Quantum Computing

Quantum States

Early 20th Century physics

• classical mechanics viewed matter as composed of particles, and light as composed of continuous electromagnetic waves
• diffraction experiment: beam of subatomic particles hitting a crystal diffract in a wave-like pattern
  • de Broglie wavelength associated with matter
• photoelectric effect: an atom hit by a beam of light may absorb it, causing electrons transition to a higher energy orbital
  • absorbed energy may be emitted as light causing electrons to transition back to the original orbital
  • light-matter transactions always occur via discrete packets of energy, i.e. photons
• further experimental evidence: old duality particle-wave theory needed to be replaced by a theory in which both matter and light can exhibit both particle- and wave-like behaviour.
• Young’s double slit experiment: shine light at a boundary with 2 very close slits, between the light source and an observing wall
  • pattern of light on the wall varies between light and dark as a result of interference between light
  • with one slit closed, no interference pattern is observed
  • remarkable results:
    • double-slit experiment can be performed with a single photon: if there is a single photon, why would there be any interference pattern?
    • can also be performed with electrons, protons, atomic nuclei, bucky balls, all of which exhibit interference behaviour
• conclusion: rigid distinction between waves and particles as a means of describing the physical world is untenable at the quantum level

Quantum States

Particle on a line

• consider a subatomic particle on a line that may only be found at one of several equally spaced points ${x_0, …, x_{n-1}}$ separated by distance $\delta x$
• describe the current state of the particle as a complex vector $[c_0, …, c_{n-1}]^T$
• denote the particle being at point $i$ as $\ket{x_i}$ (a ket)
• each basic state has an associated column vector $\ket{x_i} \rightarrow \delta_{ij} \in \mathbb{C}^n$
• note these vectors form the canonical basis of $\mathbb{C}^n$
• in quantum physics, the particle can be in a fuzzy blending of states: all vectors in $\mathbb{C}^n$ represent a legitimate physical state
• superposition: an arbitrary state $\ket \psi$ is then a linear combination of the basic states $\ket{x_i}, …, \ket{x_{n-1}}$ with complex amplitudes $c_0, …, c_{n-1}$
  • represents particle being simultaneously in all locations, a blending of all $\ket{x_i}$

\[\ket \psi = c_0\ket{x_0} + ... + c_{n-1}\ket{x_{n-1}}\]

• every state can therefore be represented as an element of $\mathbb{C}^n$ as

\[\ket \psi \rightarrow [c_0, ..., c_{n-1}]^T\]

• probability that, after observing the particle, we will detect it at point $x_i$:

\[p(x_i) = \frac{|c_i|^2}{|\ket \psi|^2} = \frac{|c_i|^2}{\sum_j{|c_j|^2}}\]

• clearly $p(x_i) \in \mathbb{R}$ and $0 \le p(x_i) \le 1$
• when $\ket \psi$ is observed, it will be found in one of the basic states
• kets can be added: $\ket \psi + \ket \psi’ = [c_0 + c_0’, …, c_{n-1}+c_{n-1}’]^T$
• a ket $\ket \psi$ and its scalar multiples $c\ket \psi$ (for some $c \in \mathbb{C}$) describe the same physical state
• the length of $\ket \psi$ doesn’t matter as far as physics goes
• it then makes sense to work with a normalised ket with length $1$:

\[\frac{\ket \psi}{|\ket \psi|}\]

• for a normalised ket, we have $p(x_i)=

  c_i

  ^2$

Spin

• property of subatomic particles which is the prototypical way to implement qubits
• Stern-Gerlach experiment: electron in presence of magnetic field observed to behave as if it were a charged spinning top, by acting as a magnet and trying to align itself with the magnetic field
  • experiment: shoot beam of electrons through a magnetic field oriented in a certain direction
  • beam is split into 2 streams with opposite spin
• differences to classical spinning top:
  • electron doesn’t have internal structure: quantum property with no classical analog
  • all electrons can be found in 1 of 2 locations, not distributed between (spin can be clockwise/anticlockwise)
• for each direction in space, there are only 2 spin states, spin up $\ket \uparrow$ and down $\ket \downarrow$
• arbitrary state is then a superposition of up and down:

\[\ket \psi = c_0\ket \uparrow + c_1 \downarrow\]

• inner product: modifies vector space into a space with geometry, adding angles, orthogonality, distance
• inner product of state space allows computation of transition amplitudes, which you can use to determine the likelihood the state of the system before a specific measurement will change to another state after measurement has occurred
• consider two normalised states $\ket \psi$, $\ket {\psi’}$
• let the start state be $\ket \psi$, and the end state a row vector with complex conjugate coordinates of $\ket{\psi’}$
• define the bra_ $\bra {\psi’} = \ket{\psi}^\dagger = [\overline{c_0’}, …, \overline{c_{n-1}’}]$
• the transition amplitude is then the inner product bra-ket

\[\braket{\psi'|\psi} = [\overline{c_0'}, ..., \overline{c_{n-1}'}] \begin{bmatrix} c_0 \\ \vdots \\ c_{n-1} \end{bmatrix}\]

• represent the start state, end state, and amplitude of going between these states as:

• bra-ket approach shifts focus from states to state transitions
• the transition amplitude between two states is zero when two states are orthogonal: orthogonal states are mutually exclusive alternatives
• e.g. an electron can be in arbitrary superposition of spin up and down, but after measurement in the z-direction, it will always be either up or down, not both up and down
  • if the electron was already in the up state before the z-direction measurement, it will never transition to a down state as a result of that measurement
• every complete measurement of a quantum system has an associated orthonormal basis of all possible outcomes
• with $\ket \psi$ in the basis ${\ket{b_0}, …, \ket{b_{n-1}}}$, i.e.

\[\ket \psi = \sum_{i=0}^{n-1}{b_i\ket{b_i}}\]

• each $

  b_i

  ^2$ is the probability of ending up in state $\ket{b_i}$ after a measurement has been made

Summary

• we can associate a vector space with a quantum system, with its dimension reflecting the number of basic states of the system
• states can be superposed by adding their representing vectors
• a state is left unchanged if its representing vector is multiplied by a complex scalar
• the state space has a geometry given by its inner product: this has a physical meaning, namely the likelihood of a given state to transition to another one after measurement
• orthogonal states are mutually exclusive

Observables

• physical quantities only make sense with respect to a quantifiable observation
• a physical system can be specified by a pair: (state space, observables)
  • state space: set of all states the system may occupy
  • observables: set of physical quantities able to be observed in each state of the state space
• each observable can be considered a question we can pose to the system: if the system is in a particular state $\ket \psi$ what values can we observe?
• Postulate: each physical observable has a corresponding hermitian operator
  • reminder: Hermitian means $A^\dagger = A$
  • an observable is a linear operator: it maps states to states
  • the application of an observable $\Omega$ to a state vector $\ket \psi$ is the resulting state $\Omega \ket \psi$
  • in general $\Omega \ket \psi$ is not a scalar multiple of $\ket \psi$; they do not represent the same state, i.e. $\Omega$ has modified the state of the system
• Postulate: let $\Omega$ be a hermitian operator associated with a physical observable. Then the eigenvalues of $\Omega$ are the only possible values the observable can take as a result of measuring it on any given state. The eigenvectors of $\Omega$ form a basis for the state space.
• so observables can be considered legitimate questions we can pose to quantum systems. The question may be answered with the eigenvalues of the observable

Position

• specific question: “where can the particle be found?”
• what’s the corresponding hermitian operator, $P$, for position?
  • how does it operate on basic states e.g. $\ket {x_i}$? $P(\ket \psi) = P(\ket {x_i}) = x_i \ket \psi$: $P$ acts as multiplication by position
  • the basic states form a basis, so for an arbitrary stat: $P(\sum{c_i \ket{x_i}}) = \sum{x_i c_i \ket{x_i}}$
• as a matrix: this is the diagonal matrix whose entries are the $x_i$ coordinates
• note:
  • $P$ is trivially hermitian
  • all diagonal elements are real
  • eigenvalues are $x_i$ values
  • normalised eigenvectors are the basic state vectors

Momentum

• specific question: “what is the particle’s momentum?”
• represented by operator $M$, proportional to the rate of change of the state vector across space

\[M(\ket \psi) = -i \hbar \frac{\partial\ket\psi}{\partial x}\]

Spin

• specific question: “for a given direction in space, in which direction is the particle spinning?”
  • e.g. up/down in z direction? left/right in x direction? in/out in y direction?
• spin operators:

\[S_z = \frac{\hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}, S_y = \frac{\hbar}{2} \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}, S_x = \frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}\]

• each spin operator has a corresponding orthonormal basis:
  • $S_z : {\ket \uparrow, \ket \downarrow}$, up and down
  • $S_y : {\ket \leftarrow, \ket \rightarrow}$, left and right
  • $S_x : {\ket \swarrow, \ket \nearrow}$, in and out

Manipulating Observables

• in physics we frequently add, multiply quantities to produce other meaningful quantities: momentum as mass*velocity, …
• to what extent can quantum observables be manipulated to obtain other observables?
• ✓ multiplication by a real scalar, $c \in \mathbb{R}, c\Omega$
  • Multiplying a hermitian matrix by a real scalar produces a hermitian matrix
• $\times$ multiplication by a complex scalar: the result may not be hermitian
• ✓ addition of two hermitian matrices $\Omega_1 + \Omega_2$
• set of hermitian matrices of fixed dimension forms a $\mathbb{R}$ vector space (but not a $\mathbb{C}$ one)
• products? e.g. $\Omega_1 \cdot \Omega_2$. Issues:
  • the order in which operators are applied to state vectors matters in general, as matrix multiplication is not generally commutative
  • the product of 2 hermitian operators is not guaranteed to be hermitian
• what does it take for the product of 2 hermitian operators to be hermitian?
  • recall $\braket{H\cdot V, W} = \braket{V, H\cdot W}$ for hermitian $H$. Accordingly for hermitian $\Omega_1, \Omega_2$:

\[\braket{\Omega_1\cdot\Omega_2\phi, \psi} = \braket{\Omega_2\phi, \Omega_1\psi} = \braket{\phi, \Omega_2\cdot\Omega_1\psi}\]

• for $\Omega_1\cdot\Omega_2$ to be hermitian, we need:

\[\braket{\Omega_1\cdot\Omega_2\phi, \psi} = \braket{\phi, \Omega_1\cdot\Omega_2\psi}\]

• which implies we need $\Omega_1\cdot\Omega_2 = \Omega_2\cdot\Omega_1$
• we define the commutator operator as:

\[[\Omega_1, \Omega_2] = \Omega_1\cdot\Omega_2 - \Omega_2\cdot\Omega_1\]

• if $[\Omega_1, \Omega_2] = 0$, then the product $\Omega_1\cdot\Omega_2 = \Omega_2\cdot\Omega_1$ is hermitian
• e.g. $[S_x, S_y] = 2iS_z$, i.e. the spin operators do not commute
• note that the product of a hermitian operator with itself always commutes, as does the exponent operation. Therefore for a single hermitian $\Omega$, we get the entire algebra of polynomials over $\Omega$, i.e. all operators of the following form commute with one another:

\[\Omega' = \alpha_0 + \alpha_1\Omega+\alpha_2\Omega^2 + ... +\alpha_{n-1}\Omega^{n-1}\]

• consequently if the commutator of 2 hermitian operators is 0 (i.e. the operators commute), you are able to to assign their product as the mathematical equivalent of the physical product of their associated observables
• if the commutator is non-zero, we get Heisenberg’s uncertainty principle

Expected Value

• hermitian operators are those which behave well with respect to the inner product: $\braket{\Omega\phi, \psi} = \braket{\phi, \Omega\psi}$ for each pair $\ket\psi, \ket\psi$
  • this means $\braket{\Omega\psi,\psi}\in\mathbb{R}$ for each $\psi$, denoted $\braket{\Omega}_\psi$
  • subscript denotes dependence on state vector
• Postulate: $\braket{\Omega}_\psi$ is the expected value of observing $\Omega$ repeatedly on the same state $\psi$
  • let $\lambda_1, …, \lambda_{n-1}$ be the eigenvalues of $\Omega$
  • prepare the system so that it is in state $\ket{\psi}$ and let us observe the value of $\Omega$: this will yield one of the $\lambda_i$
  • repeat this $n$ times, such that each $\lambda_i$ has been seen $p_i$ times
  • now compute the estimated expected value of $\Omega$ as $\frac{1}{n}\sum{\lambda_i p_i}_i$
  • if $n$ is sufficiently large, this will be very close to $\braket{\Omega\psi, \psi}$

Variance

• the variance will indicate the spread of distribution around expected value
• introduce the hermitian operator

\[\Delta_\psi(\Omega) = \Omega - \braket{\Omega}_\psi I\]

• this operates on a generic vector $\ket\phi$ as: \(\Delta_\psi(\Omega)\ket\phi = \Omega(\ket\phi) - (\braket{\Omega}_\psi)\ket\phi\)

• i.e. $\Delta_\psi(\Omega)$ substracts the mean from the result of $\Omega$

• variance of $\Omega$ at $\ket\psi$ is then the expectation value of $\Delta_\psi(\Omega)$ squared: \(Var_\psi(\Omega) = \braket{(\Delta_\psi(\Omega))\cdot(\Delta_\psi(\Omega))}_\psi\)

• note this is not too far from $Var(X) = E((X-\mu)^2)$
• the variance of the same hermitian varies from state to state: on an eigenvector of the operator, the variance is 0, and the expected value is the corresponding eigenvalue: the observable is sharp on its eigenvectors; there is no ambiguity of outcome

Heisenberg’s Uncertainty Principle

• consider observables represented by hermitians $\Omega_1, \Omega_2$ and a given state $\ket\psi$
• compute $Var_\psi(\Omega_1), Var_\psi(\Omega_2)$. Do they relate, and if so, how?
• i.e. given 2 observables we would hope to simultaneously minimise each variance such that the outcome was sharp for both
• if the variances were not correlated, you would expect a sharp measure of each observable on a convenient state
• however the variances are correlated
• Theorem: Heisenberg’s uncertainty principle the product of the variances of 2 arbitrary hermitian operators on a given state is always greater than or equal to one quarter of the square of the expected value of their commutator:

\[Var_\psi(\Omega_1)\cdot Var_\psi(\Omega_2)\ge \frac{1}{4}|\braket{[\Omega_1, \Omega_2]}_\psi^2\]

• so the commutator measures how good a simultaneous measure of 2 observables can possibly be
• if the commutator happens to be 0, there is no fundamental limit to the accuracy
• however there are plenty of operators that do not commute e.g. directional spin operators
• position-momentum also do not commute. The expression of $\ket\psi$ with respect to the eigenbasis of each observable paints markedly different stories
  • $\ket\psi$ can be expressed in the momentum eigenbasis, which treats $\ket\psi$ like a wave, decomposing it into sinusoids
  • $\ket\psi$ expressed in the position eigenbasis is made of Dirac deltas, peaks zero everywhere except at a point, i.e. decomposed into a weighted sum of peaks

Summary

• observables are represented by hermitian matrices
• the result of an observation is always an eigenvalue of the hermitian

• $\braket{\psi

  \Omega

  \psi}$ represents the expected value of observing $\Omega$ on $\ket\psi$

• observables do not commute (in general): this means the order of observation matters, and that there is a fundamental limit on our ability to simultaneously measure their values

Measurement

• measurement: act of carrying out an observation on a physical system
  • observable corresponds to specific question posed
  • measuring is the process of asking a specific question and receiving a definite answer
• classical physics made the false implicit assumptions that
  • the act of measuring does not change the state of the system
  • the result of a measurement on a well-defined state is predictable: if a state is known with certainty, the value of the observable on that state can be anticipated
• these assumptions are wrong:
  • systems are perturbed as a result of measurement
  • only the probability of observing specific values can be calculated: measurement is inherently nondeterministic
• so far we know that as the result of an observation, an observable can only assume one of its eigenvalues
• how frequently will we see a given eigenvalue $\lambda?$ What happens to the state vector if $\lambda$ is observed?
• Postulate: let $\Omega$ be an observable, and $\ket\psi$ be a state. If the result of measuring $\Omega$ is the eigenvalue $\lambda$, the state after measurement will always be the eigenvector $\ket{e}$corresponding to $\lambda$.
  • we say that the system has collapsed from $\ket\psi$ to $\ket{e}$
• what is the probability that a (normalised) start state $\ket\psi$ will transition to a specific eigenvector $\ket e$?

  • this is given by the square of the inner product of the states, $\braket{e

    \psi}^2$

  • this has the geometrical meaning of the projection of $\ket\psi$ along $\ket e$

Meaning of expected value

• remember the normalised eigenvectors of $\Omega$ form an orthogonal basis of the state space, so we can express $\ket\psi$ as a linear combination w.r.t. this basis: $\ket\psi = \sum{c_i\ket{e_i}}$
• compute the mean

\[\braket{\Omega}_\psi = \braket{\Omega\psi,\psi}=\sum{|c_i|^2\lambda_i}\]

• this is exactly the mean value of the probability distribution $(\lambda_0, p_0), …, (\lambda_{n-1}, p_{n-1})$
  • $p_i$: square amplitude of collapse into the corresponding eigenvector
• after measuring an observable, the system transitions to the corresponding eigenvector. If you ask the same question again, you will get the same answer
  • what if you change the question?

Order Matters

• consider making successive measurements for different observables
• each observable has a different set of eigenvectors to which the system will collapse
• the answer will depend on the order in which questions are posed
• e.g. polarising sheet and a beam of light
  • light can be polarised, where the wave only vibrates along a specific plane orthogonal to propogation (as opposed to all possible planes)
  • polarising sheet placed within the beam of light
    • measures polarisation of light in orthogonal basis corresponding to direction of the sheet
    • filters out photons that collapsed to one of the elements of the basis
  • adding a 2nd sheet
    • oriented in same direction: no difference whatsoever. Asking the same question repeatedly
    • rotated by 90°: no light passes through. The light that was not filtered by the first sheet is now guaranteed to be filtered by the second.
  • adding a 3rd sheet before/after 1st/2nd
    • no effect. No light permitted before, and none allowed through the additional sheet
  • placing a 3rd sheet in the middle, at 45°:
    • light passes through all three sheets:
      • left sheet: measures all light relative to up-down basis
      • light in vertical polarisation state that goes through is then in a superposition with respect to the basis of the diagonal sheet
      • the middle sheet then collapses half, filters some, and passes some through
      • the light passed through is again in a superposition with respect to the 3rd sheet, so some light again passes through
      • note: with 50% filtering by each sheet, only 1/8 of the original light passes through

Summary

• the end state of a measurement of an observable is always one of its eigenvectors
• the probability for an initial state to collapse into an eigenvector of the observable is given by the length squared of the projection
• when measuring several observables sequentially, the order of measurement matters

Dynamics

• so far we have considered static quantum systems, so we need quantum dynamics to examine how quantum systems evolve over time

Unitary Transformations

• Postulate: the evolution of a quantum system (that is not a measurement) is given by a unitary operator or transformation
  • if $U$ is a unitary matrix representing a unary operator, and $\ket{\psi(t)}$ represents a state of the system at time $t$, then:

\[\ket{\psi(t+1)} = U\ket{\psi(t)}\]

• properties of unitary transformations
  • closed under composition: the product of 2 arbitrary unitary matrices is unitary
  • closed under inverse: the inverse of a unitary matrix is unitary
  • multiplicative identity: the identity operator is trivially unitary
• the set of transformations constitutes a group of transformations with respect to composition

System evolution

• assume we have a rule $\mathfrak{U}$ that associates with each instant of time $t_i$ a unitary matrix $\mathfrak{U}[t_i]$
• initial state vector $\ket\psi$

• you can then apply each $\mathfrak{U}[t_i]$ to form a sequence of state vectors \(\mathfrak{U}[t_0]\ket\psi, ..., \mathfrak{U}[t_{n-1}]...\mathfrak{U}[t_0]\ket\psi\)

• this sequence is called an orbit of $\ket\psi$ under the action of $\mathfrak{U}[t_i]$ at time click $t_i$
• evolution is time symmetric: you can apply $\mathfrak{U}^{\dagger}[t_i]$ to undo the action of a given timestep
• quantum computation will work by
  • placing the computer in an initial state,
  • applying a sequence of unitary operators to the state
  • measuring the output and producing a final state
• the sequence of unitary matrices, i.e. the system dynamics, are determined via the Schrodinger equation
  • classical physics gave conservation of energy
  • Hamiltonian: $\mathcal{H}$ the observable for energy, with a hermitian matrix representing it
  • solution with initial conditions allows determination of system evolution

\[\frac{\partial\ket{\psi(t)}}{\partial t} = -i\frac{2\pi}{\hbar}\mathcal{H}\ket{\psi(t)}\]

Summary

• quantum dynamics is given by unitary transformations
• unitary transformations are invertible: all closed system dynamics are time reversible, provided that no measurements are involved
• concrete dynamics is given by the Schrodinger equation, which determines the evolution of a quantum system whenever its hamiltonian is specified

Assembling Quantum Systems

Assembly

• consider a system with 2 particles confined to the grid,
  • positions of particle 1 can be ${x_0, …, x_{n-1}}$
  • positions of particle 2 can be ${y_0, …, y_{m-1}}$
• assembling quantum systems means tensoring the state space of their constituents
• Postulate: assume we have 2 independent quantum systems $Q, Q’$ represented by respective vector spaces $\mathbb{V}, \mathbb{V}’$.
  The quantum system obtained by merging $Q$ and $Q’$ will have the tensor product $\mathbb{V}\otimes\mathbb{V’}$ as a state space
• using this postulate, we can assemble as many systems as we want: the tensor product is associative, so we can build progressively larger systems

\[\mathbb{V_0}\otimes ...\otimes\mathbb{V_k}\]

• by considering the electromagnetic field as a system composed of infinitely many particles, you can use this procedure to make field theory amenable to the quantum approach
• considering the confined particle example, there are $nm$ basic states: $\ket{x_i}\otimes\ket{y_j}$ means particle 1 is at $x_i$, particle 2 is at $y_j$
• you can then express the generic state vector as a superposition of the basic states:

\[\ket\psi = \sum_{i,j} c_{ij}\ket{x_i}\otimes\ket{y_j}\]

• this is a vector in the $nm$ dimensional complex space $\mathbb{C}^{mn}$

• $

  c_{ij}

  ^2$ gives the probability of finding the two particles at $x_i, y_j$ respectively

Entanglement

• the basic states of the assembled system are the tensor product of basic states of its constituents
• it would be nice if we could rewrite an arbitrary state vector as the tensor product of two states from respective subsystems
• this cannot be done (in general). Consider the following 2 particle system in state $\ket\psi$

\[\ket\psi = \ket{x_0}\otimes\ket{y_0} + \ket{x_1}\otimes\ket{y_1} \\ = 1\ket{x_0}\otimes\ket{y_0} 0\ket{x_0}\otimes\ket{y_1} + 0\ket{x_1}\otimes\ket{y_0} + \ket{x_1}\otimes\ket{y_1} \\\]

• attempt to write $\ket\psi$ as the tensor product of 2 states from respective subsystems

\[(c_0\ket{y_0}+c_1\ket{y_1})\otimes(d_0\ket{y_0}+d_1\ket{y_1}) = c_0d_0\ket{x_0}\otimes\ket{y_0} c_0d_1\ket{x_0}\otimes\ket{y_1} + c_1d_0\ket{x_1}\otimes\ket{y_0} + c_1d_1\ket{x_1}\otimes\ket{y_1}\]

• This would imply $c_0d_0=c_1d_1=1$ and $c_0d_1 = c_1d_0 = 0$, which has no solutions. Therefore $\ket\psi$ cannot be written as a tensor product
• what does this mean? If you measure the first particle, you have a 50% chance of finding it at position $x_0$. If it is found at $x_0$, then, as $\ket{x_0}\otimes\ket{y_1}$ has coefficient $0$, there is no chance of finding particle 2 at position $y_1$: i.e. particle 2 must be at position $y_0$!
• we say that the individual states of the particles are entangled
• this holds even if $x_i$ is light years away from $y_i$: regardless of spatial distance, a measurement’s outcome for one particle will always determine the measurement’s outcome for the other one
• other states are perfectly able to be decomposed into tensor products of subsystem states, and these are referred to as separable states

Spin

• there is a law of conservation of total spin of quantum system
• consider $z$ direction, and the corresponding spin basis, up and down
• consider a composite particle whose total spin is 0: the particle may split up at some point into 2 particles that have non-zero spin
• the spin states of those two particles will then be entangled: the sum of the spins must cancel each other out to conserve total spin
• if we measure the z-direction spin of the left particle in state up ($\ket{\uparrow_L}$), the spin of the right particle must be $\ket{\downarrow_R}$
• the bases for the left and right particles are
  • $\mathcal{B}_L = {\uparrow_L, \downarrow_L}$
  • $\mathcal{B}_R = {\uparrow_R, \downarrow_R}$
• the basis for the entire system is:

\[\{\uparrow_L\otimes\uparrow_R, \uparrow_L\otimes\downarrow_R, \downarrow_L\otimes\uparrow_R, \downarrow_L\otimes\downarrow_R\}\]

• the entangled particles can then be described by:

\[\frac{\ket{\uparrow_L\otimes\downarrow_R}+\ket{\downarrow_L\otimes\uparrow_R}}{\sqrt{2}}\]

• when you measure the left particle and it collapses to the state $\ket{\uparrow_L}$, instantaneously, the right particle collapses to the state $\ket{\downarrow_R}$ even if it is millions of light years away

Summary

• the tensor product allows us to build complex quantum systems out of simpler ones
• the new system is not able to be analysed simply in terms of states of the subsystems: an entire set of new states has been created which cannot in general be resolved into their constituents
• entanglement is used in quantum computing for:
  • algorithm design
  • cryptography
  • teleportation
  • decoherence

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