# Tag Info

## Hot answers tagged born-rule

68

I am late to this party here, but I can maybe advertize something pretty close to a derivation of quantum mechanics from pairing classical mechanics with its natural mathematical context, namely with Lie theory. I haven't had a chance yet to try the following on first-year students, but I am pretty confident that with just a tad more pedagogical guidance ...

37

Why would you ever try to motivate a physical theory without appealing to experimental results??? The motivation of quantum mechanics is that it explains experimental results. It is obvious that you would choose a simpler, more intuitive picture than quantum mechanics if you weren't interested in predicting anything. If you are willing to permit some ...

20

A wave function is a complex-valued function $f$ defined on ${\mathbb R}^1$ (if your electron is confined to a line) or on ${\mathbb R}^2$ (if your electron is confined to a plane) or ${\mathbb R}^3$ (if your electron ranges over three-space), and satisfying $$\int |f|^2=1$$ (where the integral is defined over the entire line or plane or 3-space). Every ...

17

As StephenG mentioned in a comment, the paper you're asking about is the subject of a commentary in arXiv:quant-ph/0509130, by Markus Bier; Li and Li attempt a rebuttal of that comment in Appendix C of v2 of their paper. The comment by Markus Bier is phrased in dry academic language, and there are certain aspects of the phrasing that simply do not fit ...

16

The immediate problem with obtaining the Born rule in the many-worlds interpretation is quite elementary: you can't even begin to attach probabilities to "worlds" (or to events within worlds), in your theory of many worlds, if the theory isn't even clear on what a world is. Physical states according to various interpretations In classical physics, a ...

16

Part of you problem is "Probability amplitude is the square root of the probability [...]" The amplitude is a complex number whose amplitude is the probability. That is $\psi^* \psi = P$ where the asterisk superscript means the complex conjugate.1 It may seem a little pedantic to make this distinction because so far the "complex phase" of the amplitudes ...

13

The theory of quantum mechanics has been developed to explain observations, i.e. measurements. Without observations it is a floating mathematical construct. One of the postulates to connect the mathematics with reality is: To every observable there corresponds an operator which operating on the state function will give an eigenvalue. So the question ...

11

Strictly speaking, the Born rule cannot be derived from unitary evolution, furthermore, in some sense the Born rule and unitary evolution are mutually contradictory, as, in general, a definite outcome of measurement is impossible under unitary evolution - no measurement is ever final, as unitary evolution cannot produce irreversibility or turn a pure state ...

11

Before trying to understand quantum mechanics proper, I think it's helpful to try to understand the general idea of its statistics and probability. There are basically two kinds of mathematical systems that can yield a nontrivial formalism for probability. One is the kind we're familiar with from everyday life: each outcome has a probability, and those ...

11

What I don't get is how he ended up with an equation that is a probability distribution. He ended up with Schroedinger's equation, but he did not think it describes probability. He thought more along the lines of de Broglie idea, that the electron is some kind of wave, and $\psi$ - solution to the equation - expresses mathematically shape and other ...

11

There is a Hilbert space H (probably of dimension 3 or more, as in Gleason's theorem?) equipped with some logical apparatus (the lattice L?). Correct, and the lattice $L(H)$ is that of orthogonal projectors/closed subspaces of a separable complex Hilbert space $H$. As a partially ordered set, the partial ordering $P\leq Q$ relation is the inclusion of ...

10

You should use history of physics to ask them questions where classical physics fail. For example, you can tell them result of Rutherford's experiment and ask: If an electron is orbiting around nucleus, it means a charge is in acceleration. So, electrons should release electromagnetic energy. If that's the case, electrons would loose its energy to collapse ...

10

Well there are multiple reasons, but a very important one is that it can be proven (from the Schrödinger equation) that $$\frac{\mathrm d}{\mathrm dt}\int \mathrm d\boldsymbol x\ |\psi(\boldsymbol x,t)|^2=0$$ so that, if at any moment in time we have $\int \mathrm d\boldsymbol x\ |\psi(\boldsymbol x,t)|^2=1$, this will remain true at any other time. On ...

9

If I would be designing an introduction to quantum physics course for physics undergrads, I would seriously consider starting from the observed Bell-GHZ violations. Something along the lines of David Mermin's approach. If there is one thing that makes clear that no form of classical physics can provide the deepest law of nature, this is it. (This does make ...

9

Those factors usually come out when separating the wave-function for a 3-dimensional problem (like the Hydrogen atom) in its radial and angular parts: $$\psi(r, \theta, \phi) = R(r) Y(\theta, \phi).$$ If you are then interested in the probability of finding the particle (or whatever you are studying) at a distance from the origin between $r$ and $r+dr$ ...

9

To be precise, it is not the complex-valued wave function $\psi(\vec{x})$ that is interpreted as a probability density, but its absolute square $|\psi(\vec{x})|^2$ is, sometimes $\psi(\vec{x})$ is called probability wave to stress the difference. This is significant, if $\psi(\vec{x})$ itself was a density, or if $|\psi(\vec{x})|^2$ itself obeyed an ...

8

Though there are many good answers here, I believe I can still contribute something which answers a small part of your question. There is one reason to look for a theory beyond classical physics which is purely theoretical and this is the UV catastrophe. According to the classical theory of light, an ideal black body at thermal equilibrium will emit ...

8

The classical limit for light is a wave theory. The (quantum) amplitude of the wavefunction wave becomes the (classical) amplitude of the wave (e.g. the magnitude of the electric field), and the (quantum) expected number of photons in a volume becomes the (classical) intensity of the wave (e.g. the squared electric field). Of course it is correct to map the ...

8

The wavefunction being zero at a single point has no physical significance because the wavefunction is a probability density (more precisely, its squared modulus is that), not a probability. All that counts are the integrals of the density over regions of non-zero measure, the values at single points are completely irrelevant. Formally, wavefunctions lie in $... 7 All the key parts of quantum mechanics may be found in classical physics. 1) In statistical mechanics the system is also described by a distribution function. No definite coordinates, no definite momenta. 2) Hamilton made his formalism for classical mechanics. His ideas were pretty much in line with ideas which were put into modern quantum mechanics long ... 7 In quantum mechanics, the amplitude$\psi$, and not the probability$\mid\psi\mid^2$, is the quantity which admits the superposition principle. Notice that the dynamics of the physical system (Schrödinger equation) is formulated in terms of and is linear in the evolution of this object. Observe that working with superposition of$\psi$also permits complex ... 7 You need quantum field theory to describe light as it is clearly a relativistic system. Photons are excitations of the electromagnetic (EM) field which is a spin 1 boson field. In the limit of large numbers of photons the usual classical limit is obtained, i.e. one can describe the field evolution classically to a good approximation. So it is not correct to ... 7 The answer is negative for two distinct reasons. (1) In QM operator means linear operator and the map$\psi \mapsto |\psi(x)|^2$is not linear, evidently. (2) Wavefunctions are elements of$L^2(\mathbb R)$and these elements are defined up to zero-measure sets. I mean that, if$\psi(x) \neq \psi'(x)$for$x\in E$where$E$has zero measure, then$\psi=\psi'...

7

The wave function is the solution to the Schrödinger equation, given your experimental situation. With a classical system and Newton's equation, you would obtain a trajectory, showing the path something would follow: the equations of motion. For a quantum mechanical system you get a wave function, and the rules it obeys over time. With this you can determine ...

7

This is because the values of the wavefunction are complex numbers. Squaring a complex number doesn't necessarily give you a positive number, or even a real one. For example, $i^2=-1$, and $(1+i)^2 = 2i$.

7

You see no formal proof of the Born rule because it is not a theorem of quantum mechanics, it is a postulate. See, for instance, this question and its answers. Normalization has nothing to do with the Born rule, since we can formulate the Born rule for arbitrary non-normalized states $\lvert \psi\rangle,\lvert \phi\rangle$ as  P(\lvert \psi\rangle,\lvert \...

6

There is a paper called Ruling Out Multi-Order Interference in Quantum Mechanics that, I think, answers your question in the negative (within a certain bound anyway). The authors show that the Born rule implies quantum interference comes only in pairs of possibilities (second order interference), and that by relaxing the Born rule one would expect higher ...

6

The use of the word "postulate" in the question may indicate an unexamined assumption that we must or should discuss this sort of thing using an imitation of the axiomatic approach to mathematics -- a style of physics that can be done well or badly and that dates back to the faux-Euclidean presentation of the Principia. If we make that choice, then in my ...

6

There is no integration of the radial part because, as you said yourself, we want the probability of finding the electron somewhere in the spherical shell between $r$ and $r+dr$ from the nucleus. (in a differential shell between $r$ and $r+dr$, and no need to integrate over $r$.)

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