# Tag Info

0

The transformation is valid for the Froze-Core Approximation. More information can be found in this text, up till section 1.3: https://arxiv.org/pdf/0910.1921.pdf

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Use the Taylor series for the exponential function, $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots.$$

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To find the action of $x,y,z$ on $|l,m\rangle$ you need to use the Clebsh-Gordon procedure for combining the $l=1$ vector defined by $x-iy \sim |l=1,m=-1\rangle$, $z\sim |l=1,m=0\rangle$, $x+iy\sim |l=1,m=+1\rangle$ to the $|l,m\rangle$ state. You will get a linear combination of $l-1$, $l$, $l+1$

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The hydrogen atom has discrete energy levels. These are the levels of the entire system of proton and electron. In an excited state also the proton wave function changes.

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Indeed, protons are quantum particles, they are fermions, and must also satisfy the Pauli exclusion principle (fun fact: that's why we exist). The interesting thing is that they also distribute in energy levels, that's what we call the Nuclear Shell Model. https://en.wikipedia.org/wiki/Nuclear_shell_model The forces are different, also the interaction, so ...

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Here is another way to look at this which might be helpful. For objects small enough that quantum mechanical effects cannot be ignored, material particles begin exhibiting wavelike properties. Whether we detect their material aspect or their wavelike aspect then starts to depend on the physical details of the detection apparatus we are using. This is because ...

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As the previous answer mentioned, there is a misprint in the problem. So the Eqn(1) in OP should really be written as $$\langle\chi|m\frac{x_{k+1}-x_k}{\epsilon}m\frac{x_k-x_{k-1}}{\epsilon}|\psi\rangle=-\hbar^2\int\chi^*(x,t)\frac{\partial^2}{\partial x^2}\psi(x,t) dx\tag{1*}.$$ Also, It might seem to be that Eq.(5) is applied to Eq.(8) directly by ...

1

Now, in order to calculate the probability current, I want to use $J(x)= \frac{\hbar}{m}|\psi(x)|^2\vec{k}$. I don't know why you think this formula is applicable here. As @Jakob already stated in his comments, this formula is not correct in general. The correct general formula can also be found e.g. in Wikipedia - Probability current: $$J(x) = \frac{\hbar}{... 1 Uhm I don't exactly know which page of Breuer&Petruccione you are referring to, but maybe the following remarks can help: In the context of GKLS (Lindblad) master equations, a stationary state is defined as an eigenvector of the Liouvillian \mathcal{L} with zero eigenvalue. The Liouvillian is defined as the generator of the dynamical semigroup (Eq.(3.... 0 A very useful property of the \delta "function" is$$\delta(f(x))=\sum_i\frac{\delta(x-x_i)}{|f'(x_i)|},$$where x_i are the zeros of f(x), i.e. f(x_i)=0, and |f'(x_i)| means the absolute value of the derivative of f evaluated at x_i. In your case, f(x) would be$$f(p_f)=\frac{p_f^2}{2\mu}-\frac{p_i^2}{2\mu},$$and zeros are \pm ... 1 The relationship between wave mechanics and matrix mechanics is so natural in the language of Dirac's bra- and ket- language, and the theory of Hilbert space, that hardly anyone pays attention to the distinction, justifiably, and most people switch dialects in the middle of their discussion. Basically, anything requiring solving differential equations is ... 0 The Condon Approximation is an approximation that is done within the Born-Oppenheimer Approximation of molecular systems. Take a look at the electronic transition dipole moment for example,$$ \vec \mu_{if}(R) = -e\int^\infty_\infty \psi_i(r,R)\sum_l^{N_{\text{electrons}}}\vec r_l \psi_f(r,R) dr $$where \psi_k is the k'th electronic adiabatic state and ... 1 An overview of experiments on the quantum zeno effect can be found on its wikipedia page. The first observation of the effect seems to be presented in Wayne M. Itano, D. J. Heinzen, J. J. Bollinger, and D. J. Wineland "Quantum Zeno effect" Phys. Rev. A 41, 2295 (1990). 0 The unitary part of the circuit (that is, the CNOT followed by the Hadamard), is the gate sending Bell states into the computational basis. More explicitly, writing the input states as$$|\psi\rangle=\psi_0|0\rangle+\psi_1|1\rangle, \qquad|\phi\rangle=\phi_0|0\rangle+\phi_1|1\rangle,$$the output state before the measurement can be written as$$ |\psi\rangle|...

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For the particular case specified in the OP (a cavity mode coupling to an external bath), the most standard form is probably the Gardiner-Collett Hamiltonian [see C. W. Gardiner and M. J. Collett Phys. Rev. A 31, 3761 (1985)] or a variation thereof. Adapted to your specific example, the interaction Hamiltonian would read $$H_\mathrm{int} = \sum_i^N g_i a_i^\... 4 It's tempting to try to interpret the mathematics of QM or QFT too literally, however this can be a dangerous game to play. Quite often you will cause more problems for yourself than you will solve by asking what exactly particles are (are they points, are they waves, are they fields, are they strings etc.), not least because different theories will give you ... 3 Emission and absorption of a photon aren't instantaneous processes as long as you don't perturb the system by measuring its state. If you do, the system collapses in a non-reversible way. Let me explain how it evolves in the case you don't measure: Spontaneous emission An atom in the excited state | a \rangle emits light in the same spatial pattern as a ... 2 What you want is not possible: You cannot represent the density operator \rho of \lvert\phi\rangle+\lvert\psi\rangle as a function of \rho_\phi and \rho_\psi. The reason is that \lvert\phi\rangle+\lvert\psi\rangle depends on the global phase of the two states -- e.g., if you change \lvert\psi\rangle\to e^{i\vartheta}\lvert\psi\rangle, you will ... 3 Nothing went wrong, the difference between both results is just a global constant phase, which has no meaning. If \psi_n(x) is an eigenfunction of the Hamiltonian, so is e^{i \alpha}\psi_n(x). In your case, e^{i\alpha}=(-1)^m. 2 In the first problem, the collision time is a random variable, and you're given a PDF telling you how this variable is distributed. The collision could take place at any time after t=0. In the second problem, the collision time is definite — we know the collision happens at a certain time t=\tau. If it helps, you could think about the collision time as a ... 2 We have to be careful when we use bra-ket notation to deal with non-hermitian operators. In these cases, it is more convinient to use (\cdot,\cdot) notation. Being a and b two states, and A an operator,$$(a,Ab)=(A^\dagger a,b),$$which in bra-ket notation would be$$\langle a|A|b\rangle=\langle a|\Big(A|b\rangle\Big)=\Big(A^\dagger|a\rangle\Big)^\...

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Yes, what you wrote is correct. One advanced note I wish someone had told me explicitly when I was learning quantum mechanics: when you get to study more complex systems (and in particular more complex spacetime symmetries), sometimes it is useful to generalize the definition of symmetry from the one you have given. In particular, rather than equating the ...

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The elctron distribution function $f(E)$ determines to which extend the electronic state with energy E is occupied. In your case Fermi-Dirac statistics apply, so we know the exact functional context. The Fermi-Dirac distribution function is given as $$f(E) = \frac{1}{e^{\frac{E-\mu}{k_{\mathrm{B}}T}} + 1}$$ , where $\mu$ is the ...

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Energy conservation always holds. The thing is, if you're trying to promote some transition in a quantum system and you're using a time-dependent oscillatory perturbation (like, say, a pulse of laser light), and your perturbation is confined within a window of finite duration, then your perturbation does not have a well-defined energy. As such, the ...

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In the time independent theory, the distinction is really the mathematical question of whether the Hamiltonian's eigenvalue spectrum is continuous (scattering) or discrete (bound), which in turn affects the kind of normalisation that can be applied to those eigenstates. In general, a hermitian operator's spectrum will have a combination of both (including ...

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It seems to me that the double-slit "mystery" is simply active observation effects on the wavefunction. There is no camera or sensor that does not "vacuum up" electrons or photons to give you a reading. Even the human eyeball or eardrum is not completely passive, they ever-so-slightly affect the light and sound waves of the observed ...

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I think your problem is simply not keeping track of the time differences for the states. That is using the notation that $t_k$ is the time when $x_k$ operates and Feynman and Hibbs notation that they used to write their Eq. 7.95 just above their statement of this problem (problem 7.15), \langle \chi|(x_{k+1}-x_k)(x_{k}-x_{k-1}|\psi\rangle = \...

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It turns out that my second hamiltonian is wrong and that if one calculates correct, the superposition of stationary states yields the original hamiltonian again. I did my calculations for a slightly changed original hamiltonian but that should not matter: $$\begin{pmatrix} E_0 + \mu \epsilon & -A \\\ -A & E_0 - \mu \epsilon \end{pmatrix}$$ $$E_1 =... 2 Elliptic regularity implies that \psi admits weak derivatives of every order which are L^2 locally. Now, Sobolev's lemma implies that these derivatives must be standard derivatives. In summary, every weak solution is just the standard solution \psi_m. 25 There are dozens of tricks to do this, but most of them reside in phase-space quantization, with which you are likely unfamiliar if you are asking this, so I'm staying out of it. Let's skip the hats for operators, set \hbar=1 for simplicity, non-dimensionalizing, and work in the coordinate representation, so, essentially, p is shorthand for p=-i\partial_x... 2 I don't think you can do it unless You are given what f(x) and g(p) are. If they are given you can use different properties for commutation relation and [x,p]=i\hbar to compute what [f(x),g(p)] will be. Consider a simple example (More complicated function may not solve this way)$$[x^2,p]=x[x,p]+[x,p]x=xi\hbar+i\hbar x=2i\hbar x$$Further ... 2 This can be done as follows$$a_+a_-|n\rangle =a_+\sqrt{n}|n-1\rangle =\sqrt{n}\ a_+|n-1\rangle =\sqrt{n}\sqrt{n}|n\rangle\boxed{a_+a_-|n\rangle =n|n\rangle}$$Note:$$a_-|n\rangle =\sqrt{n}|n-1\rangle a_+|n\rangle =\sqrt{n+1}|n+1\rangle$$Edit: In term of your notation this would go as follows, You just need to change$$|n\rangle \rightarrow \...

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There will be little two slit interference. The wave function is an expansion of Slater determinants dominated by determinants where each electron goes through a different slit, or rather, where each orbital travels through a different slit. Two slit interference results when orbitals pass through both slits. Because of Coulomb repulsion the contribution of ...

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If a valid quantum wave function in the position basis must be continuous in both the wave function, $\psi(x)$, and it first derivative, $d \psi(x)/dx$, then the same must be true of the corresponding wave function in the momentum basis. A step function in the momentum basis is not a valid quantum wave function. Thus, I assume we cannot use $\sin(x)/x$ ...

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It’s important to remember you’re dealing with particles. On the screen there will only be two electron impacts which can hardly form a pattern. Both electrons are accelerated multiple times as they travel toward the screen. Once when they are fired, once as their spins rotate while passing through a slit and many more times as they move from the slits to ...

1

I'm not too sure, what are you asking for but considering this: So what would the momentum be in either of these regions? The energy eigenfunction in position basis is given by $$\phi_n(x)\propto \sin(k_nx)$$ where $k_n=n\pi /L$. Then we know that the momentum eigenfunction has a form $$\chi(x)\propto e^{ipx/\hbar}$$ Thus I can write $$\phi_n(x)\propto (e^... 1 In terms of projective measurements, we just project the state. So consider a measurement operator \hat{A} with eigenvalues \{a\}. Let \hat{P}_a be the projector onto the space spanned by the eigenvectors with eigenvalue a. In other words, if |n_a\rangle are eigenvectors of \hat{A} with eigenvalue a, then$$ \hat{P}_a = \sum_{n} | n_a\rangle\...

2

Periodic boundary conditions state $$\psi(x)=\psi(x+L)\tag{1'}.$$ Goodstein says "[...] it (the particle) has equal probability of being anywhere on the line", which would not be true if we take $\psi(0)=\psi(L)=0$. You can see that with the condition ($1'$), $k$ takes the values that he claims. Edit: In this particular case, I don't know what are ...

0

That $h$ in the first Hamiltonian is a bit confusing, let me call it $\Omega$. Also, I guess the second term in both of them is actually $\frac{m\omega^2}{2}$. $H_1$ is the sum of a one-dimensional harmonic oscillator and a term concerning the spin. As the spin term does not affect to spatial variables, i.e. $x$ and $p$,we can treat both terms separately and ...

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In the first case you get $$[ij,kl]=[i,j]kl-kl[i,j]=\left[[i,j],kl\right]=\left[j,[kl,i]\right]+\left[i,[j,kl]\right]$$ using the Jacobi identity. So you see that assuming only $[i,j]$ to be nonzero also reduces the first result to zero.

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In your first worked-out example, you first comuuted $i$ to the right (past the $j$ to the right of it), and then commuted $j$ to the right (past the $i$ which is now at the rightmost position. In your third displayed equation, and under the assumption that $k$ and $l$ commute with everything, you'll find [ij,kl] = \underbrace{[i,j]kl}_a + j[i,k]l + jk[i,l]...

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Just diagonalize the Hamiltonians, the elements on the diagonal are the EV's. If you are unsure how check this paper(https://aip.scitation.org/doi/abs/10.1063/1.524093?journalCode=jmp). HINT: express the Hamiltonians in second quantisation basis.

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The Fermi-Dirac distribution gives the probability of an energy level being occupied. You have to combine it with the density of states $g(\epsilon)$ at that energy to get the physical properties you could be interested in, for example, the internal energy $U$ can be computed as U = \int \epsilon\, f_{FD}(\epsilon)\, g(\epsilon)\,\mathrm{d}\...

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There are some problems in trying to define a quasi-probability distribution in phase space for quantum fields, à la Wigner. The foremost complication is given by the lack of a universal reference measure in an infinite dimensional vector (e.g. Hilbert) space. For finite dimensional vector spaces, there is the Lebesgue measure, and thus it makes sense to ...

1

If you measure energy, you will put the system in an energy eigenstate. The measured value will be one of the eigenvalues. The superpostion amplitudes will help you calculate the probability of each outcome.

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when radiation (light spectrum) falls on blue colored body its internal energy get increased due radiatic interaction. This energy difference is equivalence to wavelength of blue spectrum. Consequently, it absorbs all the spectra initially blue too but eventually emits the blue spectrum only rather than other and hence object is found to be blue in color.

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Your confusing two issues. Statistical mechanics replaces the momentum of lots of classical particles i.e. a long list of - say - $p_x$ values for each particle each moving along its own trajectory, by a few average quantities, such as the average momentum, and the statistical spread of the momentum values about the mean. In this way instead of dealing ...

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When a measurement is made, the wavefunction collapses to an eigenstate of the operator corresponding to the variable being measured. While one certainly can express a wavefunction in a basis of states equal to some linear combination of position eigenstates, the elements of this basis will not be eigenstates of the position operator, so they will not be ...

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It is probably messy to implement in Fock space (number states), but can be related to the conventional squeeze operator through the methods of Nieto et al. discussed in this answer, in an elaborate CBH rebraiding. In Hilbert space, however, it is elementary, if not trivial. Let us discuss the effect of the operator $\hat S'=e^{-3\hat p^2/2}$ on a plain ...

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