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$\newcommand{\ket}[1]{\lvert #1 \rangle}$There is no such thing as "looking like a collapsed wavefunction", even if you believe there's collapse. Let's go to the finite dimensional case and have a simple two-level spin system, that is, our Hilbert space is spanned by, e.g., the definite spin states in the $z$-direction ...


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If a given state is an eigenstate of a particular observable then that observable has a standard deviation of 0, however that says nothing about the distribution of any other observables. The extreme example of this are the eigenstates of position and momentum; a momentum eigenstate is represented by delta function in momentum space but in space it is ...


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Most papers on quantum mechanics don't explain issues like interpretation clearly and non-locality clearly. The most notable exceptions to this are David Deutsch and to a lesser extent David Wallace. "The Fabric of Reality" by David Deutsch is a popular book that explains quantum mechanics, see especially chapter 2. See also "The Beginning of Infinity" by ...


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Lectures on Quantum Theory: Mathematical and Structural Foundations by Chris Isham is a thin, easy to read book. The first 6 or so chapters are a simple introduction to quantum mechanics, but from about chapter 7 or 8 he goes into the Quantum Measurement problem and various interpretations and their associated difficulties. He also discusses Bell's Theorem, ...


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First of all, I'm not a physics specialist. In my opinion, the reason why I'm typing this text at this second and the next one is that we are working inside another parameter. In other words, we are dealing with a "computer" object that is made up of other quantum objects, but that molecules are working as a computer. Once we catch computer, it is always ...


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Nothing prevents the electron's spin from being measured along a particular axis, and then subsequently measured along an axis perpendicular to the first. In this situation, however, the spins along the perpendicular axes would not be known simultaneously, so the uncertainty principle would not be violated. As an example, say that we perform our own ...


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"By the uncertainty principle" is the answer. In more detail, let's say we're talking about x and y axes. The first measurement puts the electron into an eigenstate of the spin X observable (the question of how it does this is the quantum measurement problem). Whichever of the two X eigenstates this "collapse" ends up in, it not an eigenstate of the spin Y ...


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Your linked article ('How do probabilities emergeā€¦') only seems to explain why each universe is internally consistent and acts according to what we would statistically expect. The argument pretty much goes that it is of course entirely possible to get weird behavior, but it is precisely as likely to occur as in a single universe obeying the known physical ...


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Suppose the particles are initially in the (entangled) state $$A\otimes B+C\otimes D$$ where $A$ and $C$ are position eigenstates for particle 1 and $B$ and $D$ are position eigenstates for particle 2. Note that this state is the same as $$X\otimes Y+Z\otimes W$$ where $X=(1/2)(A+C)$ and $Y=(1/2)(A-C)$ are momentum eigenstates for particle 1 and ...


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I think the key point that you're missing is that as soon as you make a measurement, the entanglement between the two particles is broken. It should also be noted that the original particle also obeyed the uncertainty principle, and that at a quantum level there is no direct relationship between position, momentum and time. Another confusing factor is that ...


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Ok. Let's suppose that the initial state of the two particle are an eigenstate of the momentum operator (momentum is well defined). Quantum mechanics tell us that the position of the center of mass is not well defined. If we measure the position of the particle 1 (electron), then we do two thing in the system: We apply a measurement in a part of the whole ...


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Since you're an electronics student, I'll speak your language. Think of momentum and position as parameters in time and frequency domain of a signal rather than classical observable that are well defined. If you do so, you can easily realize that your frequency isn't well defined if you don't do an infinitely long measurement. This is simply due to the wave ...


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The problem is that you propose to make the two measurements "at time $t$ simultaneously". Measuring the particle's momentum cannot be done instantaneously; the more precisely you want to measure it, the longer the minimum required observation time becomes. (Rougly speaking that's because knowing the particle's momentum is equivalent to measuring its ...


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What is wrong with this logic is that you are supposing a particle has simultaneously well-defined position and momentum. This is not true - a state localized in real space is delocalized in momentum space, and vice versa. The classical conservation laws hold on the quantum level as operator laws, not as laws on the states.


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When you measure an observable over a certain state, any possible outcome lies within the physical spectrum of the observable itself (which can be shown to coincide with the algebraic notion of spectrum for linear operators). So after the measurement has given you a value, say, $\lambda$, any other measurement on the system will give you $\lambda$ again with ...


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This phenomenon is called the collapse of the wave function. It is one of the tenets of the Copenhagen interpretation of quantum mechanics. The eigenstates $|\xi_i\rangle$ of the $\Xi$ operator form a complete set. From linear algebra we have $$I=\sum_i|\xi_i\rangle\langle \xi_i|$$ where $I$ is the identity operator. We apply this to the state vector ...


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It does not matter whether or not you look. What matters is that it is possible to look in principle, because the quantum particle interacts with the LED and therefore becomes entangled with it (your unitary matrix acts on both the particle and LED). So the outcome will always be the same as experiment 1. The technical term for this is "which-way" ...


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An observable that has a definite value is an eigenvalue of the operator. If $A$ is a hermitian matrix of which $|\psi\rangle$ is an eigenstate, we have $$\tag{1}A|\psi,a\rangle=a|\psi,a\rangle$$ You asked about the wave function, not the state vector, though. We can still get all the information we need from the wave function $\psi(x)=\langle ...


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I post this answer to check my understanding. Imagine a wavefunction in 1 dimensions with a known energy and momentum it's wavefunction will be: $$\Psi(x, t) = e^{i(kx-\omega t)} = e^{i(px-E t)/\hbar}$$ With some calculus and algebra you can derive the momentum operator and get this: $$-i\hbar \partial_x \Psi = p \Psi$$ There $-i\hbar \partial_x$ is ...


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I'd say, "observer" is an arbitrary entity which converts quantum information into classical. Classical information is, roughly, anything that can be duplicated without distortion and transmitted. The need for classicality is anthropic: we are conditioned by evolution to share information for survival and to value "rational thought" whose main defining ...



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