# Tag Info

4

It seems to me that you're looking for the Boltzmann transport equation: $$\frac{\partial f}{\partial t}+\frac{\mathbf p}{m}\cdot\nabla f+\mathbf F\cdot\frac{\partial f}{\partial\mathbf p}=Q+\left(\frac{df}{dt}\right)_{\rm coll}$$ with $f$ the distribution in phase-space, $\mathbf p$ the particle momentum, $Q$ some source term, and the RHS an interaction ...

0

If the speed is random but constrained to 0-1 m/s, you have to guess what is meant. I'm guessing it is any distribution with linear shape, and of course area 1. It could go from 0 to 2, from 2 to 0, from 0.5 to 1.5, or simply be the uniform distribution from 1 to 1. In any case, it's got a Y-intercept (the pdf of speed 0) and a slope. If it has a non-zero ...

-2

In this case, linear refers to the axis for speed being linear (not exponential or logarithmic). A probability density function is described as the normal distribution of an event. See the wiki page.

2

A chi-squared test is used to compare binned data (e.g. a histogram) with another set of binned data or the predictions of a model binned in the same way. A K-S test is applied to unbinned data to compare the cumulative frequency of two distributions or compare a cumulative frequency against a model prediction of a cumulative frequency. Both chi-squared ...

1

Short answer The reason why a physical quantity such as probability is given by $|\Psi|^2$ rather than some other function of $\Psi$ is geometry, namely Pythagoras' theorem. If you have a vector which points from the origin to the $(\hat x,\hat y,\hat z)$ coordinates $(x,y,z)$, then the length $\ell$ is given by $\ell^2=x^2+y^2+z^2$. Why is this the ...

3

Let us consider the famous double-slit experiment with photons. With the usual set-up, we denote the number of photons passing through by $N$ and we will denote the number of photons which hit the film between $y$ and $y + \Delta y$ by $n(y)$. The probability that a photon will be detected between $y$ and $y+ \Delta y$ at a time $t$ is given by: ...

3

I think the answer is "because it works". Early in the development of QM, that interpretation was given to the wave function, and over the decades it has proven to be a useful interpretation. It works. Additionally, the fact that it is possible to define an associated current, and that there is a quantum mechanical expression that guarantees that ...

0

What you are missing is that it is not a viable strategy to get the probability by counting worlds. The only viable way of calculating the probability in MWI is the Born rule. See this paper especially section 9: http://arxiv.org/abs/0906.2718.

0

Your observation is correct. If tossing a coin were a quantum measurement, in the many-worlds interpretation of quantum mechanics (MW), there would be a branch of "worlds" in which the outcome was always heads. This would not violate anything we know about probability or quantum mechanics. The full ensemble of worlds would have the expected binomial ...

0

Without entering the quantum mechanics of the situation, we can see that each toss is a new world. The next toss is another world, so the series of heads do not add in the way you think to make a world of all heads. Each world deserted by each new toss will have the usual probabilities of heads or tails. A world of all heads is possible with sequential ...

1

Note: ChocoPouce's answer is the same as mine but is more mathematical. You have a (spherically symmetric) probability density distribution $\rho$ in space (which we get from the square of the amplitude). The "radial probability density" is roughly the chance that the electron is at a given radius, say $r = 0.1\mathrm{nm}$? In other words, how much of this ...

1

The link uses a spherical shell element which is $4\pi r^2 \mathrm{d}r$ and has the dimension of a volume ($r^2$ is a surface and $\mathrm{d}r$ is a length. The wavefunction of the ground state is spherical, if it weren't the calculation should have been made using a spherical volume element such as $r^2\mathrm{d}r\mathrm{d}\theta\mathrm{d}\varphi$. There ...

0

One finds a full explanation of this formula and much more about the connection between random walks and electric networks in the little gem Random walks and electric networks by Peter G. Doyle and J. Laurie Snell, freely available.

0

Once you measure a single electron's property, it collapses. It is a quantum randomness. Thus there is no classical analogy. There are two kinds of randomness: classical randomness, unknown because we are lazy. (contribute to entropy) quantum randomness, unknown because the God plays dice. (does not contribute to entropy) Classical randomness ...

1

The current density formula you consider is not right, because $\frac{d(\psi\psi^*)}{dt}$ is always 0. The current density is defined as $\bf{j}=(i\hbar/2m)(\psi\nabla\psi^*-\psi^*\nabla\psi)$. In your one dimensional case change $\nabla$ to $\partial/\partial x$, you will get your answer. A more physical definition of current density operator is this: ...

0

Oh, I think I've figured it out... I sort of realised this just as I posted my question : From my earlier intros to quantum mechanics I know that p = mv = $\hbar k$. By simple substitution I can then obtain a final answer of : j = $v\lvert{A}\rvert^{2}$ I'll wait for someone to confirm my line of thinking.

2

Yes, it is possible. The simplest qualitative answer to this is that, at the microscopic level, the electrons in a conductor are dictated by quantum mechanics, which is inherently probabilistic. Velocities and positions are rarely ever totally excluded from a given value; it's just insanely unlikely for a single electron to attain that given value. ...

1

For any function of $x$ and $t$ that depends on the combination $x\pm vt$ (for constant $v$ represents a wave with a fixed shape that travels in the $\mp x$ direction with speed $v$. That is to say, $$x\pm vt={\rm constant}$$ In your wave function, $$\psi(x)=Ae^{i(kx-\omega t)}+Be^{-i(kx+\omega t)}\tag{0},$$ the first term represents a right-moving wave ...

0

The terms in $A$ and $B$ represent waves traveling in opposite directions. Their values are set by the initial conditions of the problem. Some problems will involve waves traveling to the right, others to the left, others a combination of the two. Evidently your book chose to focus on a wave traveling to the right for simplicity. Note that this does ...

4

To add to @Luboš Motl's great answer, I just want to mention the connection to the electric current and electric charge density from electrodynamics, which you may be more familiar with. Note that probability is unitless, so probability density has units of 1/Volume and probability current has units of 1/area*time. These are the same units as electric ...

12

The total probability of all mutually excluding alternatives must always be 100%, so it is conserved. The conservation law in the spacetime tend to be "local", so just like for the charge conservation, we may derive the conservation of the probability in Schrödinger's equation from the local continuity equation $$\frac{\partial \rho}{\partial t} + \mathbf ... 1 You are absolutely right that the limit in which this approximation holds is$$\beta(\epsilon - \mu) \gg 1 \,, which is not trivially the 'high-temperature limit', and indeed looks rather like the low temperature limit. However, it also looks like the limit of large negative $\mu$. If we want to know how temperature will affect the exponent, we need to ...

1

There are two misconceptions in your examples and there is another misconception in the reply to your question by anna v. The first problem concerns what happens when a person chooses whether to turn left or right. If the person in question has a reason to go to a specific place and he has to turn left to do so, then he will turn left with extremely high ...

1

The many worlds interpretation of quantum mechanics takes the Platonic ideal to the extreme Pythagorean one "God always geometrizes": That it is mathematics that creates reality and not reality that is being described by mathematics. They have taken the mathematics of quantum mechanics, a theory we believe describes the microcosm and from which the ...

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