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## Hot answers tagged quantum-mechanics

20

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.

15

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 ...

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 ...

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.

6

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: ...

5

Schrodinger's equation cannot be derived. It was thought up using logical arguments and so far it has seemed to work experimentally. The equations is essentially a re-write up for energy conservation: $$E = T + V$$ Where $T$ is the Kinetic Energy and $V$ is the potential. However, to be more explicit we must work with operators (if you are unsure what ...

4

You might be interested in this "elementary" derivation of the free particle Schroedinger equation from Maxwell's equations. It seems to be in the same spirit as Schroedinger's original reasoning. The niceness of this approach is that if you also include special relativity, it nets you both the free particle Schroedinger equation and its relativistic ...

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 ...

3

From Feynman's lectures :) Where did we get that from? It's not possible to derive it from anything you know. It came out of the mind of Schrödinger. ADDITION Though a postulate such as Schrödinger equation cannot be proven, one can notice that in QM a state vector $\Psi$ is said to give the most complete description of a state of a system. So it is ...

3

If you represent the wave function $\psi(x)$ with it's fourier transform, \begin{eqnarray*} \psi(x) &=& \frac{1}{\sqrt{2\pi \hbar}}\int \tilde{\psi}(p)e^{\frac{ipx}{\hbar}}dp\\ \psi(x)^\star &=& \frac{1}{\sqrt{2\pi \hbar}} \int \tilde{\psi}^\star(q)e^{\frac{-iqx}{\hbar}}dq \end{eqnarray*} (where p and q are almost like "dummy" momenta), ...

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 $\Psi(r,t)$ ...

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 ...

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 ...

2

If you don't want the answer you could look in the wiki Group Velocity. In particular: One derivation of the formula for group velocity is as follows. Consider a wave packet as a function of position $x$ and time $t$: $α(x,t)$. Let $A(k)$ be its Fourier transform at time $t=0$: $\alpha(x,0)= \int_{-\infty}^\infty dk \, A(k) e^{ikx}$, By the superposition ...

2

Wow that's an interesting question. I believe the answer is no, and the reason is that you can have light, or no light, but you cannot have "not light." If that makes sense. Darkness is the absence of light, so you cannot really have a material darkness. A similar idea is that cold is just a lack of heat. If you make something cold, it is not because ...

2

All of them. Even molecules show their wave-like nature, as does, in principle, every object. Speaking of these topics an interesting read about diffraction of C60 molecules is: http://www.univie.ac.at/qfp/research/matterwave/c60/ The point is that the wave-like nature of objects can only be observed at lengths comparable the object's De Broglie wavelength, ...

1

Black isn't a color. It's just absence of light. So, a torch can't project light having such color. However, for the second part (dis-illuminate everything), I have an answer: Light is wave and waves can cancel each other. Projecting specially programmed adjustable light wave which create fully destructive interference with other available light waves can ...

1

The answer to your first question is no. Consider the three-dimensional Hilbert space $\mathbb C^4$, and let $B_1$ be the canonical basis and $$B_2=\left\{ \frac{1}{2}\begin{pmatrix}1\\1\\1\\1\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1\\ i\\-1\\-i\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1 \\ -1\\1\\-1\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1\\ ... 1 If your desired basis is the set {|n\rangle}, then the completeness relation tells you: \hat{O} = \sum_a \sum_b \langle a|\hat{O}|b \rangle |a \rangle \langle b|. Ideally, we prefer to do this in the orthonormal basis in which the operator \hat{O} is diagonal, in which case this becomes \hat{O} = \sum_a \langle a|\hat{O}|a \rangle |a \rangle \langle ... 1 The time dependent Schrodinger equation is one of 5 (or 6) postulates of quantum mechanics. It is not proper to say that it is derived, unless you have a different set of postulates. for example, in the references below, the time dependent Schrodinger equation is the 5th postulate. http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html ... 1 Quantum mechanically bound particles (like electrons in an atom or molecule) do not have well defined momentum.1 What they do have is a well defined distribution of momenta. This is one of the reasons that we say electrons are bound in "orbitals" and not in "orbits". That distribution is the same for the orbital around the same type of atom (or in the ... 1 Heisenberg's uncertainty principle is$$\Delta x \Delta p \geq \hbar/2.$$Since the well is of width L, you have a measure for the uncertainty on the position \Delta x. Then assume the lowest possible value for \Delta p, i.e. the one for which the above inequality becomes an equality. Lastly, use E = \dfrac{p^2}{2m} to find an expression for E. ... 1 Consider a potential, which approximately can be described by two harmonic oscillators with different base frequencies, for example (working in dimensionless units)$$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 ...

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 ...

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 ...

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