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You may have seen the reasoning to follow in most textbooks already but apparently it is not emphasized enough so I will say it again here. The crucial starting point is the second law of thermodynamics that claims that the entropy change of the universe $\Delta S_{univ}$ is either zero or strictly positive for any physical change that occurs in it. I ...

0

If I have understood correct your question(and the discussion below) : In Quantum Mechanics we get some probabilities(in theory) for measuring certain eigenvalues of physical observables, that are, operators. Let's say you are making only one measurement and you get a certain price for that measurement. Can you say that the probabilities have been verified? ...

0

$\newcommand{\ket}[1]{\lvert #1 \rangle}$*Unitarity* is a central feature of all quantum theories, fields or no fields. Unitarity is simply the demand that the time evolution operator $U(t_1,t_2)$ from any time $t_1$ to any time $t_2$ be unitary, i.e. preserve the inner product $(\dot{},\dot{})$ of the Hilbert space of states $\mathcal{H}$, i.e. for any ...

0

Unitarity stems for "conservation of probability". It should be quite clear that quantum-mechanical evolution, as prescribed by Schroedinger equation, is due to the strongly unitary group generated by the Hamiltonian (when time-independent) or, more generally, by the corrisponding unitary propagator (when the Hamiltonian is time-dependent). In each case, ...

1

Yes, you can find a relation between $E_F$ and other functions. However remember that this value is a parameter, not a function. The average number of particles in state $k$ in Fermi-Dirac statistic is $$<n_k> = \frac{1}{e^{\beta(E-\mu)}+1}$$ With $\beta = 1/kT$. At a very low temperatures, you find that $<n_k>$ can only have two values: 0, ...

2

It depends on which logarithm you choose. To demonstrate (but not prove) the information theory equation: the definition $S = \log W$ (uncertainty is the logarithm of multiplicity, Boltzmann's famous relation) becomes equivalent, if $p_i = 1/W$ ($W$ equally probable outcomes) to $$S = -\sum_{i=1}^W p_i \log p_i.$$But since $\log_a(b) = \log_c(b) / \log_c(a)$ ...

3

Given the wavefunction $\psi(x)$, the probability to find any particle within an interval $[x_0,x_0+\Delta x]$ is $$P([x_0,x+\Delta x_0]) = \int_{x_0}^{x_0+\Delta x} \lvert\psi(x)\rvert^2\mathrm{d}x$$ i.e. $\lvert\psi(x)\rvert^2$ is not a probability, but a probability density that has to be integrated over a set of non-zero measure to yield a notion of ...

5

The square of the wavefunction, $|\Psi(x,t)|^2$, is the probability density function for finding the particle. This means that the probability of finding the particle in an interval of (infinitesimal) width $\mathrm dx$ at position $x$ equals $|\Psi(x,t)|^2\mathrm dx$. On occasion, however, authors will drop the $\mathrm dx$ if it is convenient and does not ...

3

Unitary dynamics means that the evolution of quantum states are describen by unitary operators. Physically this means that there is no dissipation in the system. This is important, because inherently quantum phenomena needed for quantum computation eg. entanglement and quantum superposition are only observable for a long period if no dissipation or other ...

-1

you can make any state if you properly choose bases as, $|\Psi> = \alpha|0> + \beta |1>$ where $|0>$ and $|1>$ are assumed as the complete bases. In this case, $\alpha$ and $\beta$ are the probability amplitude to observe 0 or 1 respectively. Similarly if the system can be written by a continuous bases, like space $|x>$, any state can be ...

2

In the case of an infinite superposition of eigenstates it becomes more complicated but we can still write a general expression for it. If $$\psi(x) = \sum_{n=0}^\infty a_n \phi_n(x)$$ where the $\phi_n$ are the eigenstates of the Hamiltonian. The time dependent wavefunction will look like: $$\Psi(x,t) = \sum_{n=0}^\infty a_n \phi_n(x) T_n(t)$$ where $T_n = ... -1 In that case, you have a sum of overlaps between a pair of functions having same eigenvalue index. The cross overlap will be zero because of the orthogonality of basis set which is very important. -1 If you have a quantum state in which more than one of the possible outcomes of a particular measurement has non-zero amplitude (an unsharp state, as opposed to a sharp state in which there is only one outcome), then the MWI says that there will be multiple versions of you, and each version will see one possible outcome. In the standard (non-MWI) way of ... 2 So, I suppose that$Φ(k)$is the probability density of the momentum. Is this true? Almost.$\Phi(k)$is the probability amplitude for the momentum of the particle. The probability density is obtained as usual by squaring the amplitude, giving$|\Phi(k)|^2$. For a free particle, all values of momentum are always allowed, which enables the superposition ... 2 First, realize we are doing an approximation when we are evaluating the coefficient$A_p$in the Fourier series $$\psi(x) = \sum_p A_p \psi_p(x)$$ by the integral $$A_p = \int_{-\infty}^\infty \psi_p^*(x)\psi(x)\mathrm{d}x$$ since the limits should really be the boundary of the interval on which the Fourier series is periodic. Furthermore,$\psi(x) = ...

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