QUANTUM PHYSICS:

Whimsical Laws of Nature at the Most Fundamental Level

 

 

Sean Bae

Quantum Mechanics in 3 Minutes

  • Wave-particle Duality
  • State and Measurement
  • Applications
F = ma
F=maF = ma

Newtonian Mechanics

F = ma
F=maF = ma

Newtonian Mechanics

F = ma
F=maF = ma

Newtonian Mechanics

Wave-Particle Duality

Superposition

  • An electron can exist in two places at the same time
  • Quantum state can exist in multiple states simultaneously as long as they are not observed

Measurement

  • Superposition collapses
  • Forces the electron to take a stand and exist in only one position
  • Complete knowledge of a system is forbidden
    • Heisenberg's Uncertainty Principle
|\text{cat}\rangle = \frac{1}{\sqrt{2}}(|\text{dead}\rangle + |\text{alive}\rangle)
cat=12(dead+alive)|\text{cat}\rangle = \frac{1}{\sqrt{2}}(|\text{dead}\rangle + |\text{alive}\rangle)
  1. Trap a cat in a box
  2. A radioactive material in the box will decay with 50% probability and kill the cat in 1 hour
  3. Wait for 1 hour
  4. Is the cat is both dead and alive??

Schrödinger's Cat

Quantum Entanglement

100,000 light years

Milky Way

COLLEGE PARK, MD

Milky Way

ME

YOU

ME

YOU

ME

YOU

ME

YOU

|\downarrow\rangle
|\downarrow\rangle
|\uparrow\rangle
|\uparrow\rangle

ME

YOU

|\downarrow\rangle
|\downarrow\rangle
|\uparrow\rangle
|\uparrow\rangle

ME

YOU

Information travelled faster than the speed of light?

"Spukhafte Fernwirkung!"

("Spooky action at a distance")

Albert Einstein

"God does not play dice with the universe"

Albert Einstein

John Bell

1964

Practical use of entanglement?

Measurement

\text{Systems do not have definite properties until measured}
Systems do not have definite properties until measured\text{Systems do not have definite properties until measured}

von Neumann

Many worlds

Copenhagen

(\sum_i c_i|a_i\rangle)\otimes|\text{measurement device ready}\rangle\otimes |\text{observer ready}\rangle
(iciai)measurement device readyobserver ready(\sum_i c_i|a_i\rangle)\otimes|\text{measurement device ready}\rangle\otimes |\text{observer ready}\rangle
\text{No collapse of wave function}
No collapse of wave function\text{No collapse of wave function}
(\sum_i c_i|a_i\rangle)\otimes|\text{measurement device shows i}\rangle\otimes |\text{observer sees i}\rangle
(iciai)measurement device shows iobserver sees i(\sum_i c_i|a_i\rangle)\otimes|\text{measurement device shows i}\rangle\otimes |\text{observer sees i}\rangle
\text{but does not define what constitutes a measurement}
but does not define what constitutes a measurement\text{but does not define what constitutes a measurement}

Many Worlds Interpretation

i\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V\Psi
iΨt=22m2Ψx2+VΨi\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V\Psi

Schrödinger Equation in 1D

i\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V\Psi
iΨt=22m2Ψx2+VΨi\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V\Psi

Assumption: 

\Psi \text{ is real}
Ψ is real\Psi \text{ is real}

Complex

i\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V\Psi
iΨt=22m2Ψx2+VΨi\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V\Psi

Assumption: 

\Psi \text{ is real}
Ψ is real\Psi \text{ is real}

Complex

Real

i\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V\Psi
iΨt=22m2Ψx2+VΨi\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V\Psi

Assumption: 

\Psi \text{ is real}
Ψ is real\Psi \text{ is real}

Complex

Real

Contradiction!

Dimension of Wave Function

\text{Recall in functional space, }\langle\psi |\psi\rangle = \int_{-\infty}^{+\infty} \psi^* \psi dx = 1
Recall in functional space, ψψ=+ψψdx=1\text{Recall in functional space, }\langle\psi |\psi\rangle = \int_{-\infty}^{+\infty} \psi^* \psi dx = 1
\text{Dimension of }[\psi] = [\frac{1}{\sqrt{L}}]
Dimension of [ψ]=[1L]\text{Dimension of }[\psi] = [\frac{1}{\sqrt{L}}]

How can we mathematically manipulate qubits to do interesting computation?

Logic Gates

Logic Circuit

\text{Any Turing computable function can be converted into the form:}
Any Turing computable function can be converted into the form:\text{Any Turing computable function can be converted into the form:}
f : \{0,1\}^n \rightarrow \{0,1\}^m
f:{0,1}n{0,1}mf : \{0,1\}^n \rightarrow \{0,1\}^m

Logic Circuit

\text{Any Turing computable function can be converted into the form:}
Any Turing computable function can be converted into the form:\text{Any Turing computable function can be converted into the form:}
\text{NAND and FANOUT are universal}
NAND and FANOUT are universal\text{NAND and FANOUT are universal}

Quantum Gates

|0100\rangle
0100|0100\rangle
|\text{result}\rangle
result|\text{result}\rangle

Gate

|0\rangle\otimes |1\rangle\otimes |0\rangle = |0\rangle|1\rangle|0\rangle = |010\rangle
010=010=010|0\rangle\otimes |1\rangle\otimes |0\rangle = |0\rangle|1\rangle|0\rangle = |010\rangle

Simplified Notation

Quantum Gates

|0100\rangle
0100|0100\rangle
|\text{result}\rangle
result|\text{result}\rangle

Gate

|0\rangle\otimes |1\rangle\otimes |0\rangle = |0\rangle|1\rangle|0\rangle = |010\rangle
010=010=010|0\rangle\otimes |1\rangle\otimes |0\rangle = |0\rangle|1\rangle|0\rangle = |010\rangle

Simplified Notation

Evolution operator U act for finite time according to the Schrödinger Equation

Quantum Gates

|000\rangle + |001\rangle +
000+001+|000\rangle + |001\rangle +
|\text{result}\rangle
result|\text{result}\rangle

Gate

|0\rangle\otimes |1\rangle\otimes |0\rangle = |0\rangle|1\rangle|0\rangle = |010\rangle
010=010=010|0\rangle\otimes |1\rangle\otimes |0\rangle = |0\rangle|1\rangle|0\rangle = |010\rangle

Simplified Notation

One gate operation acts on all components

|010\rangle + |100\rangle + |110\rangle
010+100+110|010\rangle + |100\rangle + |110\rangle

Quantum Gates

|\psi\rangle
ψ|\psi\rangle
|\text{result}\rangle
result|\text{result}\rangle

Gate

|0\rangle\otimes |1\rangle\otimes |0\rangle = |0\rangle|1\rangle|0\rangle = |010\rangle
010=010=010|0\rangle\otimes |1\rangle\otimes |0\rangle = |0\rangle|1\rangle|0\rangle = |010\rangle

Simplified Notation

\text{Can act on }2^{N} \text{ basis states in 1 clock cycle}
Can act on 2N basis states in 1 clock cycle\text{Can act on }2^{N} \text{ basis states in 1 clock cycle}

Qubit

\mathcal{H} \cong \mathbb{C}^2
HC2\mathcal{H} \cong \mathbb{C}^2
|\psi\rangle = \alpha |0\rangle + \beta |1\rangle
ψ=α0+β1|\psi\rangle = \alpha |0\rangle + \beta |1\rangle
|\alpha |^2 + |\beta |^2 = 1
α2+β2=1|\alpha |^2 + |\beta |^2 = 1

State Space

Generalized Qubit

Normalization

Quantum Gates

X = \text{X gate (NOT)}
X=X gate (NOT)X = \text{X gate (NOT)}
H = \text{Hadamard gate}
H=Hadamard gateH = \text{Hadamard gate}
Z = \text{Z-gate}
Z=Z-gateZ = \text{Z-gate}
X|0\rangle = |1\rangle
X0=1X|0\rangle = |1\rangle
X|1\rangle = |0\rangle
X1=0X|1\rangle = |0\rangle
Z|0\rangle = |0\rangle
Z0=0Z|0\rangle = |0\rangle
Z|1\rangle = -|1\rangle
Z1=1Z|1\rangle = -|1\rangle
H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H0=12(0+1)H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H1=12(01)H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)

Quantum Teleportation Protocol

|\Psi\rangle = |\psi\rangle_t\otimes|\psi^{-}\rangle_{a,b}
Ψ=ψtψa,b|\Psi\rangle = |\psi\rangle_t\otimes|\psi^{-}\rangle_{a,b}
|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle)
ψ±=12(01±10)|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle)
|\phi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle)
ϕ±=12(00±11)|\phi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle)

Quantum Teleportation Protocol

|\Psi\rangle = |\psi\rangle_t\otimes|\psi^{-}\rangle_{a,b}
Ψ=ψtψa,b|\Psi\rangle = |\psi\rangle_t\otimes|\psi^{-}\rangle_{a,b}
=\frac{1}{\sqrt{2}}(\alpha |0\rangle_{t0} + \beta |1\rangle_{t1})\otimes(|01\rangle_{a,b} - |10\rangle_{a,b})
=12(α0t0+β1t1)(01a,b10a,b)=\frac{1}{\sqrt{2}}(\alpha |0\rangle_{t0} + \beta |1\rangle_{t1})\otimes(|01\rangle_{a,b} - |10\rangle_{a,b})
|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle)
ψ±=12(01±10)|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle)
|\phi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle)
ϕ±=12(00±11)|\phi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle)
\text{Expand the terms}
Expand the terms\text{Expand the terms}

Quantum Teleportation Protocol

|\Psi\rangle = |\psi\rangle_t\otimes|\psi^{-}\rangle_{a,b}
Ψ=ψtψa,b|\Psi\rangle = |\psi\rangle_t\otimes|\psi^{-}\rangle_{a,b}
=\frac{1}{\sqrt{2}}(\alpha |0\rangle_{t0} + \beta |1\rangle_{t1})\otimes(|01\rangle_{a,b} - |10\rangle_{a,b})
=12(α0t0+β1t1)(01a,b10a,b)=\frac{1}{\sqrt{2}}(\alpha |0\rangle_{t0} + \beta |1\rangle_{t1})\otimes(|01\rangle_{a,b} - |10\rangle_{a,b})
=\frac{1}{2}[(\alpha |1\rangle_{b1} - \beta |0\rangle_{b0} ) |\phi^+\rangle_{t,a} + (\alpha |1\rangle_{b1} + \beta |0\rangle_{b0} )|\phi^-\rangle_{t,a}
=12[(α1b1β0b0)ϕ+t,a+(α1b1+β0b0)ϕt,a=\frac{1}{2}[(\alpha |1\rangle_{b1} - \beta |0\rangle_{b0} ) |\phi^+\rangle_{t,a} + (\alpha |1\rangle_{b1} + \beta |0\rangle_{b0} )|\phi^-\rangle_{t,a}
+ (\beta |1\rangle_{b1} - \alpha |0\rangle_{b0} )|\psi^+\rangle_{t,a} - (\alpha |0\rangle_{b0} + \beta |1\rangle_{b1} )|\psi^-\rangle_{t,a}]
+(β1b1α0b0)ψ+t,a(α0b0+β1b1)ψt,a]+ (\beta |1\rangle_{b1} - \alpha |0\rangle_{b0} )|\psi^+\rangle_{t,a} - (\alpha |0\rangle_{b0} + \beta |1\rangle_{b1} )|\psi^-\rangle_{t,a}]
|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle)
ψ±=12(01±10)|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle)
|\phi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle)
ϕ±=12(00±11)|\phi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle)
\text{Alice makes a measurement in Bell basis}
Alice makes a measurement in Bell basis\text{Alice makes a measurement in Bell basis}

Quantum Teleportation Protocol

|\Psi\rangle = |\psi\rangle_t\otimes|\psi^{-}\rangle_{a,b}
Ψ=ψtψa,b|\Psi\rangle = |\psi\rangle_t\otimes|\psi^{-}\rangle_{a,b}
=\frac{1}{\sqrt{2}}(\alpha |0\rangle_{t0} + \beta |1\rangle_{t1})\otimes(|01\rangle_{a,b} - |10\rangle_{a,b})
=12(α0t0+β1t1)(01a,b10a,b)=\frac{1}{\sqrt{2}}(\alpha |0\rangle_{t0} + \beta |1\rangle_{t1})\otimes(|01\rangle_{a,b} - |10\rangle_{a,b})
=\frac{1}{2}[(\alpha |1\rangle_{b1} - \beta |0\rangle_{b0} ) |\phi^+\rangle_{t,a} + (\alpha |1\rangle_{b1} + \beta |0\rangle_{b0} )|\phi^-\rangle_{t,a}
=12[(α1b1β0b0)ϕ+t,a+(α1b1+β0b0)ϕt,a=\frac{1}{2}[(\alpha |1\rangle_{b1} - \beta |0\rangle_{b0} ) |\phi^+\rangle_{t,a} + (\alpha |1\rangle_{b1} + \beta |0\rangle_{b0} )|\phi^-\rangle_{t,a}
+ (\beta |1\rangle_{b1} - \alpha |0\rangle_{b0} )|\psi^+\rangle_{t,a} - (\alpha |0\rangle_{b0} + \beta |1\rangle_{b1} )|\psi^-\rangle_{t,a}]
+(β1b1α0b0)ψ+t,a(α0b0+β1b1)ψt,a]+ (\beta |1\rangle_{b1} - \alpha |0\rangle_{b0} )|\psi^+\rangle_{t,a} - (\alpha |0\rangle_{b0} + \beta |1\rangle_{b1} )|\psi^-\rangle_{t,a}]
|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle)
ψ±=12(01±10)|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle)
|\phi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle)
ϕ±=12(00±11)|\phi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle)
\text{Bob applies the correct gates to retrieve Alice's qubit}
Bob applies the correct gates to retrieve Alice's qubit\text{Bob applies the correct gates to retrieve Alice's qubit}

Deutsch-Josza Algorithm

\text{Oracle } f : \{0,1\}^{n} \rightarrow \{0,1\}
Oracle f:{0,1}n{0,1}\text{Oracle } f : \{0,1\}^{n} \rightarrow \{0,1\}
x f0 f1 f2 f3
0 0 0 1 1
1 0 1 0 1
\text{Is } f \text{ constant or balanced?}
Is f constant or balanced?\text{Is } f \text{ constant or balanced?}
n = 1
n=1n = 1

Deutsch-Josza Algorithm

Classical: 2 queries

Quantum: 1 query

U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
Ufxy=xyf(x)U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle

Deutsch-Josza Algorithm

U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
Ufxy=xyf(x)U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
0112(0+1)12(01)|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H0=12(0+1)H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H1=12(01)H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
\text{Apply the Hadamard gates}
Apply the Hadamard gates\text{Apply the Hadamard gates}

Deutsch-Josza Algorithm

U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
Ufxy=xyf(x)U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
0112(0+1)12(01)|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H0=12(0+1)H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H1=12(01)H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
12(00f(0)+10f(1)+01f(0)+11f(1))\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
\text{Apply the Oracle gate}
Apply the Oracle gate\text{Apply the Oracle gate}

Deutsch-Josza Algorithm

U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
Ufxy=xyf(x)U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
0112(0+1)12(01)|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H0=12(0+1)H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H1=12(01)H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
12(00f(0)+10f(1)+01f(0)+11f(1))\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
\text{Reorganize the terms}
Reorganize the terms\text{Reorganize the terms}
\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)(-1)^{f(0)}\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
12(0+(1)f(0)f(1)1)(1)f(0)12(01)\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)(-1)^{f(0)}\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)

Deutsch-Josza Algorithm

U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
Ufxy=xyf(x)U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
0112(0+1)12(01)|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H0=12(0+1)H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H1=12(01)H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
12(00f(0)+10f(1)+01f(0)+11f(1))\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
\text{Discard the second qubit}
Discard the second qubit\text{Discard the second qubit}
\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)(-1)^{f(0)}\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
12(0+(1)f(0)f(1)1)(1)f(0)12(01)\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)(-1)^{f(0)}\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)
12(0+(1)f(0)f(1)1)\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)

Deutsch-Josza Algorithm

U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
Ufxy=xyf(x)U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
0112(0+1)12(01)|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H0=12(0+1)H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H1=12(01)H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
12(00f(0)+10f(1)+01f(0)+11f(1))\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)(-1)^{f(0)}\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
12(0+(1)f(0)f(1)1)(1)f(0)12(01)\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)(-1)^{f(0)}\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)
12(0+(1)f(0)f(1)1)\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)
\rightarrow \frac{1}{2}(|0\rangle + |1\rangle + (-1)^{f(0)\oplus f(1)}|0\rangle - (-1)^{f(0)\oplus f(1)}|1\rangle)
12(0+1+(1)f(0)f(1)0(1)f(0)f(1)1)\rightarrow \frac{1}{2}(|0\rangle + |1\rangle + (-1)^{f(0)\oplus f(1)}|0\rangle - (-1)^{f(0)\oplus f(1)}|1\rangle)
\text{Apply the Hadamard gates}
Apply the Hadamard gates\text{Apply the Hadamard gates}

Deutsch-Josza Algorithm

U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
Ufxy=xyf(x)U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
0112(0+1)12(01)|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H0=12(0+1)H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H1=12(01)H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
12(00f(0)+10f(1)+01f(0)+11f(1))\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
\rightarrow \frac{1}{2}(|0\rangle + |1\rangle + (-1)^{f(0)\oplus f(1)}|0\rangle - (-1)^{f(0)\oplus f(1)}|1\rangle)
12(0+1+(1)f(0)f(1)0(1)f(0)f(1)1)\rightarrow \frac{1}{2}(|0\rangle + |1\rangle + (-1)^{f(0)\oplus f(1)}|0\rangle - (-1)^{f(0)\oplus f(1)}|1\rangle)
\rightarrow \frac{1}{2}[(1 + (-1)^{f(0)\oplus f(1)})|0\rangle + (1 - (-1)^{f(0)\oplus f(1)})|1\rangle]
12[(1+(1)f(0)f(1))0+(1(1)f(0)f(1))1]\rightarrow \frac{1}{2}[(1 + (-1)^{f(0)\oplus f(1)})|0\rangle + (1 - (-1)^{f(0)\oplus f(1)})|1\rangle]
\text{Reorganize the terms}
Reorganize the terms\text{Reorganize the terms}
\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)(-1)^{f(0)}\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
12(0+(1)f(0)f(1)1)(1)f(0)12(01)\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)(-1)^{f(0)}\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)
12(0+(1)f(0)f(1)1)\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)

Deutsch-Josza Algorithm

U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
Ufxy=xyf(x)U_f |x\rangle |y\rangle = |x\rangle |y\oplus f(x)\rangle
|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
0112(0+1)12(01)|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H0=12(0+1)H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)
H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
H1=12(01)H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
12(00f(0)+10f(1)+01f(0)+11f(1))\rightarrow \frac{1}{2}(|0\rangle |0\oplus f(0)\rangle + |1\rangle |0\oplus f(1)\rangle + |0\rangle |1\oplus f(0)\rangle + |1\rangle |1\oplus f(1)\rangle)
\rightarrow \frac{1}{2}(|0\rangle + |1\rangle + (-1)^{f(0)\oplus f(1)}|0\rangle - (-1)^{f(0)\oplus f(1)}|1\rangle)
12(0+1+(1)f(0)f(1)0(1)f(0)f(1)1)\rightarrow \frac{1}{2}(|0\rangle + |1\rangle + (-1)^{f(0)\oplus f(1)}|0\rangle - (-1)^{f(0)\oplus f(1)}|1\rangle)
\rightarrow \frac{1}{2}[(1 + (-1)^{f(0)\oplus f(1)})|0\rangle + (1 - (-1)^{f(0)\oplus f(1)})|1\rangle]
12[(1+(1)f(0)f(1))0+(1(1)f(0)f(1))1]\rightarrow \frac{1}{2}[(1 + (-1)^{f(0)\oplus f(1)})|0\rangle + (1 - (-1)^{f(0)\oplus f(1)})|1\rangle]
\text{if measured } |0\rangle \text{, then } f \text{ is constant, and balanced otherwise}
if measured 0, then f is constant, and balanced otherwise\text{if measured } |0\rangle \text{, then } f \text{ is constant, and balanced otherwise}
\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)(-1)^{f(0)}\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
12(0+(1)f(0)f(1)1)(1)f(0)12(01)\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)(-1)^{f(0)}\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)
12(0+(1)f(0)f(1)1)\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + (-1)^{f(0)\oplus f(1)}|1\rangle)

How do you find a needle

in a haystack?

How do you find a needle

in a haystack?

Unstructured Search

Classical

T(N) = \mathcal{O}(N)
T(N)=O(N)T(N) = \mathcal{O}(N)

Quantum

T(N) = \mathcal{O}(\sqrt{N})
T(N)=O(N)T(N) = \mathcal{O}(\sqrt{N})

Unstructured Search

Classical

T(N) = \mathcal{O}(N)
T(N)=O(N)T(N) = \mathcal{O}(N)

Quantum

T(N) = \mathcal{O}(\sqrt{N})
T(N)=O(N)T(N) = \mathcal{O}(\sqrt{N})

Our discussion of qubit has been abstract

Our discussion of qubit has been abstract

 

How can we realize a physical qubit?

p

e-

|0\rangle
0|0\rangle

Hydrogen

p

e-

|1\rangle
1|1\rangle

Hydrogen

p

e-

e-

|\psi\rangle
ψ|\psi\rangle

Hydrogen

Particle in a Box

m
mm
0
00
l
ll
|\psi (t)\rangle = \text{ }?
ψ(t)= ?|\psi (t)\rangle = \text{ }?

Schrödinger Equation

i\hbar\frac{d}{dt}|\psi\rangle = \hat{H} |\psi\rangle
iddtψ=H^ψi\hbar\frac{d}{dt}|\psi\rangle = \hat{H} |\psi\rangle

Schrödinger Equation

i\hbar\frac{d}{dt}|\psi\rangle = \hat{H} |\psi\rangle
iddtψ=H^ψi\hbar\frac{d}{dt}|\psi\rangle = \hat{H} |\psi\rangle
|\psi (t)\rangle = e^{-i\frac{\hat{H}}{\hbar}t} |\psi (0)\rangle
ψ(t)=eiH^tψ(0)|\psi (t)\rangle = e^{-i\frac{\hat{H}}{\hbar}t} |\psi (0)\rangle
\text{General solution for differential equation}
General solution for differential equation\text{General solution for differential equation}

General Solution

|\psi (t)\rangle = e^{-i\frac{\hat{H}}{\hbar}t} |\psi (0)\rangle
ψ(t)=eiH^tψ(0)|\psi (t)\rangle = e^{-i\frac{\hat{H}}{\hbar}t} |\psi (0)\rangle

General Solution

|\psi (t)\rangle = e^{-i\frac{\hat{H}}{\hbar}t} |\psi (0)\rangle
ψ(t)=eiH^tψ(0)|\psi (t)\rangle = e^{-i\frac{\hat{H}}{\hbar}t} |\psi (0)\rangle
|\psi (0)\rangle = \sum_n c_n|\psi_n \rangle
ψ(0)=ncnψn|\psi (0)\rangle = \sum_n c_n|\psi_n \rangle
\text{WLOG, initial state is a superposition of arbitrary basis states}
WLOG, initial state is a superposition of arbitrary basis states\text{WLOG, initial state is a superposition of arbitrary basis states}

General Solution

|\psi (t)\rangle = e^{-i\frac{\hat{H}}{\hbar}t} |\psi (0)\rangle
ψ(t)=eiH^tψ(0)|\psi (t)\rangle = e^{-i\frac{\hat{H}}{\hbar}t} |\psi (0)\rangle
|\psi (0)\rangle = \sum_n c_n|\psi_n \rangle
ψ(0)=ncnψn|\psi (0)\rangle = \sum_n c_n|\psi_n \rangle
|\psi (t)\rangle = \sum_n c_n e^{-iE_n t/\hbar}|\psi_n \rangle
ψ(t)=ncneiEnt/ψn|\psi (t)\rangle = \sum_n c_n e^{-iE_n t/\hbar}|\psi_n \rangle
\text{Plug the second equation into the first equation}
Plug the second equation into the first equation\text{Plug the second equation into the first equation}

Particle in a Box

|\psi_n\rangle = \sqrt{\frac{2}{l}}\sin{(\frac{n\pi}{l}x)}
ψn=2lsin(nπlx)|\psi_n\rangle = \sqrt{\frac{2}{l}}\sin{(\frac{n\pi}{l}x)}
E_n = \frac{n^2\pi^2 \hbar^2}{2ml^2}
En=n2π222ml2E_n = \frac{n^2\pi^2 \hbar^2}{2ml^2}

Particle in a Box

|0\rangle = \sqrt{\frac{2}{l}}\sin{(\frac{\pi}{l}x)}
0=2lsin(πlx)|0\rangle = \sqrt{\frac{2}{l}}\sin{(\frac{\pi}{l}x)}
E_1 = \frac{\pi^2 \hbar^2}{2ml^2}
E1=π222ml2E_1 = \frac{\pi^2 \hbar^2}{2ml^2}
|1\rangle = \sqrt{\frac{2}{l}}\sin{(\frac{2\pi}{l}x)}
1=2lsin(2πlx)|1\rangle = \sqrt{\frac{2}{l}}\sin{(\frac{2\pi}{l}x)}
E_2 = \frac{2\pi^2 \hbar^2}{ml^2}
E2=2π22ml2E_2 = \frac{2\pi^2 \hbar^2}{ml^2}

Particle in a Box

|\psi(0)\rangle = \alpha |0\rangle + \beta |1\rangle
ψ(0)=α0+β1|\psi(0)\rangle = \alpha |0\rangle + \beta |1\rangle
|\psi (t)\rangle = \alpha |0\rangle e^{-\frac{iE_1 t}{\hbar}} + \beta |1\rangle e^{-\frac{iE_2 t}{\hbar}}
ψ(t)=α0eiE1t+β1eiE2t|\psi (t)\rangle = \alpha |0\rangle e^{-\frac{iE_1 t}{\hbar}} + \beta |1\rangle e^{-\frac{iE_2 t}{\hbar}}
|\psi (t)\rangle = e^{-\frac{iE_1 t}{\hbar}} (\alpha |0\rangle + \beta |1\rangle e^{-\frac{i(\Delta E)t}{\hbar}})
ψ(t)=eiE1t(α0+β1ei(ΔE)t)|\psi (t)\rangle = e^{-\frac{iE_1 t}{\hbar}} (\alpha |0\rangle + \beta |1\rangle e^{-\frac{i(\Delta E)t}{\hbar}})
\Delta E \approx 10 \text{ eV}
ΔE10 eV\Delta E \approx 10 \text{ eV}
\nu = \frac{\Delta E}{h} = 2.5 \times 10^{15} \text{ Hz}
ν=ΔEh=2.5×1015 Hz\nu = \frac{\Delta E}{h} = 2.5 \times 10^{15} \text{ Hz}
\therefore \text{Atomic qubits are controlled optically via interactions with light pulses}
Atomic qubits are controlled optically via interactions with light pulses\therefore \text{Atomic qubits are controlled optically via interactions with light pulses}

State-of-the-art

Trapped Atomic Ion

  • very long (>>1s) memory
  • <20 coherent qubits
  • engineering needed
  • connection reconfigurable

Superconducting Loop

  • short (10^-6s) memory
  • <10 coherent qubits
  • printable circuits and VLSI
  • not reconfigurable

Q & A

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