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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 its 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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Your integration is wrong. The probability density function measures the probability of finding the particle between $x$ and $x+dx$. If you integrate over $0 \leq x \leq L$ you don't get a function of $x$ but a number instead (one for this interval). If you state that $P_n$ is the integrated density, then it should depend on the interval in which you did the ...


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It must not be greater than 1. To find the probability function you must integrate the probability density, $\psi^* \psi$, over the region in which you want to calculate the probability: $$P_n(x_1,x_2)=\int_{x_1}^{x_2} \psi^*(x)\psi(x)\,dx \, .$$


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"What aspect of quantum mechanics makes it natural to say that probability and 'non-determinism' should take a central role?" The fundamental reason is philosophical and comes from the Teleportation thought experiment, which raises the question of what happens to subjective experience after an hypothetical teleportation device avoids deleting the ...


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A wavefunction of a quantum system is the system's state written in a particular form - no more nor any less than that. Here the word state has an exactly analogous meaning to the state of a classical system insofar that the state at any time uniquely defines the state at any other time and contrariwise. The state's evolution with time in either a quantum or ...


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The "wave function of an orbital" is just referring to "the wave function for electrons the corresponding state". When speaking of electron levels in an atom, it is commonly accepted the Copenhagen interpretation of Quantum Mechanics, where a wave function of an isolated system, contains all the information relevant for its full description. Rigorously you ...


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In quantum mechanics we use a wave function to describe the quantum state that an electron is in. It basically tells you where you are likely to find a particle. Notice how it is typically introduced as a function of position. That is because by taking the value at a certain location and squaring it we get the probability of finding it there.


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There exists a valid derivation the Born rule, but here you start from a weaker assumption, it's not the sort of an ab initio derivation from only the other postulates as the MWI advocates would like to have. This argument works as follows. We start by asking how we know that the Born rule is in fact valid. The answer must involve doing experiments, ...


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You are imagining the particle in the well as a classical system i.e. a point particle moving to and fro in the well. However this is not a good description of the system. A quantum particle does not have a position. By this I mean that it is meaningless to ask what the position of the particle is because position, in the sense we normally use the term, is ...



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