# Tag Info

1

I think what you are asking whether the relationship $$\mathrm{normalizable} \iff \mathrm{continuous}$$ holds, which is utterly wrong! The wave function has to be continuous*. Notwithstanding take $\psi(x)=H(x-1/2)-H(x+1/2)$, where $H(x)$ is the Heaviside step function. $$\implies \int_{-\infty}^\infty \mathrm{d}x \,\, ||{\psi(x)}||^2 = 1$$ (Area ...

1

My question is, why am I doing this? Becase, by convention, we want the probability account for all possible outcomes to sum to unity. The fact that we will get some outcome at a measurement, is guaranteed, and with this normalization it reflects a total probability of 1.

0

Consider 2D (XY) case for simplicity, $\vec x = x\vec e_x + y\vec e_y$. $\vec e_x$ is unit vector along x-axis. By definition, gradient is the following operator upon scalar: $\nabla=(\frac{\partial}{\partial x}\vec{e_x} + \frac{\partial}{\partial y}\vec{e_y})$ Apply this operator to $\vec p\cdot\vec x$: $\nabla(\vec{p}\cdot \vec ... 0 The reason why they put equations in is because physics is best described in terms of equations. They tried to remove randomness with equation, ultimately believing that QM is not that random, that it is behaving according to certain parameters. They tried to set the boundaries in the randomness/unpredictability, but the equations still couldn't remove the ... 0 First of all, I'm not a physics specialist. In my opinion, the reason why I'm typing this text at this second and the next one is that we are working inside another parameter. In other words, we are dealing with a "computer" object that is made up of other quantum objects, but that molecules are working as a computer. Once we catch computer, it is always ... 0 Since the two slit experiment is a bit complicated for what I'm about to discuss, allow me to consider a simplified toy model. Consider an$N$component state vector$| \alpha \rangle $, about to be acted on by some operation, and subsequently measured. This evolution can be represented by: $$| \beta \rangle = U |\alpha\rangle$$ The important bit here is ... 1 Assuming our only aim is to solve double slit experiment (or other problems that can be mapped into that). Actually the double slit experiment for electrons is a derivative/prediction from the quantum mechanical theory as it started with the Schrodinger equation ,its wavefunction solutions and the interpretation of differential operators with energy ... 3 You say that we are only interested in the probability distribution on the screen,$\rho(x,t) = \lvert \psi(x,t) \rvert^2$, which is essentially correct. So, why do we have$\psi(x,t) = \lvert\psi(x,t)\rvert\mathrm{e}^{\frac{\mathrm{i}}{\hbar}S(x,t)}$? Well, looking at the time evolution equation for the probability density, the continuity equation of ... 3 The non-negative real probability distribution can't interfere like a complex wave function can. To produce interference phenomena it is necessary for quantum mechanics to deal with probability amplitudes, not just probabilities. 0 Your linked article ('How do probabilities emerge…') 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 ... 1 We cannot know where the particle is with certainty. The particle, in general, does not have a definite location to know. Under certain conditions, we can know and predict future probability distributions. The evolution of the state is determined by the Hamiltonian (in the Schrodinger picture). The problem is that we don't know the Hamiltonian ... 1 Correct, expect just don't say "on average," just we will find the particle at different locations more often than others in accordance with the probability distribution described by a wave function. Incorrect, you may always calculate the evolution of an initial state if you know the effective Hamiltonian or effective Action the state is subject to. ... 0 As a chemical reaction approaches equilibrium, one of forward or backward reactions dominate the other. Yes. But note the domination of one reaction (amount of its products increasing) is already implied by the assumption that system approaches chemical equilibrium. If none reaction dominated, there would already be chemical equilibrium. According ... 0 In chemical thermodynamics, it is derived from 2nd law of thermodynamics that in case the final temperature and pressure are the same as the initial ones, one equilibrium state can change into another (say, by removal of a partition) only if the final value of the Gibbs energy is not greater than the initial value. When the reaction is done in a reversible ... 0 Mostly likely you would see $$|01\rangle+|10\rangle$$ vs. $$|01\rangle-|10\rangle$$ since they have same Eigenvalue for energy and hard to distinguished unless there are another measure to tell the difference.$|00\rangle$and$|11\rangle$are very different in this sense. 0 The difference between those two states is in their phase$|\psi(\alpha)\rangle=\frac{1}{\sqrt{2}}(|11\rangle+\exp[i\alpha]|00\rangle)$for$\alpha=2\pi$you have the first one and for$\alpha=\pi$you have the second one with the minus sign. Phsycally the difference lies in the interference pattern you might see once you compute the probability amplitudes, ... 1 When you define a base of vectors, in your case$\{ |1,1\rangle, |1,0\rangle, |0,1\rangle, |0,0\rangle \} $you assign to each vector some phase, e.g. the vector$|1,1\rangle$has a certain phase that you don't mention explicitly outside the bra-kets. Now, in your calculi you may have superpositions of these vectors of the form$|\psi\rangle = ae^{i\alpha} ...

21

This question gets to the heart of what makes quantum mechanical amplitudes different from classical probabilities. It is true that if you make measurements in the basis of states $\{\lvert 00 \rangle ,\lvert 11 \rangle\}$ then the two states have the same measurement statistics, and so cannot be distinguished. The interesting thing is that it is possible to ...

0

[...]when the incoming radiation's energy comes closer to the binding energy of the electrons, then the probability of fluorescence increases. Is this correct? When the incident energy is close to the binding energy there is a sharp increase in the absorption. See, for example, Figure 1-5 in the following document: ...

2

This is a bit of a negative answer, but consider instead an expectation of some other quantity, such as the energy: $$\langle E \rangle = \sum_i p_i E_i.$$ Now, it's obvious what $E_i$ means - it's the energy of state $i$ - but what does $p_iE_i$ mean? The answer is not very much really - it's the contribution of state $i$ to the expected energy, but it's ...

0

I believe this relates to density of states and the shape of the bands. When you try to excite an electron from one state to another, there needs to be an available state "upon arrival" - and the probability of the transition is therefore directly related to the density of states. If I'm not mistaken, this is the reason that the excitation is strongest when ...

Top 50 recent answers are included