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in quantum physics position X and momentum P can’t be simultaneously measured. But, linear combinations aX+bP 𝘢𝘳𝘦 observable, and their distributions can be used to back out the joint X,P probability density. The Wigner distribution but, it is 𝘸𝘦𝘪𝘳𝘥

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Do you actually know what convex optimization is in the geometric, guarantee-theoretic sense or have you only met it through solvers and loss curves? Convexity is rare comfort in optimization...there are no spurious local minima, no surprise traps, and inequalities you can use like tools instead of prayers. So, what is this convexity? Let x = (x₁, x₂) and let f(x) be convex. Plot the surface z = f(x). Pick a contact point x₀. The local slope is the gradient p = ∇f(x₀). That p is exactly the data that defines the supporting plane: z = f(x₀) + p · (x − x₀). Thus, f is said to be convex because for every x, f(x) ≥ f(x₀) + p · (x − x₀). So the plane at x₀ can slide under the surface, but it never slices through it. Not near the point...everywhere. Now for here is the interesting part: The slope becomes a coordinate system! Rewrite the same plane as z = p · x − b, where b is the offset. Because the plane passes through (x₀, f(x₀)), the offset is forced to be b = p · x₀ − f(x₀). And that number isn’t just geometry trivia. It’s the convex conjugate: f*(p) = sup over x ( p · x − f(x) ). At a differentiable contact point, the supporting plane touches f tightly enough that the supremum is achieved at x₀, giving the identity f*(p) = p · x₀ − f(x₀) when p = ∇f(x₀). So one moving contact point gives two linked readouts: primal position x₀ dual position (slope) p = ∇f(x₀) dual offset f*(p) One surface. Two worlds. #ConvexOptimization #Optimization #MachineLearning #SignalProcessing #AppliedMath #Engineering

Mathelirium

38,506 views • 5 months ago

Quantum mechanics has a reputation for being mystical mainly because people skip the rules and jump to interpretations. In this lecture series, we’re doing the opposite. We start from the rules, follow the algebra, and let the picture be the calculation. Classical Probability Theory combines alternatives by adding their probabilities. Quantum Theory combines them one step earlier…add complex amplitudes first, then square at the end. That swap in order is everything. Expand |a₁ + a₂|² and you don’t just get |a₁|² + |a₂|²…you get a cross-term, 2 Re(a₁ a₂*). Its sign is set by phase, so the same two contributions can reinforce or cancel. Interference is just the algebra of squaring a sum. In the 3D render, the surface height is proportional to |a(x)| (so peaks become bright bands after squaring), while the surface skin is colored by the local phase arg(a(x)). As the phase knob φ(t) is swept on path 2, the cross-term oscillates, and you literally watch the interference ridges slide across the screen. We model a detector screen with coordinates x in R² (think x = (x,y)). A quantum state assigns a complex amplitude a(x). The rule for outcomes is p(x) = |a(x)|² Now the key situation: two coherent alternatives contribute to the same outcome x. Let their amplitudes be a₁(x) and a₂(x). Quantum says a(x) = a₁(x) + a₂(x) So the probability density becomes p(x) = |a₁(x) + a₂(x)|² Expand it (this is the whole episode): p(x) = (a₁ + a₂)(a₁* + a₂*) = |a₁|² + |a₂|² + a₁ a₂* + a₁* a₂ = |a₁|² + |a₂|² + 2 Re(a₁ a₂*) That last term is the interference term. It can be positive or negative. To see phase explicitly, write each contribution in polar form: a₁(x) = r₁(x) exp(i θ₁(x)) a₂(x) = r₂(x) exp(i θ₂(x)) Then a₁ a₂* = r₁ r₂ exp(i(θ₁ − θ₂)) So the cross-term is 2 Re(a₁ a₂*) = 2 r₁ r₂ cos(θ₁(x) − θ₂(x)) That’s the fringe engine: p(x) = r₁² + r₂² + 2 r₁ r₂ cos(Δθ(x)) Now the phase knob we animate: Add a controllable phase shift φ to path 2: a₂(x) → a₂(x) exp(i φ) Then Δθ(x) → Δθ(x) − φ, so p(x; φ) = r₁² + r₂² + 2 r₁ r₂ cos(Δθ(x) − φ) As φ changes smoothly, the bright/dark pattern slides continuously. Same setup, same geometry, same magnitudes r₁,r₂, only phase changed. #QuantumMechanics #WaveInterference #ComplexAmplitudes #DoubleSlit #Physics #Mathematics

Mathelirium

81,405 views • 5 months ago

Lecture 2 on our Quantum Mechanics Series Schrödinger’s equation doesn’t start from mystery. It starts from a very specific bet…the state of a particle is a complex field ψ(x,t), and whatever dynamics we write down must move ψ forward in time in a way that preserves total probability. We ask a basic question…what equation should ψ satisfy so that |ψ|² behaves like a conserved density, the way mass density does in fluid flow? What is ψ? Think of ψ(x,t) as the amplitude assigned to “the particle is at position x at time t”. It’s not a probability. It’s the object you add first, and only at the end do you square p(x,t) = |ψ(x,t)|² Because ψ is complex, it has magnitude and phase. Write it in polar form ψ(x,t) = r(x,t) exp(i θ(x,t)) Then r² = |ψ|² is the density, and θ will end up controlling flow (the probability current). Where does Schrödinger’s equation come from? Start with two empirical inputs about waves and particles: E = ħ ω p = ħ k Here ħ (“h-bar”) is Planck’s constant divided by 2π. It’s the unit conversion factor between the wave description (frequency ω, wavevector k) and the particle description (energy E, momentum p). In units, ħ has units of joule-seconds, so multiplying ω (1/seconds) gives energy (joules), and multiplying k (1/meters) gives momentum (kg·m/s). It’s the number that tells you how much energy or momentum you get per unit frequency or wavenumber. A plane wave with angular frequency ω and wavevector k is ψ(x,t) = A exp(i(k·x − ω t)) Now notice what derivatives do to this wave: ∂ψ/∂t = −i ω ψ ∇ψ = i k ψ ∇²ψ = −|k|² ψ Multiply those identities by ħ: i ħ ∂ψ/∂t = ħ ω ψ = E ψ −i ħ ∇ψ = ħ k ψ = p ψ −ħ² ∇²ψ = ħ² |k|² ψ = p² ψ So for plane waves, the operators Ê = i ħ ∂/∂t p̂ = −i ħ ∇ act like energy and momentum! Now use the classical nonrelativistic energy relation: E = p²/(2m) + V(x) This is bookkeeping for a particle moving slow enough that relativity can be ignored. The term p²/(2m) is kinetic energy. If p = mv, then p²/(2m) = (m²v²)/(2m) = (1/2)mv². The term V(x) is potential energy. It depends on position because forces come from spatially varying energy. A slope in V pushes the particle. Examples: for a charged particle in an electric potential φ(x), V(x) = q φ(x). Near Earth, V(z) = mgz. The point is total energy equals kinetic plus potential. Turn that into an equation for ψ by replacing E and p with the operators above: Ê ψ = (p̂²/(2m) + V) ψ Compute p̂² = (−i ħ ∇)·(−i ħ ∇) = −ħ² ∇², so we get i ħ ∂ψ/∂t = ( −ħ²/(2m) ∇² + V(x) ) ψ That is the time-dependent Schrödinger equation. The derivation here is a controlled heuristic: we matched the plane-wave identities to the measured relations E = ħω and p = ħk, then imposed the same energy bookkeeping as classical mechanics. Why this equation is the right kind of rule If ψ is the state, we need a rule that preserves total probability: ∫ |ψ(x,t)|² dx = 1 Schrödinger evolution does. You can see it by deriving a continuity equation. Let ρ(x,t) = |ψ|² = ψ* ψ. Take a time derivative: ∂ρ/∂t = ψ* ∂ψ/∂t + ψ ∂ψ*/∂t Use Schrödinger and its complex conjugate: ∂ψ/∂t = (1/(i ħ)) ( −ħ²/(2m) ∇²ψ + Vψ ) ∂ψ*/∂t = (−1/(i ħ)) ( −ħ²/(2m) ∇²ψ* + Vψ* ) Plug in. The V terms cancel exactly, and what remains can be rearranged into a divergence: ∂ρ/∂t + ∇·j = 0 where the probability current is j = (ħ/(2mi)) ( ψ* ∇ψ − ψ ∇ψ* ) This is the best way to explain ehat ψ is: |ψ|² behaves like a conserved density, and the phase of ψ is what drives the current j. So in this series, ψ isn’t a slogan. It’s the object whose modulus squared is the density, whose phase generates flow, and whose time evolution is fixed (up to V) by matching wave relations to energy bookkeeping: i ħ ∂ψ/∂t = ( -ħ²/(2m) ∇² + V ) ψ #QuantumMechanics #SchrodingerEquation #WaveFunction #BornRule #Physics #MathematicalPhysics

Mathelirium

40,809 views • 5 months ago

Why Does Quantum Mechanics Use a Complex Wavefunction? Schrödinger’s equation doesn’t start from mystery. It starts from a very specific bet. The state of a particle is a complex field ψ(x,t), and whatever time-evolution rule we choose has to move ψ forward while preserving total probability. So the basic question is simple. What equation should ψ satisfy so that |ψ|² behaves like a conserved density, the way mass density does in fluid flow? What is ψ? Think of ψ(x,t) as an amplitude attached to the statement the particle is at position x at time t. It’s not a probability. It’s the thing you add first, and only at the end do you square it: p(x,t) = |ψ(x,t)|² Because ψ is complex, it has magnitude and phase. Write it as ψ(x,t) = r(x,t) exp(i θ(x,t)) Then r² = |ψ|² is the density, and the phase θ ends up controlling the flow through the probability current. Where does Schrödinger’s equation come from? Start with two empirical inputs that tie waves to particles: E = ħ ω p = ħ k Here ħ is Planck’s constant divided by 2π. It’s the conversion factor between frequency and energy, and between wavenumber and momentum. A plane wave with angular frequency ω and wavevector k is ψ(x,t) = A exp(i(k·x − ωt)) Now watch what derivatives do to this wave: ∂ψ/∂t = −i ω ψ ∇ψ = i k ψ ∇²ψ = −|k|² ψ Multiply by ħ and you get: i ħ ∂ψ/∂t = ħ ω ψ = E ψ −i ħ ∇ψ = ħ k ψ = p ψ −ħ² ∇²ψ = ħ² |k|² ψ = p² ψ So for plane waves, the operators Ê = i ħ ∂/∂t p̂ = −i ħ ∇ act like energy and momentum. Now bring in the classical, nonrelativistic energy bookkeeping: E = p²/(2m) + V(x) Kinetic plus potential. That’s it. Turn it into an equation for ψ by replacing E and p with the operators above: Ê ψ = (p̂²/(2m) + V) ψ Since p̂² = (−i ħ ∇)·(−i ħ ∇) = −ħ² ∇², this becomes i ħ ∂ψ/∂t = ( −ħ²/(2m) ∇² + V(x) ) ψ That’s the time-dependent Schrödinger equation. This derivation is a controlled heuristic. Match the plane-wave identities to the measured relations E = ħω and p = ħk, then impose the same energy bookkeeping you trust in classical mechanics. Why this is the right kind of rule If ψ is the state, we need a rule that preserves total probability: ∫ |ψ(x,t)|² dx = 1 Schrödinger evolution does, and you can see it by deriving a continuity equation. Let ρ(x,t) = |ψ|² = ψ*ψ. Differentiate: ∂ρ/∂t = ψ* ∂ψ/∂t + ψ ∂ψ*/∂t Use Schrödinger and its complex conjugate. The potential terms cancel, and what’s left can be rearranged into ∂ρ/∂t + ∇·j = 0 with probability current j = (ħ/(2mi)) ( ψ* ∇ψ − ψ ∇ψ* ) That’s the cleanest way to say what ψ is. |ψ|² behaves like a conserved density, the phase drives a current, and the time evolution is fixed, up to V, by combining wave relations with energy bookkeeping: i ħ ∂ψ/∂t = ( −ħ²/(2m) ∇² + V ) ψ #QuantumMechanics #SchrodingerEquation #WaveFunction #BornRule #Physics #MathematicalPhysics

Mathelirium

19,836 views • 4 months ago

Warmup to Statistical Mechanics What Exactly is a Hamiltonian A System? In ordinary Mechanics, you might begin with position and velocity. Hamiltonian Mechanics rewrites the same motion in a different language. Instead of position and velocity, it uses position and momentum. We write the position variables as q and the momentum variables as p. Then the full state of the system at one instant is (q, p) That pair is one point in phase space. Why do we do this? Because in these variables, the equations of motion take a remarkably clean form. Everything is generated by one single function, the Hamiltonian H(q, p) and in the simplest cases this Hamiltonian is just the total energy written in terms of position and momentum. So if you know H, you know the dynamics. You might wonder, but how can one function generate motion? The rule is dqᵢ/dt = ∂H/∂pᵢ dpᵢ/dt = −∂H/∂qᵢ These are Hamilton’s equations. Now read them slowly 😄 The rate of change of position comes from differentiating H with respect to momentum. The rate of change of momentum comes from differentiating H with respect to position, with a minus sign. This constitutes the whole engine. A simple example makes this less abstract: Take one particle of mass m moving in a potential V(q). Then the Hamiltonian is H(q, p) = p²/(2m) + V(q) The first term is kinetic energy. The second term is potential energy. Now apply Hamilton’s equations. First, dq/dt = ∂H/∂p = p/m So momentum tells you how position changes. Second, dp/dt = −∂H/∂q = −dV/dq Thus, momentum changes because of force. If you now combine these two equations, you recover ordinary Newtonian mechanics. Since p = m dq/dt, we get m d²q/dt² = −dV/dq So, Hamiltonian mechanics is not a different theory. It is the same mechanics, written in a form that exposes its geometric structure much more clearly. The animation The full 3D surface is the Hamiltonian itself, the energy landscape H(q, p). The floor underneath is phase space, marked by energy contours and the local flow field. The bright moving point is one actual state (q(t), p(t)) evolving under Hamilton’s equations. Its trail shows that the motion is not arbitrary. It is guided everywhere by the geometry of the same single function H. The render is doing more than illustrating a particle moving, it is showing how one function organizes the whole phase-space motion. The math breakdown: Start with one degree of freedom. The state is described by position q and momentum p. So the system lives in a two-dimensional phase space with coordinates (q, p) Now choose a Hamiltonian H(q, p) Think of H as the energy function. In many standard systems, H(q, p) = kinetic energy + potential energy For a particle of mass m in a potential V(q), this becomes H(q, p) = p²/(2m) + V(q) Hamilton’s equations say dq/dt = ∂H/∂p dp/dt = −∂H/∂q Now substitute this specific H. First compute the p derivative: ∂H/∂p = ∂/∂p (p²/(2m) + V(q)) = p/m So dq/dt = p/m Now compute the q derivative: ∂H/∂q = ∂/∂q (p²/(2m) + V(q)) = dV/dq So dp/dt = −dV/dq These two first-order equations completely determine the motion. Now, connect this back to Newton’s law. From dq/dt = p/m we get p = m dq/dt Differentiate both sides with respect to time: dp/dt = m d²q/dt² But Hamilton’s second equation gives dp/dt = −dV/dq So , together they imply m d²q/dt² = −dV/dq This is exactly Newton’s second law for motion in the potential V(q). Thus, Hamilton’s equations do not replace mechanic, they reorganize it. #HamiltonianMechanics #PhaseSpace #ClassicalMechanics #MathematicalPhysics #DifferentialEquations #Mathematics #Physics

Mathelirium

50,370 views • 2 months ago