On the Formulations of Quantum Mechanics

Argument: Quantum Theory \(=\) Classical Probability Theory with \(2\)-norm and continuity axiom (complex numbers).

  1. Why \(2\)-norm and why not \(p\)-norm?

    Because if there are any linear transformations other than these trivial ones that preserve the \(p\)-norm, then either \(p = 1\) or \(p=2\).

    If \(p=1\) we get classical probability theory, while if \(p =2\) we get quantum mechanics.

    Also all transformations must be norm preserving. So what kind of matrix or linear transformation preserves the \(2\)-norm? Unitary Matrices \((UU^\dagger = I)\).

  2. Why complex numbers and not real numbers?

    Axiom of continuity: There exists a continuous reversible transformation on a system between any two pure states of that system.

    Thus, we need our field associated with vector spaces to be algebraically closed. Hence, we need complex numbers.

    If you want every unitary operation to have a square root, then you have to go to the complex numbers.

  3. Why do transformations need to be linear?

    If quantum mechanics were nonlinear, then one could build a computer to solve NP-complete problems in polynomial time. (Abrams and Lloyd).

History of Quantum Computing

Models of Classical Computation

The following serve as universal models of computation:

  • Turing Machines
  • \(\lambda\)-calculus
  • Circuits

Super-Universal

Circuits can describe computations which are beyond what a Turing machine can do.

Apparently, we can solve the halting problem under some circuit model. But, what is that supposed to mean?

Quantum Computation

The idea of a QC model is to have information in superposition. If we want Turing Machines to be part of the quantum computational model then we will somehow have to make the read/write head to be in a superposition. But that is not possible (doubt). Hence we use a circuit where we keep the bits in a superposition and operate on them with gates.

History of Quantum Mechanics and Computation

Major questions in QM started arising with the EPR paradox in 1936 when faster than light travel was questioned.

Here's the paradox: Consider \(2\) particles that are moving away. Their positions are \(x\) and \(-x\) and momenta are \(p\) and \(-p\). Now we can't measure the exact position and momentum of one particle due to the uncertainty principle. But what if we measure the position of one particle and momentum of the other at the same time (Like a split second difference)? Assuming there is no interaction between the particles, we would in principle know the position and momentum of one particle at a time (cause we can know position or momentum of particle \(1\) if we measure that quantity on the second particle cause its just the negative of it). This would violate the uncertainty principle. Hence, there has to be some sort of interaction between the two particles that makes the wavefunction of particle \(2\) also collapse instantaneously when I measure position or momentum of particle \(1\). This mean that the particles were somehow connected. This was the paradox. But Schrodinger said this is possible by entanglement.

In 1964 Bell gave his Theorem and then it was later verified too which proved that Quantum Mechanics is cool with EPR pair. But there was another problem that how is faster than light communication possible? It was later proved that faster than light communication is not possible indeed as you cannot actually send 'information' with the entanglement coordination. Basically you cannot manipulate the particles to actually transmit any useful information.

Non-cloning theorem: In physics, the no-cloning theorem states that it is impossible to create an independent and identical copy of an arbitrary unknown quantum state, a statement which has profound implications in the field of quantum computing among others.

Quantum Theory from 5 reasonable Axioms

State

Mathematical object that can be used to deter-mine the probability associated with the out-comes of any measurement that may be per-formed on a system prepared by the given preparation.

However, we do not need to measure all possible probability measurements to determine the state of a system.

  • K (degrees of freedom): minimum number of probability measurements required to determine the state of a system, i.e., number of real parameters re-quired to specify the state.
  • N (dimension): maximum number of states that can be reliably distinguished from one another in a single shot measurement.

Axioms

  1. Probabilities
  2. Simplicity: \(K = K(N)\)
  3. Subspaces: If state of a system \(A\) \(\in M_{dim}\) subspace \(\implies dim(A) = M\)
  4. Composite systems: \(N = N_aN_b \text{ and } K = K_aK_b\)
  5. Continuity: \(\exists\) a continuous reversible transformation on a system between anytwo pure states of that system

Landauer's Principle

Landauer's Principle explains that when info is erased, it requires work.

Each bit erased \(\implies \Delta S = -K \ln(2)\). Hence, heat associated is given by \(Q = -KT \ln(2)\).

Hence to keep constant temp, that amount of work needs to be put in. This generalises to reversible processes requiring work.

Postulates of Quantum Mechanics

I shall state here the four major generic postulates of Quantum Mechanics stated in terms of both state-vector formalization and density-matrix formalization.

State is a vector

  1. Isolated physical system is given by its state vector operating on a certain Hilbert space.

  2. Evolution of a closed quantum system is given by a unitary transformation.

    In its physical interpretation we have this postulate governed by the Schrodinger Equation, as stated.

    \[ i{\hbar}\frac{d\vert\psi\rangle}{dt} = H\vert\psi\rangle \]

    The Hamiltonian is a hermitian operator and has a spectral decomposition, \(H = \sum E\vert E\rangle\langle E\vert\).

  3. The state space of a composite physical system is the tensor product of the state spaces of the component systems.

    \[ \vert\psi\rangle = \vert\psi_1\rangle\otimes...\otimes\vert\psi_n\rangle \]

  4. Quantum measurements are described by a collection \(\{M_m\}\) of measurement operators acting on the state space of the system.

    Probability that upon measurement the outcome is \(m = p(m) = \langle\psi\vert M_m^\dagger M_m\vert\psi\rangle\) and the state of the system becomes as follows.

    \[ \vert\psi\rangle \xrightarrow{\text{on measuring}} \frac{M_m\vert\psi\rangle}{\sqrt{p(m)}} = \frac{M_m\vert\psi\rangle}{||M_m\vert\psi\rangle||} \]

    Measurement operators also follow the completeness equation, \(\sum_m M_m^\dagger M_m = I\).

State is a density matrix

  1. Isolated physical system is given by its density matrix operating on a certain Hilbert space.

  2. Evolution of a closed quantum system is given by a unitary transformation as \(\rho \xrightarrow{U} U\rho U^\dagger\).

  3. The state space of a composite physical system is the tensor product of the state spaces of the component systems.

    \[ \rho = \rho_1\otimes...\otimes\rho_n \]

  4. Quantum measurements are described by a collection \(\{M_m\}\) of measurement operators acting on the state space of the system.

    Probability that upon measurement the outcome is \(m = p(m) = tr(M_m^\dagger M_m\rho)\) and the state of the system becomes as follows.

    \[ \rho \xrightarrow{\text{on measuring}} \frac{M_m\rho M_m^\dagger}{p(m)} \]

    Measurement operators also follow the completeness equation, \(\sum_m M_m^\dagger M_m = I\).

Measurement

Postulate

Quantum Measurements are descrivbed by a collection of measurement operators \(\{M_m\}\) with \(p(m)= \langle \psi | M_m^\dagger M_m | \psi \rangle,\ \sum{p(m)} = 1\).

\[ \vert \psi \rangle \xrightarrow{\text{on measurement}} \frac{M_m \vert \psi \rangle}{\sqrt{p(m)}} \]

If \(M_i = |i\rangle \langle i|,\ \forall i\) then we are measuring in a computational basis.

Non-distinguishability of arbitrary states

We cannot distinguish any two arbitrary non-orthogonal quantum states.

Proof: Let \(|\psi_1\rangle\) and \(|\psi_2\rangle\) be two non-orthogonal states.

Then \(\exists\ E_1, E_2\) such that \(E_1 = \sum_{j:f(j) = 1} M_j^\dagger M_j\) and similarly \(E_2\) then \(\langle \psi_1 | E_1 | \psi_1 \rangle = 1\) and \(\langle \psi_2 | E_2 | \psi_2 \rangle = 1\).

Thus, \(\sqrt{E_2}| \psi_1\rangle = 0\) and let \(\psi_2 = a\psi_1 + b \phi\) where \(\psi\) and \(\phi\) are orthogonal.

\[ \langle \psi_2 | E_2 \psi_2 \rangle = |\beta|^2 \langle \psi_2 | \phi \psi_2 \leq |\beta|^2 < 1 \]

Hence, proved by contradiction.

Projective Measurements

We can use projective measurement formalism for any general measurement too.

In case of projective measurements, \(M = \sum mP_m\) and \(p(m) = \langle \psi | P_m | \psi \rangle\).

\[ \vert \psi \rangle \rightarrow \frac{P_m \vert \psi\rangle}{\sqrt {p(m)}} \]
\[ E(M) = \sum mp(m) = \langle \psi | M | \psi \rangle = \langle M\rangle \]
\[ \Delta(M) = \sqrt{\langle M^2\rangle - \langle M \rangle^2} \]

Here, \(E(M)\) is expectation and \(\Delta(M)\) is standard deviation or the square root of variance.

POVM Measurements

POVM Measurements are a formalism where only measurement statistics matters.

\[ \{E_m\} \rightarrow \sum{E_m} = I,\ p(m) = \langle \psi | E_m |\psi \rangle \]

Here, each of \(E_m\) are hermitian.

Global Phase doesn't matter

We say \(e^{i\theta} |\psi\rangle \equiv |\psi\rangle\) but why? Because, \(\langle \psi | M_m^{\dagger}M_m | \psi \rangle = \langle \psi \vert e^{-i\theta}M_m^\dagger M_m e^{i\theta}\vert \psi \rangle\).

However, be aware that the global phase is quite different from the relative phase.

Density Matrices

We can represent a system as an ensemble of pure states \(\{p_i, \psi_i\}\). Now, if you have exact knowledge of the system then it is for sure in a pure state, i.e., \(\rho = |\psi\rangle\langle\psi|\). However, if we have classical uncertainty amongst the possible states. The system can be represented as a mixed state \(\rho = \sum_{i} p_i\psi_i\) where the probabilities \(p_i\) are classical in nature.

This formulation helps us a lot in dealing with quantum information, noisy systems and helps us represent measurements better, as well. Why? Because it provides a convenient means for describing quantum systems whose state is not completely known.

Postulates in Density Matrices formulation

  1. Isolated physical system is given by its density matrix operating on a certain Hilbert space.

  2. Evolution of a closed quantum system is given by a unitary transformation as \(\rho \xrightarrow{U} U\rho U^\dagger\).

  3. The state space of a composite physical system is the tensor product of the state spaces of the component systems.

    \[ \rho = \rho_1\otimes...\otimes\rho_n \]

  4. Quantum measurements are described by a collection \(\{M_m\}\) of measurement operators acting on the state space of the system.

    Probability that upon measurement the outcome is \(m = p(m) = tr(M_m^\dagger M_m\rho)\) and the state of the system becomes as follows.

    \[ \rho \xrightarrow{\text{on measuring}} \frac{M_m\rho M_m^\dagger}{p(m)} \]

    Measurement operators also follow the completeness equation, \(\sum_m M_m^\dagger M_m = I\).

Properties

Theorem 1: An operator \(\rho\) is a density operator if and only if it is both positive semi-definite \((\rho = \rho^\dagger\) and non-negative eigenvalues\()\) and \(tr(\rho) = 1\).

Converse is easy to prove. If an operator is both positive and has trace as one, then it shall have a spectral decomposition of the form \(\sum_i\lambda_i|i\rangle\langle i|\).

For the direct proof, let us consider \(tr(\rho) = \sum_i p_i tr(|\psi_i\rangle\langle\psi_i|) = 1\).

Theorem 2: The sets \(|\tilde\psi\rangle\) and \(|\tilde\phi\rangle\) generate the same density matrix if and only if,

\[ |\tilde\psi_i\rangle = \sum_j u_{ij} |\tilde\phi_j\rangle \]

where \(u_{ij}\) is a unitary matrix of complex numbers, with indices \(i\) and \(j\), and we 'pad' whichever set of vectors \(|\tilde\psi_i\rangle\) or \(|\tilde\phi_j\rangle\) is smaller with additional vectors \(0\) so that the two sets have the same number of elements.

As a consequence of the theorem, note that \(\rho = \sum_i p_i |\psi_i\rangle\langle\psi_i| = \sum_j q_j |\phi_j\rangle\langle\phi_j|\) if and only if we have the following as true for some unitary matrix \(u_{ij}\).

\[ \sqrt p_i |\psi_i\rangle = \sum_j u_{ij} \sqrt q_j |\phi_j\rangle \]

Theorem 3: If \(\rho\) is a density operator, then \(\rho\) is a pure state if and only if \(tr(\rho^2) = 1\) and mixed state if and only if \(tr(\rho^2) < 1\).

Theorem 4: Observable \(M\) has expectation \(\sum_x \langle \psi_x|M|\psi_x\rangle = tr(M\rho)\).

Reduced Density Operator

This is the single-most important application of density operator formulation is the existence of reduced density operator. It is defined as follows.

\[ \rho_A = tr_B(\rho_{AB}) \]

This allows us to talk about sub-systems of a composite system.

\[ tr_B(|a_1\rangle\langle a_2| \otimes |b_1\rangle\langle b_2|) = |a_1\rangle\langle a_2| tr(|b_1\rangle\langle b_2|) \]

Schmidt Decomposition

Suppose \(|\psi\rangle\) is a pure state of a composite system, \(AB\). Then, there exists orthonormal states \(|i_A\rangle\) for system \(A\), and orthonormal states \(|i_B\rangle\) of system \(B\) such that

\[ |\psi\rangle = \sum_i \lambda_i |i_A\rangle |i_B\rangle \]

where \(\lambda_i\) are non-negative real numbers satisfying \(\sum_i \lambda_i^2 = 1\) known as Schmidt co-efficients.

Purification

Suppose we have a mixed state \(\rho_A\) for a system \(A\). Then, we can introduce another system \(R\) such that \(AR\) forms a pure state \(|AR\rangle\) and \(\rho_A = tr_R(|AR\rangle\langle AR|)\).

Given \(\rho_A = \sum_i p_i |i_A\rangle\langle i_A|\), we shall have the following where \(|i_R\rangle\) are orthonormal basis states.

\[ |AR\rangle = \sum_i \sqrt{p_i}|i_A\rangle|i_R\rangle \]

Bloch Sphere and Rotations

We can represent any \(\rho\) (density matrix) as \(\frac 1 2 (I + \vec{r}\cdot\vec{\sigma})\)

Single Qubit Operations

\[ e^{iAx} = cos(x) + isin(x)A \]

Thus, \(R_x(\theta) = e^{i\theta X/2}\), \(R_y(\theta) = e^{i\theta Y/2}\) and \(R_z(\theta) = e^{i\theta Z/2}\).

\[ R_{\hat{n}}(\theta) = e^{i\theta \hat n\cdot\vec\sigma/2} \]

In general, we have the above equation where \(\vec\sigma = X\hat i + Y\hat j + Z\hat k\).

Some Algebra

\[ X^2 = Y^2 = Z^2 = -iXYZ = I \]
\[ R_{\hat n}(\alpha) = R_z(\phi)R_y(\theta)R_z(\alpha)R_y(-\theta)R_z(-\phi)\\ = R_z(\phi)R_y(\theta)R_z(\alpha)R_y(\theta)^\dagger R_z(\phi)^\dagger \]

Theorems

  1. Any arbitrary single qubit unitary operator can be written in the form \(U = e^{i\alpha}R_{\hat n}(\theta)\).
  2. Suppose \(U\) is a unitary operation over a single qubit then \(\exists\ \alpha, \beta, \gamma, \delta\) such that \(U = e^{i\alpha}R_{\hat n}(\beta)R_{\hat m}(\gamma)R_{\hat n}(\delta)\).
  3. There exists unitaries \(A, B, C\) for any given unitary \(U\) such that \(ABC = I\) and \(U = e^{i\alpha} AXBXC\).

Proof that Bloch Sphere Unitaries as Rotations

A single qubit operator can be represented as \(U = a_0I + a_1X + a_2Y +a_3Z\).

Also, such a unitary can also be represented this way,

\[ U = \begin{bmatrix} a & b\\ c & d \end{bmatrix} \]

and thus, we obtain the following equivalences.

\[ a_0 = (a+d)/2,\ a_1 = (b+c)/2, \]
\[ a_2 = (c-b)/2i,\ a_3 = (d-a)/2 \]

Also from \(UU^\dagger = I\) we get,

\[ |a_0|^2 + |a_1|^2 + |a_2|^2 + |a_3|^2 = 1 \]
\[ a^∗_0a_1+a^∗_1a_0+ia^∗_2a_3−ia^∗_3a_2 = 0 \]
\[ a^∗_0a_2−ia^∗_1a_3+a^∗_2a_0+ia^∗_3a_1 = 0 \]
\[ a^∗_0a_3+ia^∗_1a_2−ia^∗_2a_1+a^∗_3a_0 = 0 \]

where we define \(|a_0| = cos(\theta/2)\) then \(|a_1|^2 + |a_2|^2 + |a_3|^2 = |sin(\theta/2)|\).

Then define,

\[ n_x = |a_1|/|sin(\theta/2)| \]
\[ n_y = |a_2|/|sin(\theta/2)| \]
\[ n_z = |a_3|/|sin(\theta/2)| \]

and further we get \(n_x^2 + n_y^2 + n_z^2 = 1\).

Now, we define \(exp(iα) =a_0/cos(θ/2)\) and denote the phase of \(a_1,a_2,a_3\) as \(α_1,α_2,α_3\) respectively.

By putting these in the other constraints we get, \(α_1=α_2=α_3=α−π/2\).

\[ a_0 = e^{i\alpha}cos(\theta/2), \]
\[ a_1 = -ie^{i\alpha}sin(\theta/2)n_x, \]
\[ a_2 = -ie^{i\alpha}sin(\theta/2)n_y, \]
\[ a_3 = -ie^{i\alpha}sin(\theta/2)n_z \]
\[ U= e^{iα}(\cos(\frac{θ}{2})I − i\sin(\frac{\theta}{2})(n_xX+n_yY+n_zZ))= e^{iα}R_{\hat n}(θ) \]

CHSH Inequality

In a classical experiment, we have the following setup where Alice can choose to measure either Q or R and Bob chooses either S or T. The measurements are performed imultaneously and far off from each other.

\[ E(QS) + E(RS) + E(RT) - E(QT) = E(QS + RS + RT-QT) \]
\[ \implies E(QS) + E(RS) + E(RT) - E(QT) = \sum_{q,r,s,t}p(q, r, s, t)(qs + rs + rt - qt) \]
\[ \implies E(QS) + E(RS) + E(RT) - E(QT) \leq \sum_{q,r,s,t}p\times 2 = 2 \]

and thereby we obtain the inequality \(E(QS) + E(RS) + E(RT) - E(QT) \leq 2\).

This is one of the set of Bell inequalities, the first of which was found by John Bell. This one in particular is named CHSH inequality.

Quantum Anomaly

In the quantum case, let us consider the measurements to be based on the following observables over the EPR pair \(|\psi\rangle = \frac{|01\rangle - |10\rangle}{\sqrt 2}\).

\[ Q = Z_1, R= X_1, \]
\[ S = \frac{-Z_2-X_2}{\sqrt 2}, \]
\[ T = \frac{Z_2 - X_2}{\sqrt 2} \]

Then, we have the following result.

\[ E(QS) + E(RS) + E(RT) - E(QT) = \frac{1}{\sqrt 2} + \frac{1}{\sqrt 2} + \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2} = 2\sqrt 2 > 2 \]

Thus, in other words, CHSH inequality doesn't hold.

Interpretation

The fact that CHSH doesn't hold in the quantum scenario implies that two of the major assumptions about nature is wrong in case of the classical experiment.

The assumptions are:

  • Realism: Q, R, S, T are physical quantities which have defininte values irrespective of observation.
  • Locality: Alice's measurement doesn't influence that of Bob's.

Thus, the result of CHSH being false when accounted for the quantum mechanical properties of nature (we can perform the experiment in a lab with particles) suggests that nature cannot be locally real and neither can any true mathematical representation of it be locally real.