Ch.10: Quantum mechanics

Chapter 10 of the book "No Bullshit Guide to Linear Algebra" by Ivan Savov Introduction Polarizing lenses experiment Dirac notation for vectors Quantum information processing Postulates of quantum mechanics Polarizing lenses experiment revisited Quantum physics is not that weird Quantum mechanics applications

167 flashcards · Mathematics · by lakesare

Part of Textbooks › No Bullshit Guide to Linear Algebra (Ivan Savov) · 10 courses

[Introduction] How is the state of a particle moving in one dimension described in a classical model, and in a quantum model?
Classical: position and velocity (x(t),v(t))(x(t), v(t))(x(t),v(t)) . Quantum: a wave function ∣ψ(x,t)⟩|\psi(x, t)\rangle∣ψ(x,t)⟩ , a complex-valued function of position xxx and time ttt . At a fixed time ttt , the quantum state is a function ∣ψ(x)⟩∈{R→C}|\psi(x)\rangle \in \{\mathbb{R} \to \mathbb{C}\}∣ψ(x)⟩∈{R→C} . Position and momentum are not definite values: both are computed from ∣ψ(x)⟩|\psi(x)\rangle∣ψ(x)⟩ .
[Introduction] Why is the wave function ∣ψ(x)⟩|\psi(x)\rangle∣ψ(x)⟩ of a particle also called the probability-amplitude function?
The probability of finding the particle at position xax_axa is proportional to ∣ψ(xa)∣2|\psi(x_a)|^2∣ψ(xa)∣2 . Pr⁡({particle position=xa})∝∣ψ(xa)∣2\Pr(\{\text{particle position} = x_a\}) \propto |\psi(x_a)|^2Pr({particle position=xa})∝∣ψ(xa)∣2
[Introduction] What is a bit, and what is a qubit?
Bit: x∈Z2={0,1}x \in \mathbb{Z}_2 = \{0, 1\}x∈Z2={0,1} . Qubit: a unit-length vector ∣x⟩=α∣0⟩+β∣1⟩|x\rangle = \alpha|0\rangle + \beta|1\rangle∣x⟩=α∣0⟩+β∣1⟩ in C2\mathbb{C}^2C2 . Unit length means ∣α∣2+∣β∣2=1|\alpha|^2 + |\beta|^2 = 1∣α∣2+∣β∣2=1 .Example: ∣x⟩=12∣0⟩+i2∣1⟩|x\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{i}{\sqrt{2}}|1\rangle∣x⟩=21∣0⟩+2i∣1⟩ is a qubit: 12+12=1\frac{1}{2} + \frac{1}{2} = 121+21=1 .
[Introduction] Which two questions about quantum mechanics interest philosophers?
How deterministic laws fit with probabilistic outcomes, and whether the state ∣ψ⟩|\psi\rangle∣ψ⟩ really exists or only represents our knowledge. Most scientists leave them aside: different interpretations cannot be tested by experiment, so they are outside the scope of physics research.
[Introduction] What is Savov's verdict on "quantum psychology" books, which apply quantum principles to the human mind?
Speculation, by people who know little quantum mechanics. His rule: quantum vocabulary used outside atomic-scale phenomena is a reason to stop listening.
[Introduction] What is the difference between the real world and the mathematical world of a physical model?
Experiments are performed in the real world; the mathematical world is a theoretical construct that models some aspects of the real world. A model is good if its predictions correspond closely to the behaviour of real systems.
[Introduction] When is one physics model more "true" than another?
Only when it predicts the outcomes of experiments better. No model is "true" or "real": a model consists of formulas, not atoms and lasers. Any mathematical model that predicts experiments correctly will be used.
[Introduction] What are the two kinds of mathematical models of the real world?
Classical models Describe the world with positions, velocities and similar concepts; states evolve by deterministic laws. Example: a ball thrown in the air, described by its trajectory x(t)x(t)x(t) . Quantum models Describe systems by vectors in complex vector spaces; states evolve by deterministic laws, but are observed by measurements with nondeterministic outcomes. Example: an electron described by a wave function ∣ψ(r⃗,t)⟩|\psi(\vec{r}, t)\rangle∣ψ(r,t)⟩ .
[Introduction] For each classical model, what is the corresponding quantum model? x(t)∈{R→R}x(t) \in \{\mathbb{R} \to \mathbb{R}\}x(t)∈{R→R} , r⃗(t)∈{R→R3}\ \vec{r}(t) \in \{\mathbb{R} \to \mathbb{R}^3\} r(t)∈{R→R3} , i∈Z2\ i \in \mathbb{Z}_2 i∈Z2 , j∈Zd\ j \in \mathbb{Z}_d j∈Zd
∣ψ(x,t)⟩∈{R×R→C}\boldsymbol{|\psi(x, t)\rangle \in \{\mathbb{R} \times \mathbb{R} \to \mathbb{C}\}}∣ψ(x,t)⟩∈{R×R→C} , ∣ψ(r⃗,t)⟩∈{R3×R→C}\ \boldsymbol{|\psi(\vec{r}, t)\rangle \in \{\mathbb{R}^3 \times \mathbb{R} \to \mathbb{C}\}} ∣ψ(r,t)⟩∈{R3×R→C} , ∣x⟩∈C2\ \boldsymbol{|x\rangle \in \mathbb{C}^2} ∣x⟩∈C2 , ∣y⟩∈Cd\ \boldsymbol{|y\rangle \in \mathbb{C}^d} ∣y⟩∈Cd Position in 1D, position in 3D, a bit (qubit), and a ddd -level system.
[Introduction] A photon goes through an optical circuit of lenses and mirrors, and a detector at position xfx_fxf at the end either clicks (photon arrives) or not.What does a classical model predict, and what does a quantum model predict?
Classical: click if xf=x(t)x_f = x(t)xf=x(t) for some time ttt , else no click. Quantum: with ∣ψ⟩=α∣0⟩+β∣1⟩|\psi\rangle = \alpha|0\rangle + \beta|1\rangle∣ψ⟩=α∣0⟩+β∣1⟩ , Pr⁡(click)=∣β∣2\ \Pr(\text{click}) = |\beta|^2 Pr(click)=∣β∣2 , Pr⁡(noclick)=∣α∣2\ \Pr(\text{noclick}) = |\alpha|^2 Pr(noclick)=∣α∣2 . ∣1⟩|1\rangle∣1⟩ describes the paths that arrive at the detector, ∣0⟩|0\rangle∣0⟩ the paths that don't, and ∣α∣2+∣β∣2=1|\alpha|^2 + |\beta|^2 = 1∣α∣2+∣β∣2=1 .Example: α=12, β=i2\alpha = \frac{1}…
[Introduction] A photon goes through an optical circuit to a detector, described classically by a trajectory x(t)x(t)x(t) and quantumly by ∣ψ⟩=α∣0⟩+β∣1⟩|\psi\rangle = \alpha|0\rangle + \beta|1\rangle∣ψ⟩=α∣0⟩+β∣1⟩ ( ∣1⟩|1\rangle∣1⟩ : paths that reach the detector, ∣0⟩|0\rangle∣0⟩ : paths that don't).How do the two models' assumptions about the photon differ, and how do we decide between them?
Classical: the photon follows a single path. Quantum: the photon takes several paths at once. We compare their predictions with experiment. Neither model is more "real": only the photon is real.
[Introduction] Why are quantum models less intuitive than classical models?
Quantum states are not accessible to any of our senses, and they are complex vectors, which are hard to visualize. Variables in classical models (position, velocity) correspond to things we can measure directly, e.g. with measuring tape.
[Introduction] What is a quantum measurement?
The process that maps a quantum state ∣ψ⟩|\psi\rangle∣ψ⟩ to a classical outcome. A measurement asks the particle a question; the outcome is the answer.Example: position(∣ψ⟩)=x∈R\text{position}(|\psi\rangle) = x \in \mathbb{R}position(∣ψ⟩)=x∈R momentum(∣ψ⟩)=p∈R\text{momentum}(|\psi\rangle) = p \in \mathbb{R}momentum(∣ψ⟩)=p∈R spin↑↓(∣ψ⟩)=s∈{↑,↓}\text{spin}_{\uparrow\downarrow}(|\psi\rangle) = s \in \{\uparrow, \downarrow\}spin↑↓(∣ψ⟩)=s∈{↑,↓}
[Polarizing lenses experiment] What is the polarization of light?
The orientation of the electric field E⃗\vec{E}E of the propagating electromagnetic wave. A photon travelling in the xxx -direction with electric component E⃗\vec{E}E (blue) along the zzz -axis and magnetic component B⃗\vec{B}B (red).
[Polarizing lenses experiment] When is light unpolarized, and when is it polarized?
Unpolarized: a mixture of photons with random orientations of the electric field. Polarized: all photons have the same orientation of the electric field. Light is normally unpolarized; light reflected from flat surfaces (a lake, a window) becomes polarized.
[Polarizing lenses experiment] Which photons pass through a polarizing filter, and which are reflected?
Polarization perpendicular to the filter's conducting bands: passes. Parallel to the bands: reflected. The coating conducts electricity in one direction only. A vertically polarizing filter VVV has horizontal bands. Only the vertical component of the electric field comes out of VVV .
[Polarizing lenses experiment] What comes out of each setup of polarizing lenses ( VVV vertical, HHH horizontal)?
unpolarized light →V\to V→V Vertically polarized light. unpolarized light →V→V\to V \to V→V→V Vertically polarized light: all light that passes the first VVV also passes the second. unpolarized light →V→H\to V \to H→V→H No light ( ∅\emptyset∅ ): vertically polarized light does not pass a horizontally polarizing lens. An HHH lens is a VVV lens rotated by 90∘90^\circ90∘ .
[Polarizing lenses experiment] Why don't polarized sunglasses (vertically polarizing lenses) show the sun's reflection on a lake?
The reflection from the water is horizontally polarized, and vertically polarizing lenses filter out horizontally polarized light.
[Polarizing lenses experiment] Which two assumptions of the classical physics paradigm (19th century) matter for the polarizing lenses experiment?
Determinism Outcomes depend on definite properties of the particles; an unpredictable outcome only means some property was unknown. Example: whether a photon passes a lens is decided by its polarization. Immutable properties Experiments are passive observations that do not change the particles' properties. Example: measuring a photon at one lens does not change its polarization.
[Polarizing lenses experiment] In the classical view, each photon carries a polarization tag "H" or "V". What does a VVV -polarizing lens do, written as an SQL query?
SELECT photon FROM photons WHERE tag="V"; Whether a photon passes is predetermined and depends only on its tag. An HHH lens is the same query with tag="H".
[Polarizing lenses experiment] In the polarizing lenses experiment, what does the optical power PPP at a point measure, and where is P=1P = 1P=1 ?
The brightness of the light at that point (the spot on a sheet of paper placed there); P=1P = 1P=1 right after the first lens. P=0.5P = 0.5P=0.5 is half as bright, P=0P = 0P=0 means no photons arrive.
[Polarizing lenses experiment] Light goes through an HHH -polarizing lens and then a VVV -polarizing lens. What is the optical power after each lens ( P=1P = 1P=1 after the first)?
P=1P = 1P=1 after HHH , P=0P = 0P=0 after VVV . The photons leaving HHH are horizontally polarized, and a VVV lens rejects them. Not surprising.
[Polarizing lenses experiment] A diagonally polarizing lens DDD (rotated 45∘45^\circ45∘ ) is inserted between the HHH and the VVV lens. What is the optical power after each lens ( P=1P = 1P=1 after HHH )?
P=1P = 1P=1 after HHH , P=0.5P = 0.5P=0.5 after DDD , P=0.25P = 0.25P=0.25 after VVV . Any middle orientation different from HHH and VVV lets some light through; the exact angle only changes the amount.
[Polarizing lenses experiment] Light through lenses H→VH \to VH→V : no light comes out. Light through H→D→VH \to D \to VH→D→V ( DDD diagonal): a quarter of the light after HHH comes out.Why is this result hard to explain classically?
Adding a filter increases the light, and photons that passed HHH (so are not vertical) later pass VVV . With fixed tags "H" and "V", a third filter could only remove more photons.
[Polarizing lenses experiment] Light through lenses H→VH \to VH→V : no light comes out. Light through H→D→VH \to D \to VH→D→V ( DDD diagonal): a quarter of the light after HHH comes out.What does the experiment show about measurement?
Measurement changes the state of the photon: measuring with the DDD lens changes its "H" and "V" tags. This contradicts the classical assumption that measurements are passive observations. A vector state for the polarization predicts P=0.25P = 0.25P=0.25 correctly.
[Dirac notation for vectors] A complex vector space such as Cd\mathbb{C}^dCd is called a Hilbert space.
Named after David Hilbert.
[Dirac notation for vectors] In Dirac notation, what is the standard basis of Cd\mathbb{C}^dCd ?
{∣0⟩,∣1⟩,…,∣d−1⟩}\boldsymbol{\{|0\rangle, |1\rangle, \dots, |d-1\rangle\}}{∣0⟩,∣1⟩,…,∣d−1⟩} , where ∣i⟩|i\rangle∣i⟩ has a 1 in entry iii (counting from 0) and 0 elsewhere. The zero-based indexing makes links between quantum theory and computer science more apparent.Example: in C3\mathbb{C}^3C3 , ∣2⟩=(0,0,1)T|2\rangle = (0, 0, 1)^{\mathsf{T}}∣2⟩=(0,0,1)T .
[Dirac notation for vectors] What advantage of Dirac notation shows in ∣a,b,c⟩|a, b, c\rangle∣a,b,c⟩ ?
No subscripts are needed: ∣a,b,c⟩|a, b, c\rangle∣a,b,c⟩ instead of v⃗a,b,c\vec{v}_{a,b,c}va,b,c for a vector with properties a,b,ca, b, ca,b,c .
[Dirac notation for vectors] What is a ket ∣v⟩|v\rangle∣v⟩ in C2\mathbb{C}^2C2 , as a column vector?
∣v⟩=α∣0⟩+β∣1⟩ ⇔ [αβ]\boldsymbol{|v\rangle = \alpha|0\rangle + \beta|1\rangle \ \Leftrightarrow \ \begin{bmatrix} \alpha \\ \beta \end{bmatrix}}∣v⟩=α∣0⟩+β∣1⟩ ⇔ [αβ] , with α,β∈C\alpha, \beta \in \mathbb{C}α,β∈C ∣0⟩=(1,0)T|0\rangle = (1, 0)^{\mathsf{T}}∣0⟩=(1,0)T and ∣1⟩=(0,1)T|1\rangle = (0, 1)^{\mathsf{T}}∣1⟩=(0,1)T form the standard basis.
[Dirac notation for vectors] What is the bra ⟨v∣\langle v|⟨v∣ of the ket ∣v⟩=α∣0⟩+β∣1⟩|v\rangle = \alpha|0\rangle + \beta|1\rangle∣v⟩=α∣0⟩+β∣1⟩ ?
Its Hermitian transpose: ⟨v∣=α‾⟨0∣+β‾⟨1∣ ⇔ [α‾,β‾]\boldsymbol{\langle v| = \overline{\alpha}\langle 0| + \overline{\beta}\langle 1| \ \Leftrightarrow \ [\overline{\alpha}, \overline{\beta}]}⟨v∣=α⟨0∣+β⟨1∣ ⇔ [α,β] Hermitian transpose = transpose and complex conjugation of every entry: ⟨v∣⇔v⃗†=v⃗‾T\langle v| \Leftrightarrow \vec{v}^\dagger = \overline{\vec{v}}^{\mathsf{T}}⟨v∣⇔v†=vT .Example: ∣v⟩=∣0⟩+i∣1⟩|v\rangle = |0\rangle + i|1\rangle∣v⟩=∣0⟩+i∣1⟩ gives ⟨v∣=⟨0∣−i⟨1∣⇔[1,−i]\langle v| = \langle…
[Dirac notation for vectors] For ∣v⟩=v0∣0⟩+v1∣1⟩|v\rangle = v_0|0\rangle + v_1|1\rangle∣v⟩=v0∣0⟩+v1∣1⟩ and ∣w⟩=w0∣0⟩+w1∣1⟩|w\rangle = w_0|0\rangle + w_1|1\rangle∣w⟩=w0∣0⟩+w1∣1⟩ , what is the braket ⟨v∣w⟩\langle v|w\rangle⟨v∣w⟩ ?
The inner product: ⟨v∣w⟩=v0‾w0+v1‾w1=v⃗†w⃗\boldsymbol{\langle v|w\rangle = \overline{v_0}w_0 + \overline{v_1}w_1 = \vec{v}^\dagger\vec{w}}⟨v∣w⟩=v0w0+v1w1=v†w The bra ⟨v∣\langle v|⟨v∣ already contains the complex conjugation, so the inner product needs no dagger symbol: put the bra next to the ket.Example: ∣v⟩=∣0⟩+i∣1⟩|v\rangle = |0\rangle + i|1\rangle∣v⟩=∣0⟩+i∣1⟩ , ∣w⟩=i∣0⟩+∣1⟩|w\rangle = i|0\rangle + |1\rangle∣w⟩=i∣0⟩+∣1⟩ : ⟨v∣w⟩=1⋅i+(−i)⋅1=0\langle v|w\rangle = 1 \cdot i + (-i) \cdot 1…
[Dirac notation for vectors] For ∣v⟩=α∣0⟩+β∣1⟩|v\rangle = \alpha|0\rangle + \beta|1\rangle∣v⟩=α∣0⟩+β∣1⟩ , what is ⟨v∣v⟩\langle v|v\rangle⟨v∣v⟩ , and why?
⟨v∣v⟩=∣α∣2+∣β∣2\boldsymbol{\langle v|v\rangle = |\alpha|^2 + |\beta|^2}⟨v∣v⟩=∣α∣2+∣β∣2 Expanding gives α‾α⟨0∣0⟩+α‾β⟨0∣1⟩+β‾α⟨1∣0⟩+β‾β⟨1∣1⟩\overline{\alpha}\alpha\langle 0|0\rangle + \overline{\alpha}\beta\langle 0|1\rangle + \overline{\beta}\alpha\langle 1|0\rangle + \overline{\beta}\beta\langle 1|1\rangleαα⟨0∣0⟩+αβ⟨0∣1⟩+βα⟨1∣0⟩+ββ⟨1∣1⟩ , and ⟨0∣1⟩=⟨1∣0⟩=0\langle 0|1\rangle = \langle 1|0\rangle = 0⟨0∣1⟩=⟨1∣0⟩=0 , ⟨0∣0⟩=⟨1∣1⟩=1\ \langle 0|0\rangle = \langle 1|1\rangle = 1 ⟨0∣0⟩=⟨1∣1⟩=1 .Exampl…
[Dirac notation for vectors] In Dirac notation, how do you write the coefficients of ∣v⟩|v\rangle∣v⟩ with respect to the standard basis {∣0⟩,∣1⟩}\{|0\rangle, |1\rangle\}{∣0⟩,∣1⟩} ?
As brakets ⟨i∣v⟩\langle i|v\rangle⟨i∣v⟩ : ∣v⟩=⟨0∣v⟩ ∣0⟩+⟨1∣v⟩ ∣1⟩\ \boldsymbol{|v\rangle = \langle 0|v\rangle\,|0\rangle + \langle 1|v\rangle\,|1\rangle} ∣v⟩=⟨0∣v⟩∣0⟩+⟨1∣v⟩∣1⟩ No variables v0,v1v_0, v_1v0,v1 need to be defined: ⟨i∣v⟩\langle i|v\rangle⟨i∣v⟩ is the iii -th coefficient.Example: ∣v⟩=2∣0⟩−i∣1⟩|v\rangle = 2|0\rangle - i|1\rangle∣v⟩=2∣0⟩−i∣1⟩ : ⟨0∣v⟩=2\langle 0|v\rangle = 2⟨0∣v⟩=2 , ⟨1∣v⟩=−i\langle 1|v\rangle = -i⟨1∣v⟩=−i .
[Dirac notation for vectors] What is the Hadamard basis Bh={∣+⟩,∣−⟩}B_h = \{|+\rangle, |-\rangle\}Bh={∣+⟩,∣−⟩} of C2\mathbb{C}^2C2 ?
∣+⟩=12∣0⟩+12∣1⟩,∣−⟩=12∣0⟩−12∣1⟩\boldsymbol{|+\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle, \quad |-\rangle = \frac{1}{\sqrt{2}}|0\rangle - \frac{1}{\sqrt{2}}|1\rangle}∣+⟩=21∣0⟩+21∣1⟩,∣−⟩=21∣0⟩−21∣1⟩ An orthonormal basis: ⟨+∣+⟩=⟨−∣−⟩=1\langle +|+\rangle = \langle -|-\rangle = 1⟨+∣+⟩=⟨−∣−⟩=1 , ⟨+∣−⟩=⟨−∣+⟩=0\ \langle +|-\rangle = \langle -|+\rangle = 0 ⟨+∣−⟩=⟨−∣+⟩=0 . The book describes it as the standard basis rotated by 45∘45^\circ45∘ .
[Dirac notation for vectors] How do you write any ∣v⟩∈C2|v\rangle \in \mathbb{C}^2∣v⟩∈C2 in the Hadamard basis {∣+⟩,∣−⟩}\{|+\rangle, |-\rangle\}{∣+⟩,∣−⟩} , in Dirac notation?
∣v⟩=⟨+∣v⟩ ∣+⟩+⟨−∣v⟩ ∣−⟩\boldsymbol{|v\rangle = \langle +|v\rangle\,|+\rangle + \langle -|v\rangle\,|-\rangle}∣v⟩=⟨+∣v⟩∣+⟩+⟨−∣v⟩∣−⟩ This holds because {∣+⟩,∣−⟩}\{|+\rangle, |-\rangle\}{∣+⟩,∣−⟩} is orthonormal ( ∣±⟩=12∣0⟩±12∣1⟩|\pm\rangle = \frac{1}{\sqrt{2}}|0\rangle \pm \frac{1}{\sqrt{2}}|1\rangle∣±⟩=21∣0⟩±21∣1⟩ ).Example: ⟨+∣0⟩=⟨−∣0⟩=12\langle +|0\rangle = \langle -|0\rangle = \frac{1}{\sqrt{2}}⟨+∣0⟩=⟨−∣0⟩=21 , so ∣0⟩=12∣+⟩+12∣−⟩|0\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}…
[Dirac notation for vectors] Why do brakets like ⟨+∣v⟩\langle +|v\rangle⟨+∣v⟩ make coefficients clearer than a coefficient vector like (v+,v−)BhT(v_+, v_-)^{\mathsf{T}}_{B_h}(v+,v−)BhT ?
The basis vector appears in the expression itself, so no subscript is needed to say which basis is meant. Standard basis coefficients: ⟨0∣v⟩,⟨1∣v⟩\langle 0|v\rangle, \langle 1|v\rangle⟨0∣v⟩,⟨1∣v⟩ . Hadamard basis coefficients: ⟨+∣v⟩,⟨−∣v⟩\langle +|v\rangle, \langle -|v\rangle⟨+∣v⟩,⟨−∣v⟩ .
[Dirac notation for vectors] What is the change-of-basis matrix Bh[1 ⁣ ⁣1]Bs {}_{B_h}[\mathrm{1\!\!1}]_{B_s}Bh[11]Bs from the standard basis Bs={∣0⟩,∣1⟩}B_s = \{|0\rangle, |1\rangle\}Bs={∣0⟩,∣1⟩} to the Hadamard basis Bh={∣+⟩,∣−⟩}B_h = \{|+\rangle, |-\rangle\}Bh={∣+⟩,∣−⟩} ?
[⟨+∣0⟩⟨+∣1⟩⟨−∣0⟩⟨−∣1⟩]=[121212−12]\boldsymbol{\begin{bmatrix} \langle +|0\rangle & \langle +|1\rangle \\ \langle -|0\rangle & \langle -|1\rangle \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}}[⟨+∣0⟩⟨−∣0⟩⟨+∣1⟩⟨−∣1⟩]=[212121−21] Its columns are the coefficients of ∣0⟩|0\rangle∣0⟩ and ∣1⟩|1\rangle∣1⟩ with respect to BhB_hBh ( ∣±⟩=12∣0⟩±12∣1⟩|\pm\rangle = \frac{1}{\sqrt{2}}|0\rangle \pm \frac{1}{\sqrt{2}…
[Dirac notation for vectors] ∣v⟩=v0∣0⟩+v1∣1⟩|v\rangle = v_0|0\rangle + v_1|1\rangle∣v⟩=v0∣0⟩+v1∣1⟩ . What are its Hadamard-basis coefficients v+=⟨+∣v⟩v_+ = \langle +|v\ranglev+=⟨+∣v⟩ and v−=⟨−∣v⟩v_- = \langle -|v\ranglev−=⟨−∣v⟩ , and how does Dirac notation find them without a matrix?
v+=12(v0+v1),v−=12(v0−v1)\boldsymbol{v_+ = \frac{1}{\sqrt{2}}(v_0 + v_1), \quad v_- = \frac{1}{\sqrt{2}}(v_0 - v_1)}v+=21(v0+v1),v−=21(v0−v1) Substitute and use linearity: ⟨+∣v⟩=v0⟨+∣0⟩+v1⟨+∣1⟩\langle +|v\rangle = v_0\langle +|0\rangle + v_1\langle +|1\rangle⟨+∣v⟩=v0⟨+∣0⟩+v1⟨+∣1⟩ , with ⟨±∣0⟩=12\langle \pm|0\rangle = \frac{1}{\sqrt{2}}⟨±∣0⟩=21 , ⟨±∣1⟩=±12\langle \pm|1\rangle = \pm\frac{1}{\sqrt{2}}⟨±∣1⟩=±21 .Example: ∣v⟩=∣1⟩|v\rangle = |1\rangle∣v⟩=∣1⟩ gives v+=12v_+ = \frac{1}{…
[Dirac notation for vectors] What matrix is the outer product ∣0⟩⟨0∣|0\rangle\langle 0|∣0⟩⟨0∣ , and what does it do?
[1000]\boldsymbol{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}}[1000] : the projection onto the subspace spanned by ∣0⟩|0\rangle∣0⟩ . ∣0⟩⟨0∣(α∣0⟩+β∣1⟩)=α∣0⟩⟨0∣0⟩+β∣0⟩⟨0∣1⟩=α∣0⟩|0\rangle\langle 0|(\alpha|0\rangle + \beta|1\rangle) = \alpha|0\rangle\langle 0|0\rangle + \beta|0\rangle\langle 0|1\rangle = \alpha|0\rangle∣0⟩⟨0∣(α∣0⟩+β∣1⟩)=α∣0⟩⟨0∣0⟩+β∣0⟩⟨0∣1⟩=α∣0⟩ Example: the projection onto the ∣+⟩|+\rangle∣+⟩ -subspace is ∣+⟩⟨+∣=12[1111]|+\rangle\langle +| = \frac{1}{2}\begin{bmatrix} 1 & 1 \\ 1…
[Dirac notation for vectors] How do you write a matrix A=[a00a01a10a11]A = \begin{bmatrix} a_{00} & a_{01} \\ a_{10} & a_{11} \end{bmatrix}A=[a00a10a01a11] (standard basis) in Dirac notation?
As a linear combination of outer products: A=a00∣0⟩⟨0∣+a01∣0⟩⟨1∣+a10∣1⟩⟨0∣+a11∣1⟩⟨1∣\boldsymbol{A = a_{00}|0\rangle\langle 0| + a_{01}|0\rangle\langle 1| + a_{10}|1\rangle\langle 0| + a_{11}|1\rangle\langle 1|}A=a00∣0⟩⟨0∣+a01∣0⟩⟨1∣+a10∣1⟩⟨0∣+a11∣1⟩⟨1∣ Example: [1234]=1∣0⟩⟨0∣+2∣0⟩⟨1∣+3∣1⟩⟨0∣+4∣1⟩⟨1∣\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = 1|0\rangle\langle 0| + 2|0\rangle\langle 1| + 3|1\rangle\langle 0| + 4|1\rangle\langle 1|[1324]=1∣0⟩⟨0∣+2∣0⟩⟨1∣+3∣1⟩⟨0∣+4∣1⟩⟨1∣ .
[Dirac notation for vectors] In A=a00∣0⟩⟨0∣+a01∣0⟩⟨1∣+a10∣1⟩⟨0∣+a11∣1⟩⟨1∣A = a_{00}|0\rangle\langle 0| + a_{01}|0\rangle\langle 1| + a_{10}|1\rangle\langle 0| + a_{11}|1\rangle\langle 1|A=a00∣0⟩⟨0∣+a01∣0⟩⟨1∣+a10∣1⟩⟨0∣+a11∣1⟩⟨1∣ , what does the term a10∣1⟩⟨0∣a_{10}|1\rangle\langle 0|a10∣1⟩⟨0∣ do to an input vector?
⟨0∣\langle 0|⟨0∣ selects the ∣0⟩|0\rangle∣0⟩ -coefficient of the input, and ∣1⟩|1\rangle∣1⟩ adds it, times a10a_{10}a10 , to the ∣1⟩|1\rangle∣1⟩ -component of the output. Example: a10∣1⟩⟨0∣(α∣0⟩+β∣1⟩)=a10α∣1⟩a_{10}|1\rangle\langle 0|(\alpha|0\rangle + \beta|1\rangle) = a_{10}\alpha|1\ranglea10∣1⟩⟨0∣(α∣0⟩+β∣1⟩)=a10α∣1⟩ .
[Dirac notation for vectors] How do you get the matrix entry aija_{ij}aij of an operator AAA with respect to the standard basis, in Dirac notation?
aij=⟨i∣A∣j⟩\boldsymbol{a_{ij} = \langle i|A|j\rangle}aij=⟨i∣A∣j⟩ : sandwich AAA between ⟨i∣\langle i|⟨i∣ on the left and ∣j⟩|j\rangle∣j⟩ on the right. A∣j⟩A|j\rangleA∣j⟩ is the output for the jjj -th input basis vector, and ⟨i∣\langle i|⟨i∣ takes its iii -th coefficient. So A=∑i,j⟨i∣A∣j⟩ ∣i⟩⟨j∣A = \sum_{i,j} \langle i|A|j\rangle\,|i\rangle\langle j|A=∑i,j⟨i∣A∣j⟩∣i⟩⟨j∣ .Example: for A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[1324] , A∣0⟩=∣0⟩+3∣1⟩\ A|0\rangle = |0\rangle + 3|1…
[Dirac notation for vectors] A=a00∣0⟩⟨0∣+a01∣0⟩⟨1∣+a10∣1⟩⟨0∣+a11∣1⟩⟨1∣A = a_{00}|0\rangle\langle 0| + a_{01}|0\rangle\langle 1| + a_{10}|1\rangle\langle 0| + a_{11}|1\rangle\langle 1|A=a00∣0⟩⟨0∣+a01∣0⟩⟨1∣+a10∣1⟩⟨0∣+a11∣1⟩⟨1∣ Why is ⟨1∣A∣0⟩=a10\langle 1|A|0\rangle = a_{10}⟨1∣A∣0⟩=a10 ?
Multiplying by ∣0⟩|0\rangle∣0⟩ on the right keeps only the terms ending in ⟨0∣\langle 0|⟨0∣ , and then ⟨1∣\langle 1|⟨1∣ on the left keeps only the term starting with ∣1⟩|1\rangle∣1⟩ . A∣0⟩=a00∣0⟩+a10∣1⟩A|0\rangle = a_{00}|0\rangle + a_{10}|1\rangleA∣0⟩=a00∣0⟩+a10∣1⟩ since ⟨0∣0⟩=1\langle 0|0\rangle = 1⟨0∣0⟩=1 , ⟨1∣0⟩=0\langle 1|0\rangle = 0⟨1∣0⟩=0 ; then ⟨1∣(a00∣0⟩+a10∣1⟩)=a10\langle 1|(a_{00}|0\rangle + a_{10}|1\rangle) = a_{10}⟨1∣(a00∣0⟩+a10∣1⟩)=a10 .
[Dirac notation for vectors] What is the matrix Bh[A]Bh {}_{B_h}[A]_{B_h}Bh[A]Bh of an operator AAA with respect to the Hadamard basis Bh={∣+⟩,∣−⟩}B_h = \{|+\rangle, |-\rangle\}Bh={∣+⟩,∣−⟩} ?
[⟨+∣A∣+⟩⟨+∣A∣−⟩⟨−∣A∣+⟩⟨−∣A∣−⟩]\boldsymbol{\begin{bmatrix} \langle +|A|+\rangle & \langle +|A|-\rangle \\ \langle -|A|+\rangle & \langle -|A|-\rangle \end{bmatrix}}[⟨+∣A∣+⟩⟨−∣A∣+⟩⟨+∣A∣−⟩⟨−∣A∣−⟩] Equivalently A=⟨+∣A∣+⟩∣+⟩⟨+∣+⟨+∣A∣−⟩∣+⟩⟨−∣+⟨−∣A∣+⟩∣−⟩⟨+∣+⟨−∣A∣−⟩∣−⟩⟨−∣A = \langle +|A|+\rangle|+\rangle\langle +| + \langle +|A|-\rangle|+\rangle\langle -| + \langle -|A|+\rangle|-\rangle\langle +| + \langle -|A|-\rangle|-\rangle\langle -|A=⟨+∣A∣+⟩∣+⟩⟨+∣+⟨+∣A∣−⟩∣+⟩⟨−∣+⟨−∣A∣+⟩∣−⟩⟨+∣+⟨−∣A∣−⟩∣−⟩⟨−∣ .Examp…
[Dirac notation for vectors] In Dirac notation, what do ∣v⟩|v\rangle∣v⟩ , ⟨v∣\langle v|⟨v∣ , ⟨i∣v⟩\langle i|v\rangle⟨i∣v⟩ , ⟨v∣w⟩\langle v|w\rangle⟨v∣w⟩ , ∣v⟩⟨w∣|v\rangle\langle w|∣v⟩⟨w∣ and ⟨i∣A∣j⟩\langle i|A|j\rangle⟨i∣A∣j⟩ denote?
∣v⟩|v\rangle∣v⟩ A vector (column vector). ⟨v∣\langle v|⟨v∣ Its Hermitian transpose (conjugated row vector). ⟨i∣v⟩\langle i|v\rangle⟨i∣v⟩ The iii -th coefficient of ∣v⟩|v\rangle∣v⟩ . ⟨v∣w⟩\langle v|w\rangle⟨v∣w⟩ The inner product of ∣v⟩|v\rangle∣v⟩ and ∣w⟩|w\rangle∣w⟩ . ∣v⟩⟨w∣|v\rangle\langle w|∣v⟩⟨w∣ The outer product, a matrix. ⟨i∣A∣j⟩\langle i|A|j\rangle⟨i∣A∣j⟩ The matrix coefficient aija_{ij}aij of AAA .
[Quantum information processing] What do using a digital device and using a quantum computer have in common, operationally?
Both are black boxes: you input data, the machine processes it, and you read out the answer. In both, the internal representation is unintelligible: we can't "see" quantum states, nor the raw bits of an mp3 file.
[Quantum information processing] In a sound card, what do the ADC and the DAC do?
ADC (analog-to-digital converter): digitizes sound. DAC (digital-to-analog converter): turns digital sound back into analog vibrations for the speakers. Between them, digital processing, e.g. compressing a wav file ( n=50n = 50n=50 MB) into an mp3 file ( k=5k = 5k=5 MB).
[Quantum information processing] Which steps of a quantum information processing pipeline correspond to the analog-to-digital and the digital-to-analog conversion of a digital pipeline?
State preparation corresponds to analog-to-digital conversion; quantum measurement corresponds to digital-to-analog conversion. State preparation puts information into a quantum computer, measurement reads information out.
[Quantum information processing] What are the stages of a quantum information processing pipeline, and the types of input and output?
Classical bitstring x∈Z2kx \in \mathbb{Z}_2^kx∈Z2k →\to→ state preparation →\to→ quantum processing →\to→ measurement →\to→ classical bitstring y∈Z2ℓy \in \mathbb{Z}_2^\elly∈Z2ℓ . Quantum processing uses quantum operations T:Cm→CnT: \mathbb{C}^m \to \mathbb{C}^nT:Cm→Cn on qubits, where classical processing uses functions f:Z2k→Z2ℓf: \mathbb{Z}_2^k \to \mathbb{Z}_2^\ellf:Z2k→Z2ℓ on bits.
[Quantum information processing] What are the two big differences between quantum and classical computation?
Quantum states Quantum computation processes quantum states (vectors), so it is more general than classical computation. Example: quantum operations T:Cm→CnT: \mathbb{C}^m \to \mathbb{C}^nT:Cm→Cn instead of functions f:Z2k→Z2ℓf: \mathbb{Z}_2^k \to \mathbb{Z}_2^\ellf:Z2k→Z2ℓ . One readout The output state ∣y⟩|y\rangle∣y⟩ can be measured only once, because measurement disturbs the state; and the answer depends probabilistically on ∣y⟩|y\rangle∣y⟩ . A measurement asks one question about ∣y⟩|y\ra…
[Postulates of quantum mechanics] What are the three steps of a quantum information processing pipeline, from a classical input bitstring xxx to a classical output bitstring yyy ?
State preparation xxx is used as instructions for preparing a quantum state ∣x⟩∈Cm|x\rangle \in \mathbb{C}^m∣x⟩∈Cm . Quantum operations Operations on ∣x⟩|x\rangle∣x⟩ convert it to the output state ∣y⟩∈Cn|y\rangle \in \mathbb{C}^n∣y⟩∈Cn . Measurement Measuring ∣y⟩|y\rangle∣y⟩ gives the classical bitstring yyy .
[Postulates of quantum mechanics] What do the four postulates of quantum mechanics specify?
Quantum states What a quantum state is. Quantum operations Which operations on quantum states are allowed. Measurements How information is extracted from a quantum system by measuring it. Composite systems How a system made of several quantum systems is represented. These rules hold in every field that uses quantum mechanics: physics, chemistry, engineering, quantum information.
[Postulates of quantum mechanics] Can the postulates of quantum mechanics be proved from a more basic theory?
No. Scientists take the postulates as facts and make their theories obey them.
[Postulates of quantum mechanics] Postulate 1 (quantum states). What is associated to an isolated quantum system, and what is a state?
A complex inner product space (Hilbert space), the state space; a state is a unit-length vector in it. A ddd -dimensional quantum system has state space Cd\mathbb{C}^dCd .
[Postulates of quantum mechanics] What is the global phase of a quantum state ∣ψ⟩|\psi\rangle∣ψ⟩ , and does it matter?
A factor eiθe^{i\theta}eiθ ( θ∈R\theta \in \mathbb{R}θ∈R ) multiplying the whole vector; it doesn't matter: eiθ∣ψ⟩e^{i\theta}|\psi\rangleeiθ∣ψ⟩ and ∣ψ⟩|\psi\rangle∣ψ⟩ are the same state. Example: ∣ψ⟩|\psi\rangle∣ψ⟩ , −∣ψ⟩-|\psi\rangle−∣ψ⟩ and i∣ψ⟩i|\psi\ranglei∣ψ⟩ all describe the same state ( θ=0,π,π2\theta = 0, \pi, \tfrac{\pi}{2}θ=0,π,2π ).
[Postulates of quantum mechanics] What is a qubit?
A unit-length vector ∣ψ⟩=α∣0⟩+β∣1⟩|\psi\rangle = \alpha|0\rangle + \beta|1\rangle∣ψ⟩=α∣0⟩+β∣1⟩ in C2\mathbb{C}^2C2 , with ∣α∣2+∣β∣2=1|\alpha|^2 + |\beta|^2 = 1∣α∣2+∣β∣2=1 . ∣0⟩=(1,0)T|0\rangle = (1, 0)^{\mathsf{T}}∣0⟩=(1,0)T and ∣1⟩=(0,1)T|1\rangle = (0, 1)^{\mathsf{T}}∣1⟩=(0,1)T are the standard basis of C2\mathbb{C}^2C2 . The name is short for "quantum bit", by analogy with classical bits ∈{0,1}\in \{0, 1\}∈{0,1} .Example: 12∣0⟩+32i∣1⟩\tfrac{1}{2}|0\rangle + \tfrac{\sqrt{3}}{2}i|1\rangle21∣0…
[Postulates of quantum mechanics] A qubit is a unit-length vector ∣ψ⟩=α∣0⟩+β∣1⟩∈C2|\psi\rangle = \alpha|0\rangle + \beta|1\rangle \in \mathbb{C}^2∣ψ⟩=α∣0⟩+β∣1⟩∈C2 , where ∣0⟩=(1,0)T|0\rangle = (1, 0)^{\mathsf{T}}∣0⟩=(1,0)T and ∣1⟩=(0,1)T|1\rangle = (0, 1)^{\mathsf{T}}∣1⟩=(0,1)T form the standard basis.Why can we always choose α\alphaα to be a real number?
The global phase doesn't matter: multiplying by e−iarg⁡αe^{-i\arg\alpha}e−iargα makes the coefficient of ∣0⟩|0\rangle∣0⟩ real and gives the same state. Example: i2∣0⟩+12∣1⟩\tfrac{i}{\sqrt{2}}|0\rangle + \tfrac{1}{\sqrt{2}}|1\rangle2i∣0⟩+21∣1⟩ times −i-i−i is 12∣0⟩−i2∣1⟩\tfrac{1}{\sqrt{2}}|0\rangle - \tfrac{i}{\sqrt{2}}|1\rangle21∣0⟩−2i∣1⟩ , the same state with α\alphaα real.
[Postulates of quantum mechanics] Name three physical systems that can store a qubit.
Photon polarization, electron spin, the path a photon takes at a half-silvered mirror (also: magnetic flux direction in a superconducting loop). A qubit is a device-independent representation of quantum information, as a bit is for classical information.
[Postulates of quantum mechanics] A photon meets a half-silvered mirror: ∣0⟩|0\rangle∣0⟩ means it passes through, ∣1⟩|1\rangle∣1⟩ means it is reflected. Both happen with equal probability. What is its state?
∣γ⟩=12∣0⟩+12∣1⟩\boldsymbol{|\gamma\rangle = \tfrac{1}{\sqrt{2}}|0\rangle + \tfrac{1}{\sqrt{2}}|1\rangle}∣γ⟩=21∣0⟩+21∣1⟩
[Postulates of quantum mechanics] What is quantum state preparation?
Encoding classical information xxx into a quantum system: a machine produces the state ∣x⟩|x\rangle∣x⟩ depending on its input xxx . Example: a machine producing photons with polarization ∣H⟩|H\rangle∣H⟩ (horizontal) on input x=0x = 0x=0 and ∣V⟩|V\rangle∣V⟩ (vertical) on input x=1x = 1x=1 .
[Postulates of quantum mechanics] Postulate 2 (quantum operations). How does the state of an isolated quantum system change from time t1t_1t1 to time t2t_2t2 ?
By a unitary: there is a unitary UUU with ∣ψ2⟩=U∣ψ1⟩|\psi_2\rangle = U|\psi_1\rangle∣ψ2⟩=U∣ψ1⟩ . Unitary means U†U=UU†=1 ⁣ ⁣1U^\dagger U = UU^\dagger = \mathrm{1\!\!1}U†U=UU†=11 .
[Postulates of quantum mechanics] Why does a unitary operation UUU keep a quantum state at unit length?
∥U∣ψ⟩∥2=⟨ψ∣U†U∣ψ⟩=⟨ψ∣ψ⟩=1\boldsymbol{\|U|\psi\rangle\|^2 = \langle\psi|U^\dagger U|\psi\rangle = \langle\psi|\psi\rangle = 1}∥U∣ψ⟩∥2=⟨ψ∣U†U∣ψ⟩=⟨ψ∣ψ⟩=1 So quantum operations are length-preserving, and the output is again a valid state.
[Postulates of quantum mechanics] Can every quantum operation (gate) GGG be undone?
Yes: GGG is unitary, so the inverse gate G−1=G†G^{-1} = G^\daggerG−1=G† exists and G−1G=1 ⁣ ⁣1G^{-1}G = \mathrm{1\!\!1}G−1G=11 .
[Postulates of quantum mechanics] What does the phase gate ZZZ do to ∣0⟩|0\rangle∣0⟩ and ∣1⟩|1\rangle∣1⟩ , and what is its matrix?
Z∣0⟩=∣0⟩, Z∣1⟩=−∣1⟩, Z=[100−1]\boldsymbol{Z|0\rangle = |0\rangle, \ Z|1\rangle = -|1\rangle, \ Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}}Z∣0⟩=∣0⟩, Z∣1⟩=−∣1⟩, Z=[100−1] Z=∣0⟩⟨0∣−∣1⟩⟨1∣Z = |0\rangle\langle 0| - |1\rangle\langle 1|Z=∣0⟩⟨0∣−∣1⟩⟨1∣ : it flips the sign (phase) of ∣1⟩|1\rangle∣1⟩ .Example: Z(α∣0⟩+β∣1⟩)=α∣0⟩−β∣1⟩Z(\alpha|0\rangle + \beta|1\rangle) = \alpha|0\rangle - \beta|1\rangleZ(α∣0⟩+β∣1⟩)=α∣0⟩−β∣1⟩ .
[Postulates of quantum mechanics] What does the NOT gate XXX do to ∣0⟩|0\rangle∣0⟩ and ∣1⟩|1\rangle∣1⟩ , and what is its matrix?
X∣0⟩=∣1⟩, X∣1⟩=∣0⟩, X=[0110]\boldsymbol{X|0\rangle = |1\rangle, \ X|1\rangle = |0\rangle, \ X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}}X∣0⟩=∣1⟩, X∣1⟩=∣0⟩, X=[0110] X=∣0⟩⟨1∣+∣1⟩⟨0∣X = |0\rangle\langle 1| + |1\rangle\langle 0|X=∣0⟩⟨1∣+∣1⟩⟨0∣ .Example: X(α∣0⟩+β∣1⟩)=β∣0⟩+α∣1⟩X(\alpha|0\rangle + \beta|1\rangle) = \beta|0\rangle + \alpha|1\rangleX(α∣0⟩+β∣1⟩)=β∣0⟩+α∣1⟩ .
[Postulates of quantum mechanics] What does the Hadamard gate HHH do to ∣0⟩|0\rangle∣0⟩ and ∣1⟩|1\rangle∣1⟩ , and what is its matrix?
H∣0⟩=∣+⟩, H∣1⟩=∣−⟩, H=12[111−1]\boldsymbol{H|0\rangle = |+\rangle, \ H|1\rangle = |-\rangle, \ H = \tfrac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}}H∣0⟩=∣+⟩, H∣1⟩=∣−⟩, H=21[111−1] ∣+⟩=12(∣0⟩+∣1⟩)|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)∣+⟩=21(∣0⟩+∣1⟩) and ∣−⟩=12(∣0⟩−∣1⟩)|-\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle)∣−⟩=21(∣0⟩−∣1⟩) form the Hadamard basis.
[Postulates of quantum mechanics] The Hadamard gate is H=12[111−1]H = \tfrac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}H=21[111−1] . What is H(α∣0⟩+β∣1⟩)H(\alpha|0\rangle + \beta|1\rangle)H(α∣0⟩+β∣1⟩) ?
α+β2∣0⟩+α−β2∣1⟩\boldsymbol{\tfrac{\alpha + \beta}{\sqrt{2}}|0\rangle + \tfrac{\alpha - \beta}{\sqrt{2}}|1\rangle}2α+β∣0⟩+2α−β∣1⟩ Example: H∣+⟩H|+\rangleH∣+⟩ , with α=β=12\alpha = \beta = \tfrac{1}{\sqrt{2}}α=β=21 : 2/22∣0⟩+0∣1⟩=∣0⟩\tfrac{2/\sqrt{2}}{\sqrt{2}}|0\rangle + 0|1\rangle = |0\rangle22/2∣0⟩+0∣1⟩=∣0⟩ .
[Postulates of quantum mechanics] Geometrically, what does H=12[111−1]H = \tfrac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}H=21[111−1] do to the real plane?
A reflection across the line at angle 22.5∘22.5^\circ22.5∘ ( det⁡H=−1\det H = -1detH=−1 ). It sends (1,0)(1, 0)(1,0) to (12,12)(\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}})(21,21) , as a 45∘45^\circ45∘ counter-clockwise rotation does, but sends (0,1)(0, 1)(0,1) to (12,−12)(\tfrac{1}{\sqrt{2}}, -\tfrac{1}{\sqrt{2}})(21,−21) , where the rotation gives (−12,12)(-\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}})(−21,21) .
[Postulates of quantum mechanics] Z=[100−1]Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}Z=[100−1] and X=[0110]X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}X=[0110] . What is the gate XZXZXZ , and what does it do to α∣0⟩+β∣1⟩\alpha|0\rangle + \beta|1\rangleα∣0⟩+β∣1⟩ ?
XZ=[0−110]\boldsymbol{XZ = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}}XZ=[01−10] , and XZ(α∣0⟩+β∣1⟩)=−β∣0⟩+α∣1⟩XZ(\alpha|0\rangle + \beta|1\rangle) = \boldsymbol{-\beta|0\rangle + \alpha|1\rangle}XZ(α∣0⟩+β∣1⟩)=−β∣0⟩+α∣1⟩ ZZZ acts first: α∣0⟩−β∣1⟩\alpha|0\rangle - \beta|1\rangleα∣0⟩−β∣1⟩ ; then XXX swaps the coefficients.
[Postulates of quantum mechanics] Postulate 3 (measurement). What mathematical object models a quantum measurement with nnn possible outcomes?
nnn projection operators Π1,…,Πn\Pi_1, \dots, \Pi_nΠ1,…,Πn on the state space with ∑i=1nΠi=1 ⁣ ⁣1\sum_{i=1}^n \Pi_i = \mathrm{1\!\!1}∑i=1nΠi=11 . ∑iΠi=1 ⁣ ⁣1\sum_i \Pi_i = \mathrm{1\!\!1}∑iΠi=11 is called a decomposition of the identity. The index iii labels the outcomes.Example: Π0=∣0⟩⟨0∣\Pi_0 = |0\rangle\langle 0|Π0=∣0⟩⟨0∣ , Π1=∣1⟩⟨1∣\Pi_1 = |1\rangle\langle 1|Π1=∣1⟩⟨1∣ on C2\mathbb{C}^2C2 .
[Postulates of quantum mechanics] Born's rule. A measurement {Πi}\{\Pi_i\}{Πi} is performed on the state ∣ψ⟩|\psi\rangle∣ψ⟩ . What is the probability of outcome iii ?
Pr⁡({outcome i})=∥Πi∣ψ⟩∥2=⟨ψ∣Πi∣ψ⟩\boldsymbol{\Pr(\{\text{outcome } i\}) = \|\Pi_i|\psi\rangle\|^2 = \langle\psi|\Pi_i|\psi\rangle}Pr({outcome i})=∥Πi∣ψ⟩∥2=⟨ψ∣Πi∣ψ⟩ Example: Π0=∣0⟩⟨0∣\Pi_0 = |0\rangle\langle 0|Π0=∣0⟩⟨0∣ on ∣ψ⟩=12∣0⟩+32∣1⟩|\psi\rangle = \tfrac{1}{2}|0\rangle + \tfrac{\sqrt{3}}{2}|1\rangle∣ψ⟩=21∣0⟩+23∣1⟩ : ∥12∣0⟩∥2=14\|\tfrac{1}{2}|0\rangle\|^2 = \tfrac{1}{4}∥21∣0⟩∥2=41 .
[Postulates of quantum mechanics] Π\PiΠ is a projection operator. Why is ∥Π∣ψ⟩∥2=⟨ψ∣Π∣ψ⟩\|\Pi|\psi\rangle\|^2 = \langle\psi|\Pi|\psi\rangle∥Π∣ψ⟩∥2=⟨ψ∣Π∣ψ⟩ ?
∥Π∣ψ⟩∥2=⟨ψ∣Π†Π∣ψ⟩=⟨ψ∣ΠΠ∣ψ⟩=⟨ψ∣Π∣ψ⟩\boldsymbol{\|\Pi|\psi\rangle\|^2 = \langle\psi|\Pi^\dagger\Pi|\psi\rangle = \langle\psi|\Pi\Pi|\psi\rangle = \langle\psi|\Pi|\psi\rangle}∥Π∣ψ⟩∥2=⟨ψ∣Π†Π∣ψ⟩=⟨ψ∣ΠΠ∣ψ⟩=⟨ψ∣Π∣ψ⟩ A projection is self-adjoint ( Π†=Π\Pi^\dagger = \PiΠ†=Π ) and idempotent ( Π2=Π\Pi^2 = \PiΠ2=Π ). Physicists usually write Born's rule as ⟨ψ∣Πi∣ψ⟩\langle\psi|\Pi_i|\psi\rangle⟨ψ∣Πi∣ψ⟩ , the projection "sandwiched" between two copies of the state.
[Postulates of quantum mechanics] A measurement {Π1,…,Πn}\{\Pi_1, \dots, \Pi_n\}{Π1,…,Πn} with ∑iΠi=1 ⁣ ⁣1\sum_i \Pi_i = \mathrm{1\!\!1}∑iΠi=11 is performed on a state ∣ψ⟩|\psi\rangle∣ψ⟩ . Why do the outcome probabilities ∥Πi∣ψ⟩∥2\|\Pi_i|\psi\rangle\|^2∥Πi∣ψ⟩∥2 add up to 1?
1=∥1 ⁣ ⁣1∣ψ⟩∥2=∥Π1∣ψ⟩+⋯+Πn∣ψ⟩∥2=∥Π1∣ψ⟩∥2+⋯+∥Πn∣ψ⟩∥2\boldsymbol{1 = \|\mathrm{1\!\!1}|\psi\rangle\|^2 = \|\Pi_1|\psi\rangle + \dots + \Pi_n|\psi\rangle\|^2 = \|\Pi_1|\psi\rangle\|^2 + \dots + \|\Pi_n|\psi\rangle\|^2}1=∥11∣ψ⟩∥2=∥Π1∣ψ⟩+⋯+Πn∣ψ⟩∥2=∥Π1∣ψ⟩∥2+⋯+∥Πn∣ψ⟩∥2 The last step is Pythagoras' theorem: the projections are mutually orthogonal, so the vectors Πi∣ψ⟩\Pi_i|\psi\rangleΠi∣ψ⟩ are orthogonal.
[Postulates of quantum mechanics] A measurement {Πi}\{\Pi_i\}{Πi} on the state ∣ψ⟩|\psi\rangle∣ψ⟩ gives outcome iii . What is the state afterwards?
∣ψi′⟩=Πi∣ψ⟩∥Πi∣ψ⟩∥\boldsymbol{|\psi'_i\rangle = \frac{\Pi_i|\psi\rangle}{\|\Pi_i|\psi\rangle\|}}∣ψi′⟩=∥Πi∣ψ⟩∥Πi∣ψ⟩ Project onto the image of Πi\Pi_iΠi , then normalize to unit length.Example: Π1=∣1⟩⟨1∣\Pi_1 = |1\rangle\langle 1|Π1=∣1⟩⟨1∣ on 12∣0⟩+32∣1⟩\tfrac{1}{2}|0\rangle + \tfrac{\sqrt{3}}{2}|1\rangle21∣0⟩+23∣1⟩ : (3/2)∣1⟩3/2=∣1⟩\frac{(\sqrt{3}/2)|1\rangle}{\sqrt{3}/2} = |1\rangle3/2(3/2)∣1⟩=∣1⟩ .
[Postulates of quantum mechanics] Does measuring a quantum state change it?
In general yes ( ∣ψi′⟩≠∣ψ⟩|\psi'_i\rangle \neq |\psi\rangle∣ψi′⟩=∣ψ⟩ , the state is disturbed); it stays the same exactly when ∣ψ⟩|\psi\rangle∣ψ⟩ lies in the image of the projection Πi\Pi_iΠi of the outcome. Example: measuring {∣0⟩⟨0∣,∣1⟩⟨1∣}\{|0\rangle\langle 0|, |1\rangle\langle 1|\}{∣0⟩⟨0∣,∣1⟩⟨1∣} leaves ∣0⟩|0\rangle∣0⟩ as it is, but turns ∣+⟩|+\rangle∣+⟩ into ∣0⟩|0\rangle∣0⟩ or ∣1⟩|1\rangle∣1⟩ .
[Postulates of quantum mechanics] The change of ∣ψ⟩|\psi\rangle∣ψ⟩ into the post-measurement state ∣ψi′⟩|\psi'_i\rangle∣ψi′⟩ is often described by saying the state collapses.
Savov avoids this word: it often leads to magical interpretations of measurement. Postulate 3 (measurement) is the most debated of the four postulates.
[Postulates of quantum mechanics] Without magic, how does Savov describe what happens in a quantum measurement?
An interaction between a tiny quantum system and a large classical apparatus with nnn possible states, which forces the state to align with one of them. It creates classical information (the outcome iii ) and destroys quantum information: the part of ∣ψ⟩|\psi\rangle∣ψ⟩ in the image of 1 ⁣ ⁣1−Πi\mathrm{1\!\!1} - \Pi_i11−Πi is lost.
[Postulates of quantum mechanics] A measurement is a set of projections {Πi}\{\Pi_i\}{Πi} with ∑iΠi=1 ⁣ ⁣1\sum_i \Pi_i = \mathrm{1\!\!1}∑iΠi=11 ; outcome iii on state ∣ψ⟩|\psi\rangle∣ψ⟩ has probability ∥Πi∣ψ⟩∥2\|\Pi_i|\psi\rangle\|^2∥Πi∣ψ⟩∥2 and leaves the state Πi∣ψ⟩∥Πi∣ψ⟩∥\frac{\Pi_i|\psi\rangle}{\|\Pi_i|\psi\rangle\|}∥Πi∣ψ⟩∥Πi∣ψ⟩ .A qubit ∣ψ⟩=α∣0⟩+β∣1⟩|\psi\rangle = \alpha|0\rangle + \beta|1\rangle∣ψ⟩=α∣0⟩+β∣1⟩ is measured in the standard basis, Π0=∣0⟩⟨0∣\Pi_0 = |0\rangle\langle 0|Π0…
Pr⁡(0)=∣α∣2\boldsymbol{\Pr(0) = |\alpha|^2}Pr(0)=∣α∣2 , Pr⁡(1)=∣β∣2\boldsymbol{\Pr(1) = |\beta|^2}Pr(1)=∣β∣2 ; afterwards ∣0⟩|0\rangle∣0⟩ or ∣1⟩|1\rangle∣1⟩ . ⟨ψ∣0⟩⟨0∣ψ⟩=αˉα=∣α∣2\langle\psi|0\rangle\langle 0|\psi\rangle = \bar{\alpha}\alpha = |\alpha|^2⟨ψ∣0⟩⟨0∣ψ⟩=αˉα=∣α∣2 . A photon α∣0⟩+β∣1⟩\alpha|0\rangle + \beta|1\rangleα∣0⟩+β∣1⟩ after a partly silvered mirror: the right detector clicks with probability ∣α∣2|\alpha|^2∣α∣2 , the top one with probability ∣β∣2|\beta|^2∣β∣2 .
[Postulates of quantum mechanics] A measurement is a set of projections {Πi}\{\Pi_i\}{Πi} with ∑iΠi=1 ⁣ ⁣1\sum_i \Pi_i = \mathrm{1\!\!1}∑iΠi=11 ; outcome iii on state ∣ψ⟩|\psi\rangle∣ψ⟩ has probability ∥Πi∣ψ⟩∥2\|\Pi_i|\psi\rangle\|^2∥Πi∣ψ⟩∥2 and leaves the state Πi∣ψ⟩∥Πi∣ψ⟩∥\frac{\Pi_i|\psi\rangle}{\|\Pi_i|\psi\rangle\|}∥Πi∣ψ⟩∥Πi∣ψ⟩ .Hadamard basis of C2\mathbb{C}^2C2 : ∣+⟩=12(∣0⟩+∣1⟩)|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)∣+⟩=21(∣0⟩+∣1⟩) , ∣−⟩=12(∣0⟩−∣1⟩)\ |-\rangle = \…
Pr⁡(+)=∣α+β∣22\boldsymbol{\Pr(+) = \frac{|\alpha + \beta|^2}{2}}Pr(+)=2∣α+β∣2 , Pr⁡(−)=∣α−β∣22\boldsymbol{\Pr(-) = \frac{|\alpha - \beta|^2}{2}}Pr(−)=2∣α−β∣2 ; afterwards the state is ∣+⟩|+\rangle∣+⟩ or ∣−⟩|-\rangle∣−⟩ . Π+∣ψ⟩=∣+⟩⟨+∣ψ⟩=α+β2∣+⟩\Pi_+|\psi\rangle = |+\rangle\langle +|\psi\rangle = \tfrac{\alpha + \beta}{\sqrt{2}}|+\rangleΠ+∣ψ⟩=∣+⟩⟨+∣ψ⟩=2α+β∣+⟩ , and ∥∣+⟩∥=1\||+\rangle\| = 1∥∣+⟩∥=1 .Example: ∣0⟩|0\rangle∣0⟩ ( α=1,β=0\alpha = 1, \beta = 0α=1,β=0 ) gives +++ and −-− with probabi…
[Postulates of quantum mechanics] Two classical bits combine into a bitstring with 4 possible values. What is the combined state space of two qubits, and its standard basis?
The tensor product space C2⊗C2=C4\mathbb{C}^2 \otimes \mathbb{C}^2 = \mathbb{C}^4C2⊗C2=C4 , with basis ∣00⟩,∣01⟩,∣10⟩,∣11⟩|00\rangle, |01\rangle, |10\rangle, |11\rangle∣00⟩,∣01⟩,∣10⟩,∣11⟩ . The basis consists of all combinations of basis vectors of the two qubits: ∣0⟩⊗∣0⟩|0\rangle \otimes |0\rangle∣0⟩⊗∣0⟩ , ∣0⟩⊗∣1⟩|0\rangle \otimes |1\rangle∣0⟩⊗∣1⟩ , ∣1⟩⊗∣0⟩|1\rangle \otimes |0\rangle∣1⟩⊗∣0⟩ , ∣1⟩⊗∣1⟩|1\rangle \otimes |1\rangle∣1⟩⊗∣1⟩ .
[Postulates of quantum mechanics] Postulate 4 (composite systems). Systems 1,…,n1, \dots, n1,…,n are in states ∣φ1⟩,…,∣φn⟩|\varphi_1\rangle, \dots, |\varphi_n\rangle∣φ1⟩,…,∣φn⟩ . What is the state space and the state of the composite system?
The tensor product of the individual state spaces; the state is ∣φ1⟩⊗∣φ2⟩⊗⋯⊗∣φn⟩|\varphi_1\rangle \otimes |\varphi_2\rangle \otimes \dots \otimes |\varphi_n\rangle∣φ1⟩⊗∣φ2⟩⊗⋯⊗∣φn⟩ .
[Postulates of quantum mechanics] How is the tensor product ∣a⟩1⊗∣b⟩2|a\rangle_1 \otimes |b\rangle_2∣a⟩1⊗∣b⟩2 (system 1 in state ∣a⟩|a\rangle∣a⟩ , system 2 in state ∣b⟩|b\rangle∣b⟩ ) usually abbreviated?
∣a⟩1⊗∣b⟩2=∣a⟩∣b⟩=∣ab⟩\boldsymbol{|a\rangle_1 \otimes |b\rangle_2 = |a\rangle|b\rangle = |ab\rangle}∣a⟩1⊗∣b⟩2=∣a⟩∣b⟩=∣ab⟩ Drop the subscripts and the ⊗\otimes⊗ symbol, and write the indices as a string in one ket.Example: ∣0⟩1⊗∣1⟩2=∣01⟩|0\rangle_1 \otimes |1\rangle_2 = |01\rangle∣0⟩1⊗∣1⟩2=∣01⟩ .
[Postulates of quantum mechanics] What is the tensor product (α1∣0⟩+β1∣1⟩)⊗(α2∣0⟩+β2∣1⟩)(\alpha_1|0\rangle + \beta_1|1\rangle) \otimes (\alpha_2|0\rangle + \beta_2|1\rangle)(α1∣0⟩+β1∣1⟩)⊗(α2∣0⟩+β2∣1⟩) in the basis ∣00⟩,∣01⟩,∣10⟩,∣11⟩|00\rangle, |01\rangle, |10\rangle, |11\rangle∣00⟩,∣01⟩,∣10⟩,∣11⟩ ?
(α1α2, α1β2, β1α2, β1β2)\boldsymbol{(\alpha_1\alpha_2, \ \alpha_1\beta_2, \ \beta_1\alpha_2, \ \beta_1\beta_2)}(α1α2, α1β2, β1α2, β1β2) All products of a coefficient of the first qubit with a coefficient of the second; expand the product like a product of sums.Example: ∣+⟩⊗∣0⟩=(12,0,12,0)=12(∣00⟩+∣10⟩)|+\rangle \otimes |0\rangle = (\tfrac{1}{\sqrt{2}}, 0, \tfrac{1}{\sqrt{2}}, 0) = \tfrac{1}{\sqrt{2}}(|00\rangle + |10\rangle)∣+⟩⊗∣0⟩=(21,0,21,0)=21(∣00⟩+∣10⟩) .
[Postulates of quantum mechanics] What is the state space of nnn qubits, compared with the possible values of nnn classical bits?
nnn qubits: unit vectors in C2n\mathbb{C}^{2^n}C2n ; nnn bits: one of 2n2^n2n values. Adding one bit doubles the number of values; adding one qubit doubles the dimension of the state space.Example: 3 qubits have state space C8\mathbb{C}^8C8 .
[Postulates of quantum mechanics] nnn qubits have the 2n2^n2n -dimensional state space C2n\mathbb{C}^{2^n}C2n . Why doesn't this immediately make quantum computers enormously powerful?
Noise: storing nnn qubits and protecting them from noise is hard. Readout: a measurement {Π1,…,Πm}\{\Pi_1, \dots, \Pi_m\}{Π1,…,Πm} gives only one of m≤2nm \le 2^nm≤2n classical outcomes. And what we learn by measuring a state is proportional to how much we disturb it.
[Postulates of quantum mechanics] When is a state ∣ψ⟩AB∈VA⊗VB|\psi\rangle^{AB} \in V_A \otimes V_B∣ψ⟩AB∈VA⊗VB of a two-part system entangled?
When it cannot be written as a tensor product ∣ϕ⟩A⊗∣φ⟩B|\phi\rangle^A \otimes |\varphi\rangle^B∣ϕ⟩A⊗∣φ⟩B of states of the two parts. Such a state needs a nonlocal description. From the point of view of information: we know more about the whole system than about its parts.Example: 12(∣00⟩+∣01⟩)=∣0⟩⊗∣+⟩\tfrac{1}{\sqrt{2}}(|00\rangle + |01\rangle) = |0\rangle \otimes |+\rangle21(∣00⟩+∣01⟩)=∣0⟩⊗∣+⟩ is not entangled.
[Postulates of quantum mechanics] What is the EPR state ∣Ψ−⟩|\Psi^-\rangle∣Ψ−⟩ (Einstein, Podolsky, Rosen)?
∣Ψ−⟩=12(∣01⟩−∣10⟩)\boldsymbol{|\Psi^-\rangle = \tfrac{1}{\sqrt{2}}(|01\rangle - |10\rangle)}∣Ψ−⟩=21(∣01⟩−∣10⟩) A two-qubit entangled state; usually Alice holds the first qubit and Bob the second.
[Postulates of quantum mechanics] Why is the EPR state 12(∣01⟩−∣10⟩)\tfrac{1}{\sqrt{2}}(|01\rangle - |10\rangle)21(∣01⟩−∣10⟩) entangled?
(a∣0⟩+b∣1⟩)⊗(c∣0⟩+d∣1⟩)=ac∣00⟩+ad∣01⟩+bc∣10⟩+bd∣11⟩(a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle) = ac|00\rangle + ad|01\rangle + bc|10\rangle + bd|11\rangle(a∣0⟩+b∣1⟩)⊗(c∣0⟩+d∣1⟩)=ac∣00⟩+ad∣01⟩+bc∣10⟩+bd∣11⟩ would need ac=0ac = 0ac=0 with ad≠0ad \neq 0ad=0 and bc≠0bc \neq 0bc=0 , which is impossible. ac=0ac = 0ac=0 means a=0a = 0a=0 or c=0c = 0c=0 ; a=0a = 0a=0 makes ad=0ad = 0ad=0 , and c=0c = 0c=0 makes bc=0bc = 0bc=0 .
[Postulates of quantum mechanics] The EPR state ∣Ψ−⟩AB=12∣0⟩A∣1⟩B−12∣1⟩A∣0⟩B|\Psi^-\rangle^{AB} = \tfrac{1}{\sqrt{2}}|0\rangle^A|1\rangle^B - \tfrac{1}{\sqrt{2}}|1\rangle^A|0\rangle^B∣Ψ−⟩AB=21∣0⟩A∣1⟩B−21∣1⟩A∣0⟩B : Alice holds qubit AAA , Bob holds qubit BBB .A measurement is a set of projections {Πi}\{\Pi_i\}{Πi} with ∑iΠi=1 ⁣ ⁣1\sum_i \Pi_i = \mathrm{1\!\!1}∑iΠi=11 ; outcome iii on state ∣ψ⟩|\psi\rangle∣ψ⟩ has probability ∥Πi∣ψ⟩∥2\|\Pi_i|\psi\rangle\|^2∥Πi∣ψ⟩∥2 and leaves the state Πi∣…
Outcomes 0 and 1 with probability 12\tfrac{1}{2}21 each, leaving ∣0⟩A∣1⟩B|0\rangle^A|1\rangle^B∣0⟩A∣1⟩B or ∣1⟩A∣0⟩B|1\rangle^A|0\rangle^B∣1⟩A∣0⟩B ; Bob always gets the opposite of Alice's outcome. Π0∣Ψ−⟩=12∣0⟩A∣1⟩B\Pi_0|\Psi^-\rangle = \tfrac{1}{\sqrt{2}}|0\rangle^A|1\rangle^BΠ0∣Ψ−⟩=21∣0⟩A∣1⟩B , with squared norm 12\tfrac{1}{2}21 . The outcomes are perfectly anti-correlated.
[Postulates of quantum mechanics] The EPR state ∣Ψ−⟩AB=12∣0⟩A∣1⟩B−12∣1⟩A∣0⟩B|\Psi^-\rangle^{AB} = \tfrac{1}{\sqrt{2}}|0\rangle^A|1\rangle^B - \tfrac{1}{\sqrt{2}}|1\rangle^A|0\rangle^B∣Ψ−⟩AB=21∣0⟩A∣1⟩B−21∣1⟩A∣0⟩B : Alice holds qubit AAA , Bob holds qubit BBB .Hadamard basis of C2\mathbb{C}^2C2 : ∣+⟩=12(∣0⟩+∣1⟩)|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)∣+⟩=21(∣0⟩+∣1⟩) , ∣−⟩=12(∣0⟩−∣1⟩)\ |-\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle) ∣−⟩=21(∣0⟩−∣1⟩) .What is ∣Ψ−⟩|\Psi^-…
∣Ψ−⟩=12(∣−⟩∣+⟩−∣+⟩∣−⟩)\boldsymbol{|\Psi^-\rangle = \tfrac{1}{\sqrt{2}}(|-\rangle|+\rangle - |+\rangle|-\rangle)}∣Ψ−⟩=21(∣−⟩∣+⟩−∣+⟩∣−⟩) : the same form, so the outcomes are again perfectly anti-correlated. ∣+⟩∣−⟩=12(∣00⟩−∣01⟩+∣10⟩−∣11⟩)|+\rangle|-\rangle = \tfrac{1}{2}(|00\rangle - |01\rangle + |10\rangle - |11\rangle)∣+⟩∣−⟩=21(∣00⟩−∣01⟩+∣10⟩−∣11⟩) and ∣−⟩∣+⟩=12(∣00⟩+∣01⟩−∣10⟩−∣11⟩)|-\rangle|+\rangle = \tfrac{1}{2}(|00\rangle + |01\rangle - |10\rangle - |11\rangle)∣−⟩∣+⟩=21(∣00⟩+∣01⟩−∣10⟩−∣1…
[Postulates of quantum mechanics] What did Einstein, Podolsky and Rosen call the effect that Alice's measurement of her half of the EPR state 12(∣01⟩−∣10⟩)\tfrac{1}{\sqrt{2}}(|01\rangle - |10\rangle)21(∣01⟩−∣10⟩) immediately fixes the outcome of Bob's, at any distance?
"Spooky action at a distance." If Alice finds ∣0⟩|0\rangle∣0⟩ , Bob's qubit is in state ∣1⟩|1\rangle∣1⟩ , however far away it is.
[Postulates of quantum mechanics] What is shared entanglement, and what classical resource is it compared with?
Many EPR pairs, Alice holding one half of each and Bob the other; a communication resource like a shared secret key. A shared secret key is what the one-time pad uses. Shared entanglement is a correlation between two parts of a system that is stronger than any classical correlation; many quantum protocols use it.
[Postulates of quantum mechanics] Give a physical process that creates an entangled state.
A particle with zero spin decays into two particles: conservation of angular momentum makes their spins opposite, in whichever direction they are measured. Example: the two electrons in the ground state of a helium atom (total spin zero) flying apart have opposite spins, but neither spin direction is known.
[Postulates of quantum mechanics] Which linear algebra concept models each of: a quantum state, evolution, measurement, a composite system?
Quantum state A unit vector ∣v⟩∈Cd|v\rangle \in \mathbb{C}^d∣v⟩∈Cd . Evolution A unitary operation. Measurement Projections adding up to the identity. Composite system The tensor product.
[Polarizing lenses experiment revisited] How are horizontal and vertical photon polarization modelled as vectors, and why that way?
∣H⟩=(1,0)T, ∣V⟩=(0,1)T\boldsymbol{|H\rangle = (1, 0)^{\mathsf{T}}, \ |V\rangle = (0, 1)^{\mathsf{T}}}∣H⟩=(1,0)T, ∣V⟩=(0,1)T : orthogonal, because the two polarizations are complementary.
[Polarizing lenses experiment revisited] How do you prepare a photon in the state ∣H⟩|H\rangle∣H⟩ (horizontal polarization)?
Pass unpolarized light through a horizontally polarizing lens; the photons that come out are in state ∣H⟩|H\rangle∣H⟩ . Likewise a vertically polarizing lens prepares ∣V⟩|V\rangle∣V⟩ .
[Polarizing lenses experiment revisited] Photon polarization as a qubit: ∣H⟩=(1,0)T|H\rangle = (1, 0)^{\mathsf{T}}∣H⟩=(1,0)T (horizontal), ∣V⟩=(0,1)T|V\rangle = (0, 1)^{\mathsf{T}}∣V⟩=(0,1)T (vertical).Which quantum measurement is a horizontally polarizing lens?
{Π←,Π→}\{\Pi_\leftarrow, \Pi_\rightarrow\}{Π←,Π→} with "goes through" Π→=∣H⟩⟨H∣=[1000]\Pi_\rightarrow = |H\rangle\langle H| = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}Π→=∣H⟩⟨H∣=[1000] and "is reflected" Π←=∣V⟩⟨V∣=[0001]\Pi_\leftarrow = |V\rangle\langle V| = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}Π←=∣V⟩⟨V∣=[0001] . A photon in state ∣γ⟩|\gamma\rangle∣γ⟩ goes through with probability ∥Π→∣γ⟩∥2\|\Pi_\rightarrow|\gamma\rangle\|^2∥Π→∣γ⟩∥2 and is then in state ∣H⟩|H\rangle∣H⟩ .
[Polarizing lenses experiment revisited] Photon polarization as a qubit: ∣H⟩=(1,0)T|H\rangle = (1, 0)^{\mathsf{T}}∣H⟩=(1,0)T (horizontal), ∣V⟩=(0,1)T|V\rangle = (0, 1)^{\mathsf{T}}∣V⟩=(0,1)T (vertical), ∣D⟩=(12,12)T|D\rangle = (\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}})^{\mathsf{T}}∣D⟩=(21,21)T (diagonal, 45∘45^\circ45∘ ). A lens polarizing in direction XXX is the measurement whose "goes through" projection is ∣X⟩⟨X∣|X\rangle\langle X|∣X⟩⟨X∣ ; a photon that goes through is in state ∣X⟩|X\ran…
None: ∥∣V⟩⟨V∣H⟩∥2=∥[0001][10]∥2=0\||V\rangle\langle V|H\rangle\|^2 = \left\|\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right\|^2 = 0∥∣V⟩⟨V∣H⟩∥2=∥∥[0001][10]∥∥2=0 . ∣H⟩|H\rangle∣H⟩ has zero overlap with ∣V⟩|V\rangle∣V⟩ . This matches the lab: all ∣H⟩|H\rangle∣H⟩ photons are rejected by the VVV lens.
[Polarizing lenses experiment revisited] A diagonally polarizing lens lets through photons with 45∘45^\circ45∘ polarization ∣D⟩=(12,12)T|D\rangle = (\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}})^{\mathsf{T}}∣D⟩=(21,21)T . What is its "goes through" projection?
Π→=∣D⟩⟨D∣=[12121212]\boldsymbol{\Pi_\rightarrow = |D\rangle\langle D| = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\[2pt] \frac{1}{2} & \frac{1}{2} \end{bmatrix}}Π→=∣D⟩⟨D∣=[21212121] The "is reflected" projection is 1 ⁣ ⁣1−∣D⟩⟨D∣=[12−12−1212]\mathrm{1\!\!1} - |D\rangle\langle D| = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} \\[2pt] -\frac{1}{2} & \frac{1}{2} \end{bmatrix}11−∣D⟩⟨D∣=[21−21−2121] .
[Polarizing lenses experiment revisited] Photon polarization as a qubit: ∣H⟩=(1,0)T|H\rangle = (1, 0)^{\mathsf{T}}∣H⟩=(1,0)T (horizontal), ∣V⟩=(0,1)T|V\rangle = (0, 1)^{\mathsf{T}}∣V⟩=(0,1)T (vertical), ∣D⟩=(12,12)T|D\rangle = (\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}})^{\mathsf{T}}∣D⟩=(21,21)T (diagonal, 45∘45^\circ45∘ ). A lens polarizing in direction XXX is the measurement whose "goes through" projection is ∣X⟩⟨X∣|X\rangle\langle X|∣X⟩⟨X∣ ; a photon that goes through is in state ∣X⟩|X\ran…
14\boldsymbol{\tfrac{1}{4}}41 : 12\tfrac{1}{2}21 pass DDD , and 12\tfrac{1}{2}21 of those pass VVV . ∥∣D⟩⟨D∣H⟩∥2=∣12∣2=12\||D\rangle\langle D|H\rangle\|^2 = |\tfrac{1}{\sqrt{2}}|^2 = \tfrac{1}{2}∥∣D⟩⟨D∣H⟩∥2=∣21∣2=21 , leaving state ∣D⟩|D\rangle∣D⟩ ; then ∥∣V⟩⟨V∣D⟩∥2=02+(12)2=12\||V\rangle\langle V|D\rangle\|^2 = 0^2 + (\tfrac{1}{\sqrt{2}})^2 = \tfrac{1}{2}∥∣V⟩⟨V∣D⟩∥2=02+(21)2=21 . The projections ("is reflected", "goes through") of each lens are written below it.