44

Yes, nucleus is composed of subatomic particles that have probability cloud. Protons and neutrons fill orbitals in the nucleus just like electrons in the atom do. What's more, every proton or neutron is a complex particle itself and the quarks inside have their very own probability cloud. (Quarks are simple objects that have no internal structure as far as ...


31

Complex numbers are used in all of mathematics, and therefore by extension they are used in other fields that require math; not just physics, but also engineering and other fields. Trying to assign a "physical interpretation" to a complex number would be like assigning a physical interpretation to a real number, such as the number 5. A complex number is ...


28

It is actually true, in an almost trivial way. The Ehrenfest theorem states that, \begin{equation} \frac{d}{dt}\langle x\rangle=\langle p\rangle,\quad \frac{d}{dt}\langle p\rangle =- \langle V'(x)\rangle \end{equation} However for all eigenfunctions for the Harmonic oscillator $\langle x\rangle=0$ (and therefore $\langle V'(x)\rangle=0$) and $\langle p\...


27

Complex numbers are used in "macro" physics. They are used in analysis of electrical circuits (especially when AC is involved) and in fluid dynamics. Solution of differential equations is simplified if complex numbers are used, as is Fourier analysis. Any scenario that involves periodic or cyclic functions can be modeled using complex numbers.


23

Very often indeed the nucleus is assumed motionless. It is then assumed that the motion of the nuclei and the electrons can be treated separately. This is known as the Born-Oppenheimer approximation. The reason is that solving the equations simultaneously is very difficult and would not be very efficient. Note that for the hydrogen atom this approximation ...


17

You are thinking of the photon as an extended classical wave, which is not correct within the standard model of particle physics. Photons are point like quantum mechanical particles, and quantum mechanics is all about probabilities. A photon approaching a black hole will have a calculable probability to fall through and be captured, or join the photon ...


15

The system doesn't "know" anything. The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $\delta$-function, if we ...


14

The nucleus does have a probability cloud. As the simplest example, consider the hydrogen-1 atom. Conservation of momentum requires that the center of mass of the electron and proton remain fixed. Therefore we have $$\Psi_p(\textbf{x}|=(\text{const.})\Psi_e(-\alpha\textbf{ x}),$$ where $\Psi_p$ is the wavefunction of the proton, $\Psi_e$ is the ...


14

I see the fatal flaw here. Inside the well, your solution must be a solution to the HO, BUT there are two solutions to the HO. There is the one we're all familiar with, a hermite polynomial times a gaussian, but $e^{x^2}$ is also a valid solution. For the HO, we discard that solution because it isn't normalizeable at the boundaries. In the finite well ...


13

Formally self-adjoint, but unbounded, operators can easily take a normalizable state (i.e a state in the Hilbert space) and make it unnormalizable and therefore no longer in the space. This can lead to all sorts of aparent paradoxes. For example consider the operator $p^4= \partial_x^4$ on infinite square well $[0,1]$. Let it act on the wavefunction $$ \...


12

The fundamental object in quantum mechanics is the amplitude, which encodes information about how a system transitions from one state to another state. For example, if you are doing a double slit experiment you might care about how an electron transitions from the incoming pre-slit state to a state where it hits a certain location $x$ on the detector. For ...


11

Your claim [any arbitrary] wavefunction of an electron $\psi(\vec{r},s)$ can always be written as $$\psi(\vec{r},s)=\phi(\vec{r})\zeta_{s,m_s} \tag 1$$ where $\phi(\vec{r})$ is the space part and $\zeta_{s,m_s}$ is the spin part of the total wavefunction $\psi(\vec{r},s)$ is false. It is perfectly possible to produce wavefunctions which cannot be written ...


11

Short answer: yes, but only the phase factor has the time dependence. The spatial profile is constant in time because the eigenstates of the Hamiltonian are Stationary states. Maths: The time dependent Schroedinger equation looks like this: $$ i\hbar \frac{\partial \Psi}{\partial t} = H \Psi = \left ( -\frac{\hbar^2 }{2 m}\frac{\partial^2}{\partial x^2} + ...


11

It's just notation. $H_1$ acts on Hilbert space ${\mathcal H}_1$, $H_2$ acts on Hilbert space ${\mathcal H}_2$. By $ |\psi_1,\psi_2\rangle$ they have an implicit tensor product $$ |\psi_1,\psi_2\rangle\stackrel{\rm def}{=} |\psi_1\rangle\otimes |\psi_2\rangle\in {\mathcal H}_1\otimes {\mathcal H}_2. $$ Then physiscists, usually without saying so, extend $...


10

Complex number as any number alone does not say anything about physics at all. It has to be bound to some measurement unit(s) or have a well-defined definition in physics. For example complex refractive index is defined in physics as : $$ {\displaystyle {\underline {n}}=n+i\kappa .} $$ Here imaginary part $\kappa$ is defined as attenuation coefficient - ...


9

From $\mathbf{L}=\mathbf{Q}\times \mathbf{P}$ we have $L_z=Q_xP_y-Q_yP_x$. Then, introduce the following new operators (assuming units of $\hbar=1$): \begin{align} q_1=\frac{Q_x+P_y}{\sqrt{2}},\\ q_2=\frac{Q_x-P_y}{\sqrt{2}},\\ p_1=\frac{P_x-Q_y}{\sqrt{2}},\\ p_2=\frac{P_x+Q_y}{\sqrt{2}}. \end{align} It is immediate to check that $$ [q_1,q_2]=[p_1,p_2]=0,\...


9

Welcome to SE -- good question! Decoherence does not mean that there won't be a wavefunction anymore, it just means that if the electron becomes coupled to the surrounding environment, its state will be described by a probabilistic mixture of orbital wavefunctions rather than a (coherent) superposition thereof. The electron in an atom doesn't have some "non-...


9

In order to calculate the matrix elements of an operator, you need to know how the operator acts on the basis states. This is usually something that you're given (because it's almost always how an operator is defined in the first place - by its action on basis states). So, if you want to calculate the matrix elements, you need at least that much information (...


8

Here is an argument. On one hand, in 1D the $n$th bound state has $n\!-\!1$ nodes. But a bound state in the delta function well is in a classically forbidden region (and hence exponentially decaying) for $x\in \mathbb{R}\backslash\{0\}$, so there cannot be any nodes. Hence $n\leq 1$. On the other hand, any attractive potential in 1D has a bound state, so $...


8

Introduction First of all we need to establish what the question even means (I will restrict all formulas to non-rotating Schwarzschild black holes of mass $M$ for concreteness). There are no bound states of the electromagnetic field near a black hole, even quasi-stationary states of finite wavelength near the photon radius $r = 3GM/c^2$ will decay, a part ...


8

I think your formula is confused. The wavefunction is $$ \psi(x) = \langle x\vert \psi\rangle = \int \delta(x-x') \psi(x')\,dx' $$ where $\delta(x-x')= \langle x'\vert x\rangle$ is the wavefuction of the position eigenfunction $\vert x\rangle$ in the position eigenfunction basis. This not what you have written with the "$x$" operator. For the ocillator ...


8

The answer is that the protons in the nucleus are quantum particles and don't have a well-defined position, but the uncertainty isn't a big factor in determining the potential experienced by the orbiting electrons, so we can just treat them as a fixed source of potential. That does, also, make the calculations much, much easier.


7

There's a lot you can do, but the short of it is that the eigen-energy quantization condition still boils down to a transcendental equation that must be solved numerically. Start with the wavefunction Ansatz $$ \psi(x) = \begin{cases} A \sin(kx) & 0<x<a \\ B \sin(k(L-x)) & a<x<L \end{cases} $$ where $E = \hbar^2 k^2/2m$. To find the ...


7

The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement. Always remember the wavefunction isn't physical. It'...


7

The delta function here can be thought of a position eigenstate, but as it is not square-integrable, it cannot be an actual wave function. Hence, "unrealistic". An actual, square-integrable wave function can be thought as being made of a superposition of dirac deltas, however: $$ \psi(x) = \int \psi(x_0) \delta(x - x_0) dx_0. $$ You can think of this as a ...


7

Simply, it is just a matter of definitions. Being $$ a=\frac{\hbar^24\pi\epsilon_0}{me^2}, $$ a constant generally called Bohr's radius, you can invert this formula to obtain $e^2/4\pi\epsilon_0$ instead. Then, put it in your equation and you are done.


7

Expanding, we have: $$Ae^{ikx}+Be^{-ikx}=(A+B)\sin(kx)+i(A-B)\cos(kx)$$ Setting $A'=A+B$ and $B'=i(A-B)$ we have that $$Ae^{ikx}+Be^{-ikx}=A'\sin(kx)+B'\cos(kx)$$ so the two solutions are equivalent. It really doesn't matter in the end.


6

Saying that a photon doesn't have a wavefunction can be misleading. A more accurate way to say it is that a photon doesn't have a strict position observable. A photon can't be strictly localized in any finite region of space. It can be approximately localized, so that it might as well be restricted to a finite region for all practical purposes; but the ...


6

Does the particle continue to be in the same eigenstate of the Hamiltonian? No. You've performed a binary measurement, i.e. the question "is the particle in the interval $[x_a,x_b]$?", with answers "yes" and "no" corresponding to the projection operators $$ \Pi_1 = \int_{x_a}^{x_b} |x\rangle\langle x | \,\mathrm dx $$ and $$ \Pi_0 = \mathbb I - \Pi_1 = \...


Only top voted, non community-wiki answers of a minimum length are eligible