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8

Generally speaking functions of operators are defined as spectral functions, making use of the spectral theorem machinery. The approaches based of power series usually have not a rigorous basis: Even identities like $$e^A = \sum_n \frac{1}{n!}A^n$$ are generally false if $A$ is an unbounded operator. Nevertheless in most relevant physical cases some of ...


4

Is it not possible for two electrons, even if they are at different locations, to have the same momentum and the same spin directions? A particle that is in a pure momentum state has a wavefunction that is a sinusoidal plane wave. Therefore its position is infinitely uncertain. You can also see this in the Heisenberg uncertainty relation, $\Delta p ...


4

Although we can define the momentum as a self-adjoint operator in $L^2[0,1]$ as you proposed, I think it's rather artificial to think about it as having relation to momentum in the case of $L^2(\Bbb{R})$. Realize that the operator $p_1$ with domain $D(p_1)=\{\psi\in\mathcal{H}^1[0,1]\,|\,\psi(1)=\psi(0)\}$, is related to spatial translations via the unitary ...


3

You do factor out the 100, that is $$\langle\psi | 100 |\psi\rangle = 100\langle \psi |\psi\rangle.$$ In general, if $\psi$ is a correctly normalized state we should have $\langle \psi|\psi \rangle = 1$, hence $100\langle\psi|\psi\rangle = 100(1) = 100.$ Edit: Alternatively (more physically?) you can think of $100$ as a Hermitian operator in and of ...


3

Let's go through Dirac's proof. He acts on the ket with both operators, $$\xi\eta\rvert A \rangle = \xi \eta'\lvert A \rangle$$ and we pick up the eigenvalue of $\eta$. Since it is a constant, we just ignore it now, and let the next operator act on the ket, extracting its eigenvalue: $$=\xi'\eta' \lvert A \rangle$$ Now substitute $\eta'\lvert A \rangle = ...


3

Yes, a potential that is nowhere negative must necessarily have a non-negative scattering length. This follows from the Riccatti equation describing the relationship between scattering lengths for potentials with variable cut-off ranges. The book "Variable phase method approach to potential scattering" by Calogero gives all details needed. Physically what ...


3

Why does the statement "any negative potential supports a bound state" hold in 1D, but not in 3D? In short, this is because for a bound state to occur, any positive kinetic energy needs to be fully offset by a negative potential energy. Achieving a large negative potential energy requires the particle to be localized in the volume where the potential is ...


3

If we were able to have a quantum mechanical model for nuclei that needed no experimental input, then the half lives of unstable nuclei would be computed utilizing the fully known wavefunctions. An approximation to this ideal is the shell model, and there are papers in the literature using the wavefunctions of the model to calculate lifetimes that fit the ...


3

The Hydrogen atom can be solved exactly in the path integral formalism, see for example Fortschritte der Physik 30 401. There is no conceptual problem to compute a probability amplitude for a system with bound-states in this formalism. For example, if the system is initially in a state $\psi(x)$ localized around $A=0$, the probability amplitude $a(B)$ to ...


3

If we are in the Schrödinger picture, we have $$i\hbar\frac{d}{dt}\psi(t)=H_S(t)\psi(t), \qquad\psi(0)=\psi_0$$ Then, we can describe the time evolution via a unitary propagator $U(t,0)$, such that $$\psi(t)=U(t,0)\psi_0$$ Substituting, leads to \begin{gather}&i\hbar\frac{d}{dt}U(t,0)\psi_0=H_S(t)U(t,0)\psi_0&\implies\\ ...


2

The $\phi_i(r)$ form an orthonormal basis of (square-integrable) functions on $\mathbb{R}$, i.e. you should have a relation like $$ \int dr\,\phi_i(r)\phi_j^*(r)=\delta_{ij}. $$ This is what you need in order to expand $\psi(r,t)$ and $\psi^\dagger(r,t)$ the way you did above. You can use this to write the $b_i(t)$ in terms of the $\psi(r,t)$ in the ...


2

You are working with a bound state . This means your energies are negative and the muon cannot escape the potential. You will have quantized energies in this situation. This is entirely analogous to the hydrogenic energies that are solved analytically in many textbooks. You can approach this problem by starting with the time-independent Schrodinger equation ...


2

As dmckee's comment says, your question doesn't really have an answer. However it's exactly the sort of question that fascinated me in my time as a teenage physics enthusiast, and I think it touches on some really interesting aspects of physics. To get a grip on this you need to understand how matter is described by quantum field theory. This is not ...


2

A different angle on this that I DON'T believe is in conflict with Terry Bollinger's answer: whether you express a wavefunction in position co-ordinates, or, as its Fourier transform, i.e. in momentum co-ordindates, the two models are precisely the same. So neither the expression of a position co-ordinate wavefunction (such as you find from the solution of ...


2

To study bound states, we have to find solutions to the Schrödinger time-independent equation $$-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=E\psi$$ Using separation of variables, in spherical coordinates $$\psi(r,\theta,\phi)=Y^m_l(\theta,\phi)\frac{u(r)}{r},$$ where $Y^m_l(\theta,\phi)$ are the spherical harmonics, the radial part can be shown after substitution ...


2

As @ACuriousMind pointed out in his comment, the definition of a pure state is one that is a vector of $\mathcal{H}$, in this case $\Bbb{C}^2\otimes\Bbb{C}^2$. If you take for example $\rho=|\varphi^+\rangle\langle\varphi^+|$, the reduced density matrix on the first space is \begin{align}\rho_1&=\operatorname{Tr}_2\rho\\ ...


2

The precise theorem is the following, cf. e.g. Ref. 1. Theorem 1: Given a non-positive (=attractive) potential $V\leq 0$ with negative spatial integral $$\tag{1} \int_{\mathbb{R}^n}\! d^n r~V({\bf r}) ~<~0 ,$$ then there exists a bound state$^1$ with energy $E<0$ for the Hamiltonian $$\tag{2} H~=~K+V, \qquad K~=~ -\frac{\hbar^2}{2m}{\bf ...


1

That the phase speed can have a dependence on the wavelength/frequency of the wave. For instance, a whistler mode wave can have a cubic dispersion relation at low frequencies. In this limit, the higher(smaller) frequencies(wavelengths) propagate faster than the converse. It results in a sort of "spreading out" of the wave modes. This if often seen ...


1

$$[P,H]f(x)=(PH-Hp)f(x)$$ But $$H=P^2/2m+E(x)$$ $$ =PE(x)-Hf(x)$$ $$ =E(-x)-E(-x)$$ $$ =0 $$ The parity operator therefore commutes with Hamiltonian.


1

At our current state of knowledge, we would expect that quantum mechanics applies everywhere including at the scale of nuclei. However at the macroscopic everyday scale of objects (cars trucks buses planes), what it predicts is not generally different from the predictions of Newtonian mechanics, whereas at the size of atoms and atomic nuclei, the predictions ...


1

Through the particle scattering he deduced that proton were highly charged centers and there were electron around it. Now the basic logic that comes to mind is that those electron would collapse into the proton, just like our earth would collapse into sun if it stops revolving. so to Keep itself from collapsing into the proton, it was proposed that it ...


1

Your derivation is not correct, since $c_k=\langle E_k|\psi\rangle$ is the coefficient of the expansion of $|\psi\rangle$ in the basis $\{|E_k\rangle\}$: $$|\psi\rangle=\sum_kc_k|E_k\rangle$$ and your derivation is implying that $\dfrac{dc_k}{dt}=0$, that is, $|\psi\rangle$ is a stationary state. But that cannot be true, since in general ...


1

I prefer a variant of Anonymous Coward's answer given above, by leaving out the environment. I would say that an observer is a system that interacts with the systems it observes by entangling orthogonal states of the systems under consideration with orthogonal states of itself and possibly other systems. So, I don't bring in the baggage of an environment ...


1

How about: $$ \left\langle\psi\ |100\,|\,\psi\right\rangle = 100\left\langle\psi\,|\,\psi\right\rangle $$ I admit to guessing though as I have not seen such notation. But, my guess is 100 as a scaler can be factored out, or multiply the ket $|\,\psi\rangle$ and then take the inner product of the Bra and Ket (and, the factor of 100 can of course be taken ...


1

To answer your question, one needs to understand a bit what is the Ginzburg-Landau (GL) formalism. Let us first recall the GL functional: $$F=\int dV\left[g\left|\left(\nabla-\dfrac{2\mathbf{i}e}{\hbar}A\right)\Psi\right|^{2}+a\left(T-T_{c}\right)\left|\Psi\right|^{2}+b\left|\Psi\right|^{4}+\dfrac{\left(\nabla\times A\right)^{2}}{2\mu}\right]$$ with ...


1

If the orbit is circular, then $p=\rm{const}$ and $r=\rm{const}$. $E$ is constant and negative (for a bound state) even though the orbit is not circular. So, one can determine $r$ from this equation: $$r=\rm{e}/\left(p^2/2m_e-E\right).$$ The minimum is zero (no kinetic energy, only the negative potential one), which is not supported experimentally.


1

The half life can in principle be determined using Fermi's Golden Rule. Well, this calculates transition probabilities per unit time, but the half life is simply derived from the transition probability. So if you know the initial and final wavefunctions and the appropriate operator then yes you can calculate the half life. However in practice nuclei are far ...


1

The fact that $D(p_0)$ is not big enough to define a self-adjoint operator does not mean it is not included into the domain of the self-adjoint momentum. The choice of boundary conditions for the function is a specification of the vector, not of the operator. Once you have fixed the self-adjoint extension you are considering (choosing the proper domain, ...



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