{H} are positive operators having trace equal to 1.
Pure state lemma
A state is pure if and only if Tr(\rho^2)=1.
For any mixed state \rho, Tr(\rho^2)<1.
[Proof ignored here]
Unitary freedom in the ensemble for density operators theorem
Let v_1,v_2,\cdots,v_l and w_1,w_2,\cdots,w_l be two collections of vectors in the finite dimensional Hilbert space \mathscr{H}, the vectors being arbitrary (can be zero) except for the requirement that they define the same density operator \rho.
\sum_{i=1}^l |v_i\rangle\langle v_i|=\sum_{i=1}^l |w_i\rangle\langle w_i|
Then there exists a unitary matrix U=(\mu_{ij})_{1\leq i,j\leq l} such that:
v_i=\sum_{j=1}^l \mu_{ij}w_j
The converse is also true.
If \rho is a density operator on \mathscr{H} given by: \sum_{i=1}^l |w_i\rangle\langle w_i| and vector v_i is given by: v_i=\sum_{j=1}^l \mu_{ij}w_j, then \rho_1=\sum_{i=1}^l |v_i\rangle\langle v_i| is the density operator of the subsystem \mathscr{H}_1.
[Proof ignored here]
Density operator of subsystems
Schmidt Decomposition theorem
Let |u\rangle\in \mathscr{H}_1\otimes\mathscr{H}_2 be a unit vector (pure state), then there exists orthonormal bases |v_i\rangle of \mathscr{H}_1 and |w_j\rangle of \mathscr{H}_2 and \{\lambda_k\},k\leq r, where r is the Schmidt rank of |u\rangle, such that:
|u\rangle=\sum_{k=1}^r \lambda_k|v_k\rangle\otimes|w_k\rangle
where \lambda_k are non-negative real numbers. such that \sum_{k=1}^r \lambda_k^2=1.
[Proof ignored here]
Remark: non-zero vector u\in \mathscr{H}_1\otimes\mathscr{H}_2 decomposes as a tensor product u=u_1\otimes u_2 if and only if the Schmidt rank of u is 1. A state that cannot be decomposed as a tensor product is called entangled.
Reduced density operator
In \mathscr{H}_1\otimes\mathscr{H}_2, the reduced density operator of the subsystem \mathscr{H}_1 is:
\rho_1=\operatorname{Tr}_2(\rho)=\sum_{k=1}^r \lambda_k^2|v_k\rangle\langle v_k|
where \rho is the density operator in \mathscr{H}_1\otimes\mathscr{H}_2.
Example:
Let \rho=\frac{1}{2}(|01\rangle+|10\rangle)\in \mathbb{C}^2\otimes\mathbb{C}^2,
Expand the expression of \rho in the basis of \mathbb{C}^2\otimes\mathbb{C}^2:
\rho=\frac{1}{2}(|01\rangle\langle 01|+|01\rangle\langle 10|+|10\rangle\langle 01|+|10\rangle\langle 10|)
then the reduced density operator of the subsystem \mathbb{C}^2 in first qubit is:
\begin{aligned}
\rho_1&=\operatorname{Tr}_2(\rho)\\
&=\frac{1}{2}(\langle 1|1\rangle|0\rangle\langle 0|+\langle 1|0\rangle|0\rangle\langle 1|+\langle 0|1\rangle|1\rangle\langle 0|+\langle 0|0\rangle|1\rangle\langle 1|)\\
&=\frac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|)\\
&=\frac{1}{2}I
\end{aligned}
State purification
Every mixed state can be derived as the reduction of a pure state on an enlarged Hilbert space.
State purification theorem
Let \rho be a mixed state in a finite dimensional Hilbert space \mathscr{H}, then there exists a unit vector |w\rangle\in \mathscr{H}\otimes\mathscr{H} such that:
\rho=\operatorname{Tr}_2(|w\rangle\langle w|)
Hint of proof:
Let u_1,u_2,\cdots,u_d be an orthonormal basis of \mathscr{H}, \sum_{i=1}^d p_i=1, p_i\geq 0, then:
\rho=\sum_{i=1}^d p_i|u_i\rangle\langle u_i|
Let w=\sum_{i=1}^d \sqrt{p_i}u_i\otimes u_i.
Observables
The observables in the quantum theory are self-adjoint operators on the Hilbert space \mathscr{H}, denoted by A\in \mathscr{O}
In finite dimensional Hilbert space, A can be written as \sum_{\lambda\in \operatorname{sp}{(A)}}\lambda P_\lambda, where P_\lambda is the projection operator onto the eigenspace of A corresponding to the eigenvalue \lambda. P_\lambda=P_\lambda^2=P_\lambda^*.
Effects and Busch's theorem for effect operators
Below is a section on Topic 4, about Gleason's theorem and definition of states, and Born's rule for describing the states using density operators.
Definition of states (non-commutative (quantum) probability theory)
A state on (\mathscr{B}(\mathscr{H}),\mathscr{P}) is a map \mu:\mathscr{P}\to[0,1] such that:
\mu(O)=0, whereOis the zero projection.- If
P_1,P_2,\cdots,P_nare pairwise disjoint orthogonal projections, then\mu(P_1\lor P_2\lor\cdots\lor P_n)=\sum_{i=1}^n\mu(P_i).
Where projections are disjoint if P_iP_j=P_jP_i=O.
Definition of density operator (non-commutative (quantum) probability theory)
A density operator \rho on the finite-dimensional Hilbert space \mathscr{H} is:
- self-adjoint (
A^*=A, that is\langle Ax,y\rangle=\langle x,Ay\ranglefor allx,y\in\mathscr{H}) - positive semi-definite (all eigenvalues are non-negative)
\operatorname{Tr}(\rho)=1.
If (|\psi_1\rangle,|\psi_2\rangle,\cdots,|\psi_n\rangle) is an orthonormal basis of \mathscr{H} consisting of eigenvectors of \rho, for the eigenvalue p_1,p_2,\cdots,p_n, then p_j\geq 0 and \sum_{j=1}^n p_j=1.
We can write \rho as
\rho=\sum_{j=1}^n p_j|\psi_j\rangle\langle\psi_j|
(under basis |\psi_j\rangle, it is a diagonal matrix with p_j on the diagonal)
Every basis of \mathscr{H} can be decomposed to these forms.
Theorem: Born's rule
Let \rho be a density operator on \mathscr{H}. then
\mu(P)\coloneqq\operatorname{Tr}(\rho P)=\sum_{j=1}^n p_j\langle\psi_j|P|\psi_j\rangle
Defines a probability measure on the space \mathscr{P}.
[Proof ignored here]
Theorem: Gleason's theorem (very important)
Let \mathscr{H} be a Hilbert space over \mathbb{C} or \mathbb{R} of dimension n\geq 3. Let \mu be a state on the space \mathscr{P}(\mathscr{H}) of projections on \mathscr{H}. Then there exists a unique density operator \rho such that
\mu(P)=\operatorname{Tr}(\rho P)
for all P\in\mathscr{P}(\mathscr{H}). \mathscr{P}(\mathscr{H}) is the space of all orthogonal projections on \mathscr{H}.
[Proof ignored here]
Extending the experimental procedure in quantum physics, many of the outcome probabilities are expectation of effects instead of projections. (POVMs)
Definition of effect
An effect is a positive (self-adjoint) operator E on \mathscr{H} such that 0\leq E\leq I.
The set of effects on \mathscr{H} is denoted by \mathscr{E}(\mathscr{H}).
An operator E is said to be the extreme point of the convex set \mathscr{E}(\mathscr{H}) if it cannot be written as a convex combination of two other effects.
That is, If E is an extreme point, then E=\lambda E_1+(1-\lambda)E_2 for some 0\leq \lambda\leq 1 and E_1,E_2\in \mathscr{E}(\mathscr{H}) implies E=E_1=E_2.
Effect operator lemma
The set of orthogonal projections on \mathscr{H}, \mathscr{P}(\mathscr{H}), is the set of extreme points of \mathscr{E}(\mathscr{H}).