# Tag Info

1

Heisenberg's uncertainty principle is $$\Delta x \Delta p \geq \hbar/2.$$ Since the well is of width $L$, you have a measure for the uncertainty on the position $\Delta x$. Then assume the lowest possible value for $\Delta p$, i.e. the one for which the above inequality becomes an equality. Lastly, use $E = \dfrac{p^2}{2m}$ to find an expression for $E$. ...

2

One gets there by noting that $\langle x | p \rangle = e^{i p x/\hbar}$ is a plane wave, and you have to throw on the test wave function to talk about the derivative operation. So, you are worried about $$\langle x| P | \Psi \rangle = \int dp ~p~ e^{i p x/\hbar} ~\langle p | \Psi \rangle$$ From there you note that you can get the ...

1

If $\cal H$ is a complex Hilbert space, and $A :D(A) \to \cal H$ is linear with $D(A)\subset \cal H$ dense subspace, there is a unique operator, the adjoint $A^\dagger$ of $A$ satisfying (this is its definition) $$\langle A^\dagger \psi| \phi \rangle = \langle \psi | A \phi \rangle\quad \forall \phi \in D(A)\:,\forall \psi \in D(A^\dagger)$$ with: ...

3

Suppose $a$ and $a^{+}$ operators satisfy $$\left\{ a,a\right\} =0\mbox{ and }\left[a,a^{+}\right]=1$$ We have basically $a^{2}=0$ and $aa^{+}=a^{+}a+1$. Now consider $aaa^{+}$. $$0=aaa^{+}=a\left(a^{+}a+1\right)=aa^{+}a+a=a^{+}aa+2a=2a.$$ So we get $a=0$.

0

Not exactly what you were asking for, but any force exerted perpendicular to the direction of motion does not change the magnitude of momentum -- though it does change the direction. Two examples are the force exerted by a uniform magnetic field on a moving charged particle, and the force of gravity on a satellite in a perfectly circular orbit. But ...

2

In the (incorrect) Bohr model of the hydrogen atom, for the ground state (n=1), the distance of the electron from the nucleus is 52.9 picometers (pm). The constant $a_0$ is defined to be this distance. The distance of the electron in the nth energy level is $a_0n^2$. According to the exact solution of the Schrodinger equation for the hydrogen atom, in the ...

0

Murray Gell-Mann has an interesting take on Bell's theorem which pertains directly to Stephen Wolfram's thesis on modeling physical laws with cellular automata in his tome: 'A New Kind of Science', an analysis which took him over 20 years to complete. According to Murray, elegant models of physics involve fundamental laws in addition to the results of ...

16

You can have multiple forces exerted on an object that add to zero. Then there will be no momentum change. Think of the two of us leaning against opposite sides of a door with the same force. The door does not change momentum, nor does either of us. I am exerting a force on my chair as I sit here.

2

A universal relation is that the force exerted on an object equals the time derivative of momentum. No force, no momentum change, vice versa.

7

No, all forces involve a change in momentum. In classical mechanics force is defined as a change in momentum. In quantum field theory particles interact via exchanging one or more bosons (see feyman diagrams). These bosons always have momentum and therefore the momentum of the interacting particles changes as well.

12

Actually Newton's second law is better stated as $$F=\frac{dp}{dt}$$ and this is even valid in relativity, both SR and GR, expressed in the right way $$f^\mu = \frac{dp^\mu}{d\tau}=m\frac{du^\mu}{d\tau} = m u^\nu\nabla_\nu u^\mu$$ (for massive particles) so classically forces are always imply a change in momentum. In QFT the concept of a force is no more ...

1

The experimentalist's answer: 1) Experimental physics has established with very many experiments that the underlying framework of nature is quantum mechanical, and this includes special relativity, when the energies are appropriate. It is dependent on a very small number of elementary particles out of which all matter that we have observed and experimented ...

1

Consider a potential, which approximately can be described by two harmonic oscillators with different base frequencies, for example (working in dimensionless units) $$U=1-e^{-(x-4)^2}-e^{-\left(\frac{x+4}2\right)^2}$$ It will look like Now let's look at two lowest energy states of the Hamiltonian $$H=-\frac1m \frac{\partial^2}{\partial x^2}+U,$$ taking ...

0

Often we do drop constants in potential energies. The problem here is that the Hamiltonian eigenvalue represents total energy; in an oscillator this energy is sloshing back and forth between kinetic and potential. If you dropped the constant from the Hamiltonian, you'd be changing the energy available for kinetic excitation.

0

I disagree that : Since energy can always be shifted by a constant value without changing anything, You are maybe thinking of classical potential energy , but the mass of a proton is fixed, for example, it cannot be shifted by a constant value, and at rest E=mc**2. This statement is not general and can only be true for the solutions of non ...

8

It depends what you're doing, and indeed most of the quantum optics literature dismisses the term as it does not contribute to the dynamics. However, it is important that beginning students form an intuition for how and where zero-point energies come in, and why they are necessary. Take a look at the eigenfunctions of the harmonic oscillator, in position ...

1

In Mathematica, using the formula $\sqrt{u_0^2-v^2}=\begin{cases}\\v\tan(v) \\-v\cot(v)\end{cases}$ where $u_0=mL^2V_0/(2\hbar^2)$, we can first compute $u_0$: << PhysicalConstants` u = Convert[ NeutronMass (1 Femto Meter)^2 V0 ElectronVolt/(2 \ PlanckConstantReduced^2), 1] Out: 1.20649*10^-8 V0 Since $u_0$ is typically on the order of an ...

2

First: The Schrödinger equation is a equation using functionals. Solutions to this equation are such $\Psi(r,t)$, that fulfill this equation. The finite square well 1.0 fm wide means you have a potential $V(r)$ which is zero for r<0 and r>1fm and -d in between. d is your depth. Now you have to determine d such, that only two of the resulting $\Psi(r,t)$ ...

4

Yes, $|\psi(x)|^2 \mathrm{d}x$ gives you the probability to find the particle between $x$ and $x + \mathrm{d}x$. The probability that $x$ will be in the interval $[a, b]$ is then $$P_{a\le x\le b} (t) = \int\limits_a^b d x\,|\psi(x)|^2 \, .$$ Normalization of $\psi(x)$ to one is required since the probabilities of all possible outcomes should add up to ...

1

"What does the peak stand for?": If you consider infinitesimally small ranges of wavelength values, the energy density (intensity) will be maximal at the peak. "And what does the graph tell us?": Considering a place of uniform temperature, with radiation in equilibrium with the surroundings, such as in a uniform temperature box, the graph tells us how ...

2

The equation [now corrected] in the question is incorrect by a factor of 2pi, because h-bar should just be h. The equation is not limited to excited states. E is the relativistic energy from $E^2 = p^2c^2 + m^2c^4$, where m is the rest mass and p is momentum. While the equation was originally for photons, De Broglie extended it to all particles in his ...

0

1) Uncertainty principle is momentum and position OR energy and lifetime, not energy and position. 2) If we confine the two particles in a infinite square well then they can only be in the well. Their wavefunctions go to zero at the boundaries. 3) True 4) False. Two particles can have the same energy. But thy have to be in two different states. For ...

-2

While all these answers are fundamentally correct, especially with regards to Schrodinger and the shell model of electrons, there is one very basic means of radioactive decay, that of electron capture, which has not yet been discussed. Yes indeed, electrons orbiting around the atom can be captured into the nucleus. (For reference, see ...

0

Sounds like graphene physics or something similar. You won't find it in a dictionary. In the band structures of many materials, it is common to find multiple similar points in reciprocal (momentum) space. For example, in silicon's band structure there are six distinct conduction bands that all have similar behaviour. These six points came to be known as ...

1

An S electron does have a probability of being at/in the nucleus. This is referred to as "Fermi Contact Interaction" in the literature. The effects of this interaction can be observed through NMR and EPR (ESR) spectroscopy.

4

An angular momentum eigenstate can be rotated using, $$\left| J , m \right\rangle \rightarrow e ^{ i {\vec S} \cdot {\vec \theta} } \left| J , m \right\rangle$$ where ${\vec S}$ is the $2J+1$ dimensional Pauli matrices. For spin $1/2$ for example, ${\vec S}$ are just the ordinary Pauli matrices, $\frac{1}{2} ... 3 As jerk_dadt said, protons are much more tightly bounded in the nucleus by the strong nuclear force. Furthermore, photoelectric effect happens typically on metals, where the most external electrons are "floating" in no-mans land between the atoms, not tightly bound to any of them. So, the bonding for the protons is way stronger than 137 times. But we can ... 4 No. Electrons are very loosely bound while protons are are held together by the strong force. It would take an enormous amount of energy to eject a proton and does not actually surmount the potential barrier of the strong force it merely tunnels through it (this is called proton decay). For electrons, which are bound to the nucleus by electromagnetic force, ... 1 I didn't read your answer, but let's think about just computing the operator$\partial_x^2 f$. First we need to compute the operator$\partial_x f$. Now I am saying "the operator" because we are viewing$\partial_x f$as a composition of first multiplying by$f$and then taking the derivative. By the product rule, we know$\partial_x f = (\partial_x f) + f ...

0

I haven't found a really good shortcut, but the following can make the integration much simpler in some cases. The time independent Schrodinger Equation: $$\frac{\hat{p}^2}{2m}\Psi+V\Psi=E\Psi$$ $$\frac{\hat{p}^2}{2m}\Psi=(E-V)\Psi$$ $$\hat{p}^2\Psi=2m(E-V)\Psi$$ So.... $$\langle p^2\rangle = \int\Psi^*\hat{p^2}\Psi dx = \int\Psi^*[2m(E-V)\Psi]dx$$ ...

0

This is most likely a scam. Alkaline batteries are made of carbon ('graphite'), alkaline, and almost any other metal. It is not a "superconductor".

1

You asked the exact same thing here and seemed satified with the answers and comments. I suggest you have a look at integration by substitution. But to give you peace of mind, let's spell it out. Assuming $a\in\mathbb{C}, a \neq 0$ and setting $y^2=ax^2$ $$\int_{-\infty}^{+\infty}e^{-ax^2}dx = ... 3 You need a slightly more general form for the integral, and in particular you need a form that allows for a constant in front of the x^2 in the exponent. To do this, using the substitution y=\sqrt{a}x, you do$$ \int_{-\infty}^\infty e^{-a x^2}\text dx= \int_{-\infty}^\infty e^{-y^2}\frac{\text dy}{\sqrt{a}}= \sqrt{\frac{\pi}{a}}. $$You can then do your ... 2 The relation |\langle \psi|\psi \rangle |^2=1 is the normalization condition for quantum states - so by itself, it doesn't mean anything. It only means something if you put it together with the fact that time evolution is unitary and hence preserves this norm. Then, if you define the process of measurements, it turns out that the probabilities can be ... 0  | x \rangle  is a position eigenstate, the state for a particle with definite location x. This is an abstract vector. \delta (x - x_0) is a wavefunction (or distribution) for a particle with definite location x_0. It is the state | x_0 \rangle on the position basis$$\langle x | x_0 \rangle = \delta (x - x_0)$$If \hat x is the position ... 1 First I should mention that I got a slightly different result for the correlation function, namely,  \frac{ \hbar }{ 2 m \omega } e ^{ \frac{ 3 }{ 2} \hbar \omega \tau }  (though I made have made a mistake). The correlation function, $$\left\langle 0 \right| \hat{x} ( t ) \hat{t} ( t - \tau ) \left| 0 \right\rangle$$ has ... 1 Decoherence happens because in a macroscopic system you are not able to create a small isolated system. In practice you are deal with statistical mixture and not pure state. There's a good description on Wikipedia. 0 If you put \chi_{r_0}(r)= \delta (r-r_0 ) then [ \chi_{r_0}(r)] forms a basis. In Dirac's notation: \chi_{r_0}(r) \rightarrow |r_0 \rangle and you can verify that this set is a basis because it satisfy: Orthonormality: \langle r_0 | r'_{0} \rangle = \delta (r_0 - r'_0 ) Closure relation: \int d^3 r_0 \ \ |r_0 \rangle \langle r_0|= 1 where 1 ... 1 The latter description is correct (as is described in Sakurai, Gasiorowicz, Griffiths, and probably some other books that I don't own). What it is saying is that the inner product between |x\rangle and |x_0\rangle is either 0 if x\neq x_0 or 1 if x=x_0. That is, the states are orthogonal. The momentum space description$$ \langle ...

1

Here is another hypothetical (i.e. extremely impracticable) answer to your question that is rather interesting (althgough Aksakal's Answer is likely to be a bit more practical!). You have to imagine yourself to be a very deft light-catcher with mirrors (I can't help thinking here of Mozart the Light Catcher). You trap light in the box by suddenly (within ...

0

If we stick to the Problem #2 you mentioned, then yes, there is an energy difference. Since seems there is no degeneracy from symmetry for $\psi(r-a) \pm \psi(r+a)$, unless numerically accident. $$E_{\pm} = \langle \psi(r-a) \pm \psi(r+a) | \hat{H} | \psi(r-a) \pm \psi(r+a) \rangle \tag{1}$$ $+,-$ correspond to $E_G,E_E$ in the Problem #2, ...

1

Well, without the delta potential the wave function is $$\tag{1} \psi_0(x,t)~=~\exp\left[ -\frac{iE_1 t}{\hbar}\right] \phi(x) ,$$ where $$\tag{2} \phi(x)~:=~\sqrt{\frac{2}{L}}\sin\frac{\pi x}{L}, \qquad E_1~:=~ \frac{\hbar^2}{2m}\frac{\pi^2}{L^2}.$$ Next we are supposed to incorporate the "full" effect of the delta function $\delta(t)$ (as opposed to ...

1

The derivation of the Fermi-Dirac distribution using the density matrix formalism proceeds as follows: The setup. We assume that the single-particle hamiltonian has a discrete spectrum, so the single-particle energy eigenstates are labeled by an index $i$ which runs over some finite or countably infinite index set $I$. A basis for the Hilbert space of the ...

4

I cannot understand why you wrote that $\eta$ is the rapidity: It would be the rapidity if $L$ were the boost, but this is not the case because the internal sign in the RHS of the formula defining $L$ is wrong. Your $L$ is formally an angular momentum if you do not pay attention to the weird name of the variable $t$, time? Well, in addition to the ...

0

In general every linear combinations of a separable solution is still a solution (superposition principle), so you can take the simplest separable solution, put it in a linear combination with arbitrary coefficient (complex numbers, phases) and you obtained a solution of Sc. equation NON-separable.

8

Flip back a page; Dirac uses real to mean Hermitian when talking about linear operators. So you can see that even if $A$ and $B$ are Hermitian, $AB$ won't be Hermitian unless they commute, whereas those linear combinations will be.

0

You are deal with action variables so the integrations extrem are always give by the classical trajectory. So for the hidrogen atom in bound state they are 2 times the distance between R1 and R2.

0

How do I transform equation (1) to equation (2) plug $R(r)=u(r)/r$ into (1), you'll get (2) immediately, where $k(x)$ would the expression before $R(r)$ in the second term of (2) and what do I use for the bounds of integration in equation (4) to get the energy eigenvalue? $\int_0^\infty$

0

@ChrisWhite is right I believe you don't need to worry about that since using the tensor product notation $|x s\rangle = |x \rangle \otimes |s\rangle$ and the angular momentum and spin operators just act on their part of the Hilbert space so we can write them as $l=l\oplus1_s$ and $s=1_x \oplus s$ and they would act in the following way $$l|x s\rangle = ... 0 The bounds for r should still be the classical turning points, as you mentioned for the harmonic oscillator. Presumably you're in a bound state of Hydrogen, i.e. have an energy of the form \frac{-13.6 eV}{n^2} for some integer n. The problem then reduces to finding the zeros of the equation$$\frac{-13.6 eV}{n^2} = -\frac{e}{r^2} - \frac{l(l+1) ...

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