Dork Scratchings

I'm still a bit obsessed with the GHZ experiment.  This is the one where you prepare a GHZ state $\lvert 000 \rangle + \lvert 111 \rangle$ and send one qubit to each of 3 protagonists: Alice, Bob, and Charlie.  (I will be ignoring shared normalizer constants throughout this post, as I find they don't add anything to the understanding.) If you do the maths it turns out that when all three choose to measure in the $\lvert + \rangle$, $\lvert - \rangle$ basis (shorthand for $\lvert 0 \rangle + \lvert 1\rangle$ and $\lvert 0 \rangle - \lvert 1\rangle$) then they are guarranteed to get a parity zero result.  On the other hand if only one measures in this basis and the other two measure in the $\lvert +i\rangle$, $\lvert -i\rangle$ basis (shorthand for $\lvert 0 \rangle + i\lvert 1\rangle$ and $\lvert 0 \rangle - i\lvert 1\rangle$) they are guarranteed to get a parity one result.  As I showed in an earlier post this appears to be incompatible with the outcomes being prede...

It has occurred to me that the cartoon I drew two days ago will appear to most people as absurd: Alice measures one half of an entangled pair and instantly splits into two versions of herself!  Bob measures the other half and also splits, but each Alice is paired with one of the Bobs in such a way that, should they meet later to discuss their results, everything remains consistent. Yet if you replace Alice and Bob with two qubits - to get a quantum circuit like the one below - no quantum computing expert would find this outcome absurd. The Einstein Podolsky Rosen experiment in circuit form with non-classical Alice and Bob Alice is qubit q0 and Bob is q3 .  Qubits q1 and q2 are entangled into the Bell state $(\lvert 01 \rangle + \lvert 10 \rangle)/\sqrt 2$ and then q1 is shared with Alice, and q2 with Bob.  In the standard (collapse-based, non-multiversal) description Alice and Bob measure q1 and q2 , collapse the q1q2 wavefunction, and find that one gets 0 while the...

I just read this article on phys.org which claimed to "resolve the paradox" that entanglement appears to imply faster than light communication.  I don't think it actually resolves the paradox, it just side steps it.  Here's how the argument goes (and I paraphrase): Alice and Bob share a pair of entangled qubits in a Bell state $\left( \lvert 01 \rangle + \lvert 10 \rangle \right)/\sqrt{2}$. When Alice measures hers it "collapses" the pair's state immediately even if Bob is a great distance away.  However, this does not imply faster than light communication because, although Alice now knows what Bob will measure, Bob does not know until he makes a measurement himself, or receives a message from Alice which can only travel at the speed of light. This argument misses the point somewhat.  If the state of both qubits changes instantly when Alice measures her qubit (e.g from $\left( \lvert 01 \rangle + \lvert 10 \rangle \right)/\sqrt{2}$ to $\lvert 01 \rangl...

Dork Scratchings has finally decarbonised!  The first to go was the fossil fuel investing pension (this has been replaced by PensionBee's fossil fuel free plan ), then went the petrol car (replaced by an electric 208), and this week we got rid of the boiler. The new heating system is a Mitzubishi heat pump which heats water by extracting heat from cold air.  The laws of Thermodynamics set a limit on the amount of heat energy that can be produced per unit of electrical energy and the formula is $$ \frac{T_H}{T_H-T_C} $$ Where $T_H$ is the temperature of the hot water ($50^\circ C$ in our case), and $T_C$ is the temperature of the cold air outside the house.  Note that the temperatures have to be in Kelvin to make this work, and the value drops as the outside temperature drops.  For example, for $50^\circ C$ water, it is 12.9 when the outside temperature is $25^\circ C$ but only 6.5 when the outside temperature is $0^\circ C$.   The heat produced per unit of ...

  Curated by Dorkscratchings, 2024

Why is it tides are semi-diurnal ? That is, why do they occur every 12 hours when they are caused by the gravitational pull of a moon that we turn to face once a day?  The straightforward answer is that water bulges out on both the side nearest and the side furthest from the moon.  (We'll ignore the Sun for simplicity, but adding it in doesn't change anything.) The Earth rotates while the bulges remain in place, causing tides 12 hours apart.  However, this doesn't explain why there should be two bulges.  The reason becomes clear when you change the frame of reference.  Instead of thinking about a frame in which both the moon and the Earth rotate, set the origin to be the centre of mass of the two objects, and choose a rotating frame in which the moon is stationary.  In this frame there is a "fictional" centrifugal force of $\omega^2 r$ which combines with the gravitational force from the moon, $GM_{\text{moon}}/(r-r_{\text{moon}})^2$ The two forces match a...

In 1964 John Stewart Bell proposed an experiment to determine whether the results of quantum measurements were truly random, or governed by hidden variables, i.e. state that exists prior to the measurement, but which we don’t have access to. The experiment involved creating a large number of EPR pairs, and firing them at two observers, Alice and Bob, who measure their photon’s polarisation, choosing the $\updownarrow$ direction or the $\nearrow\llap\swarrow$ direction at random. Determining the result of the experiment involves doing a complex statistical calculation to see if something called Bell’s inequality is satisfied or violated. The Bell experiment was first performed by in 1982 by Alain Aspect, and the result, as most commonly interpreted, is that hidden variables can only exist if Quantum Mechanics is non-local, i.e. if it supports faster-than-light causality! Some time after Bell proposed his experiment, Greenberger, Horne, and Zeilinger suggested an alternativ...

In this post I argue that economic growth is not the normal state of affairs, it is a blip caused by the discovery of a one-off resource of non-renewable energy.   The logistic equation $$ \frac{dy}{dt} = y(1-y) $$ Here's a simple model.  Humans discover an exploitable but limited resource and start consuming it.  The amount consumed, $y(t)$, is a function of time.  The general form of the equation is $\frac{dy}{dt} = \alpha y (\beta - y)$, but if you choose the right units $\alpha$ and $\beta$ both become $1$. Why should this work, in principle? Early on, the factor $1-y$ is approximately $1$ and can be ignored.  So the model states that annual consumption $dy/dt$ starts off proportional to $y$.  In other words, $y$ grows exponentially at first.  This could happen if exploiting the resource enables further exploitation of the resource.  For example, suppose a few humans are shipwrecked on an island with 1000 trees: they take ages to cut down the ...

My first neural net Like everybody else I've been playing a lot with ChatGPT recently and I'm gobsmacked by how good it is!  This has led me to start researching how exactly these things work.  In particular I wanted to understand how neural nets - a core component of the technology - are trained. Running a neural net is simple.  The neural net consists of layers of nodes connected by weights and biases.  The first layer is an input layer and setting the activation levels of each node in that layer causes the next layer to adopt values determined by the weights and biases.  The second layer determines the activation levels in the next layer and so on until the final layer, which is interpreted as the output. 

    I was given a new version of Dobble for Xmas called Dobble Connect.  Dobble consists of stack of cards with pictures on it where each pair of cards has one match, and each pair of symbols appears on one card.  This is a new version which has 91 symbols in total, 10 symbols per card, and 91 cards.  The fingerprint of the original Dobble was (57 symbols, 8 symbols, 57 cards). An additional innovation in Dobble Connect is that the cards are hexagonal and grouped into a colour per player.  Players build a tiling on the tabletop, placing cards next to each other as soon as they spot a matching symbol between their top card and one of the cards already placed.  The first player to get four of their own colour in a row wins. How is it possible for each pair of cards to have one and only one symbol in common, and for each pair of symbols to appear on one and only one card?  Does that have something to do with the strange number 91?  The answer i...

Rolling marbles  Here's a puzzle: Suppose you have N identical marbles rolling along a one dimensional table-top.  Each marble is randomly rolling to the left or to the right, all with the same speed.  Collisions are elastic, which means the marbles just change direction.  What is the maximum amount of time before all the marbles have rolled off the table? 8 marbles with speed 1 on a table of length 1 Answering this question is really difficult if you simply pick an individual marble and try to work out how long it might stay on the table as it bounces back and forth.  But there's a simpler way to look at it. Prior to each collision you have one marble rolling to the left and one to the right, and afterwards you still have one rolling to the left and one to the right.  If we swap labels following each collision then the labels never change direction.  Now it's easy to see that the answer is the same whether there are 100 marbles or just one.  And ...

Contents Intro Simple stratified model Alternative convection model Tropospheric temperature gradient Approach used in convection model Calculation of Radiative Forcing from CO₂ increase Calculation of Radiative Forcing from H₂O increase Calculation of zero-feedback ECS Effect of including feedbacks Effect of water vapour Effect of cloud cover change Effect of sea ice loss Calculation of ECS including feedbacks Limitations One more thing: Changes to Troposphere depth Intro The goal of this post is to see if we can come up with an estimate of Equilibrium Climate Sensitivity using a simple model in which the atmosphere is treated as well-mixed.  By "simple" I mean that using the model will not require any advanced mathematics or computation, but will still be realistic enough to come up with an estimate within the likely range of 2.5 - 4.0°C predicted by the IPCC Assessment Report 6 .  I will try to take as little on trust as possible and show how the results are arrived at....