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YES. There is a probability that a single photon that matches the energy gap of a molecule can pass through the electron probability cloud without being absorbed. But that probability is very low. The higher chances are that the photon will be absorbed. The reason behind this is that: Molecules are not isolated. There are innumerable molecules even in a ...


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A wavefunction is a function that takes a configuration of all the particles in the universe (if just one particle is in a different place it is a different configuration) and assigns a complex number to that configuration. If you take the square of the length of that complex number you get something people like to call a probability-density. An X-density ...


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The lower limit of your third integral should be $T - t_2 - t_1$. (I assume that $t' = t_1$ and $t'' = t_2$, otherwise the integrals don't make sense.)


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Ref.1 is already in eq. (3.1) considering a functional integral over the scalar field $\phi:M\to \mathbb{R}$. Here $M$ is spacetime. For a rigorous treatment of functional integrals, Ref. 1 points in the beginning of Section 3 to its Ref. [3.2]. In this answer we will just take an intuitive heuristic approach, and try to construct the functional integral as ...


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TL;DR: Substitution inside the delta function yields a Jacobian factor $$ \tag{1} \delta(f(v))~=~ \sum_{v_{(0)},f(v_{(0)})=0 }\frac{1}{| f^{\prime}(v_{(0)})|} \delta(v-v_{(0)}). $$ Here the sum is over the zeroes $v_{(0)}$ of the function $f(v)$. Let us for simplicity consider velocity $v$ rather than momentum $p=mv$. So energy conservation $$\tag{2} ...


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Probability is the chances of occurrence of an event, say for example the event is to find an electron around the nucleus of an atom. If the atom has an electron around the nucleus then the probability of finding the electron around the nucleus is one(1). but if the electron is to be sort some place away from the vicinity of the nucleus then the probability ...


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For the second case you want to divide by the force, not the velocity. You are basically computing what fraction of the time you spend at a particular point in phase space. However what you have is a probability density. So $P(z,p)$ is something you multiply by a volume in phase space to get a probability. A correct way to get it would be to consider a ...


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Given how weak gravity is compared to the other fundamental forces, in practical cases where we do quantum mechanics (subatomic, atomic, molecular physics, solid state, etc.) gravity is utterly negligible. Once we try to work in a regime (such as near a micro-black hole) where gravitational forces are comparable with other forces you are into the regime of ...


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I think you have a few misconceptions here. You start by talking about the particles in the beam "not interfering with each other" so the "wave function of each particle is lambda/p". There are at least two problems with this statement. I'll take the last part first. It looks like you are confusing "wave function" with "wave length". The wave function ...


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The anthropic principle will always guarantee that no universal catastrophe is ever experienced directly Even if the universe had a probability of 99.99999999% of decaying into a false vacuum a femtosecond after the big bang, none of these universes have observers that ponder their existence and post to Physics.SE, so the fact that we are here already skews ...


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This answer is in order to clarify what the physics is behind the scares for LHC safety and to point out that your statistical analysis is on wrong premisses. High energy cosmic rays permeate the universe, not only planets but all the stars with their order of magnitude greater masses and crossectional areas. If cosmic rays could seed black holes there ...


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Just to elaborate on Chris White's point that you can use the same argument to doubt the safety of anything, here's a slight reworking of the argument you give that shows why we should avoid breathing air: So, basically, from those two facts: For [hundreds of thousands] of years [humans have breathed air]. [There are no recorded incidents of ...


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Imagine the following scenario: out of all planets with life, 99% are destroyed by some events caused by cosmic rays before appearence of scientists on these planets who will conduct experiments at particle accelerators. Except we still have the planets in our solar system intact, and so apparently do thousands of other solar systems.


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Your reasoning demonstrates precisely why formal logic alone is insufficient to study nature. In particular, it lacks the ingredient of inductive inference that is a cornerstone of empirical science. A cosmic ray striking the Earth is not some random act of the gods that can have any imaginable consequence whatsoever. It is a cosmic ray striking the Earth. ...


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Okay, so, let's get this thing going right. The property you're talking about is called by different names that you can google: e.g. that the Hamiltonian matrix must be self-adjoint (where the adjoint is the conjugate transpose) or Hermitian. The easiest property to prove about Hermitian matrices is that they have real eigenvalues, for their defining ...


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What he actually is referring to is the mathematical fact that in order for the standard norm (length) of $n$-tuple $[C_j]_{j=1,2,...n}$ to be preserved during evolution determined by the equation $$ \frac{dC_i}{dt} = -\frac{i}{\hbar} \sum_j H_{ij} C_j $$ where $H_{ij}$ is time-independent complex matrix, this matrix needs to be hermitian, i.e. obey the ...


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If that property doesn't hold then $H$ is not Hermitian. This means its eigenvalues could be complex and the time evolution operator won't be unitary: $U(t)U(t)^\dagger=e^\frac{iHt-itH^\dagger }{\hbar} \neq1$ If it's not unitary, then it will change the length of your state vectors which you're evolving. Since the length of these($\sum_i |C_i(t)|^2$) ...


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In response to your edit: The real part of $(A^*(S)A(T))$ is equal to the real part of $(A(S)A^*(T))$, since they are just complex conjugates of each other. So he could have written either one. Concrete example: let $(A^*(S)A(T))=a+ib$ for some $a,b$. Then $(A(S)A^*(T))=a-ib$, since it's just the complex conjugate. Then $(A^*(S)A(T))+(A(S)A^*(T))=2a$. We ...


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$\mathfrak{R}e$: real part. $A^*$: complex conjugate of probability amplitude $A$.


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First off, I assume you accept that the probability density for the position of a particle is $|\psi(x)|^2$. Now, what roughly happens in real scattering is that a particle whose wavefunction looks like the below image comes in: Then, after interacting with a potential, two smaller packets come out. The packet that transmitted through to the other side ...



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