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For an example in a harmonic oscillator energy eigenstate, position x and momentum p both have Gaussian probability distributions around zero. We have tiny but not zero probality to observe any large p as well as x.

Each mometum eigenfunction has eigenvalue which is a definite value of momentum of the state.

Why don't you calculate \Delta^2 F by hand and compare it with your formula of n=2 for confirmation ? In notation \partial_j F = F_{,j} \Delta...

My complement A classical perfect reflection analogue of \alpha|p'>+\beta|-p'> where p'=\sqrt{2mE'} isn't a solution neither.

Just a quick notice that ##\nabla## here should be Laplacian ##\Delta##.

For this exercise for the biginners I prefer observing what happens in finite well potential V system in the limit of ##V \rightarrow \infty## to discussing mathematical subject of ##V = \infty##. We may keep the interval ##[-\infty,+\infty]## and apply usual Fourier transform method to get...

In free space system of zero potential energy everywhere, yes.

Yes in the well but it is zero outside. That matters. It is superposition of various mometum eigenstates, the coefficiets of which provide probability to observe specific momentum value by product with complex conjugate. Yes inside but it includes no zero potential energy outside. That matters.

I think covalent bond is the key word for you to understand the issue.

As a first try I take it easy as follows. Say B is vacuum, gas in A expands free adiabatically. Temperature does not change from ##T_A## . ##P_A## drops to ##\frac{V_A}{V_A+V_B}P_A##. Say A is vacuum, gas in B expands free adiabatically. Temperature does not change from ##T_B##. ##P_B##...

Explicitly \Delta^n F=\frac{1}{\sqrt{\vert g\vert}}\partial_{i_1}(\sqrt{\vert g\vert} g^{i_1k_1}\partial_{k_1} \frac{1}{\sqrt{\vert g\vert}}\partial_{i_2}(\sqrt{\vert g\vert} g^{i_2k_2}\partial_{k_2} ... \frac{1}{\sqrt{\vert g\vert}}\partial_{i_n}(\sqrt{\vert g\vert} g^{i_nk_n}\partial_{k_n} F...

If \Delta G=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kG\right), for any G, then choose G as G= \Delta^{n-1} F I cannot find meaning of odd power of nabla applying scalar function. For example can you show where we meet...

If you have already proved the formula in OP of \Delta F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kF\right) I think you have a good reason of recurrence formula \Delta^n F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert}...

\nabla f=[\frac{\partial }{\partial x}\mathbf{i}+\frac{\partial }{\partial y}\mathbf{j}+\frac{\partial }{\partial z}\mathbf{k}]f , a vector. Its innerproduct with itself is \nabla \cdot \nabla f= [\frac{\partial^2 }{\partial x^2}+\frac{\partial^2 }{\partial y^2}+\frac{\partial^2 }{\partial...

As for 2D lattice we can make one of the lattice points is (0,0) without losing generality. Say other points are ##(n_1,n_2),(m_1,m_2)## n_1^2+n_2^2=A m_1^2+m_2^2=A (n_1-m_1)^2+(n_2-m_2)^2=A where A is square of the side length. You will find contradiction in this set of formla.

Rather than "the nearest integer", I prefer floor function \lfloor x \rfloor , maximum integer that does not exceed x. \lfloor 1.2 \rfloor=1,\ 1.2=1+0.2 \lfloor -1.4 \rfloor=-2,\ -1.4=-2+0.6 z=(1-2i)+(0.2+0.6i)

The last term is 2. For the other sums you shall use the formula 1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}

1 Say a wall in the container moves outward for volume v. The gas expands adiabatically \frac{P_2}{P_1}=\frac{V^\gamma}{(V+v)^\gamma}<1 where \gamma = \frac{C_p}{C_v}>1 Then we insert a new partion where there was the wall. The number of gas molecules in the container is N_2=N_1\frac{V}{V+v}...

From the equation of state for ideal gas T_1=\frac{P_1 V}{N_1 k} T_2=\frac{P_2 V}{N_2 k} We may get the answer if we know changes of not only P but N.

From your sketch, say S is area we want, it seems 2S=\frac{1}{2}*20^2 (4\alpha)-4*\frac{1}{2}*15\cos \alpha*5-2*\frac{1}{2}*5^2(\pi+2\alpha) where \sin\alpha =\frac{1}{3} Is it same as your idea ?

Show us residues you calculated, please.

I corrected sign in the post recalling R>0. What you point out is mentioned as F(t).

I would make 0 = m(t)v(t) + \int_{0}^{t} -\dot m(u) ( v(u)+U(u))~du with U(u)<0 or 0 = m(t)v(t) + \int_{0}^{t} -\dot m(u) ( v(u)-U(u))~du with U(u) >0 for normal thrust. Thus \frac{d}{dt} [m(t)v(t)]=\dot m(t) v(t)+F(t) RHS first term : Minus. Loss of momentum which fuel to be burnt has hold in...

Yes, it is in the integral. My typo not dt but du, dummy variable for integration. I corrected it in the post. Thanks.

More comprehensive treatment.　Say rocket is at rest at t=0. Momentum of the system is conserved zero. 0=m(t)v(t) + R \int_0^t ( v(u)+U )du differentiating by t, \frac{d}{dt}[m(t)v(t)]+R [ v(t)+U ]=\frac{d}{dt}[m(t)v(t)]+Rv(t)-F=0 where m(t) is mass of rocket body and fuel in tank...

m(t) is defined by man made convention of removing ejected fuel mass so d/dt [m(t)v(t)] ##\neq## F does not surprise me. For simplicity say F=0, fuel are just disposed with no thrust or kept in tank but we take off account for m(t). From my post #20 \frac{d}{dt}[m(t)v(t)]+Rv(t)=0 We are...

The system consists of Rocket body and fuel. All the fuel is contained in rocket at time t. Some fuel are ejected from Rocket during t and t+##\triangle## t. All the fuel ejected before time t, which continues inertial motion of constant mometum any way, does not matter here. If we choose...

No, I meant P is just a momentum value that the system t and the system t+##\triangle## t share by momentum conservation law.

At time t P=m(t)v(t) At time t + ##\triangle t## P=m(t+\triangle t) v(t+\triangle t) + R\triangle t \{v(t)+U\} where U is speed of ejected fuel relative to rocket. U < 0, m(t)=m(0)-Rt RHS first term is momentum rocket body and remaing fuel hold. RHS second term is momentum that ejected fuel...

If we omit ##-B^2 y## in RHS, we can solve the simplified ODE, y=B\int_0^t du \ e^{\frac{iA}{\omega}\cos \omega u} We may be able to expect that in a short time the solution of the original ODE does not so much different from it. I would appreciate it If you could check the difference with...

Classical Poisson bracket { } , https://en.wikipedia.org/wiki/Poisson_bracket, corresponds with quantum commutator [ ] with \frac{[\ \ ]}{i\hbar} \rightarrow \{\ \ \} in classical limit. ##\frac{\partial V}{\partial x}## comes from classical Poisson bracket.

Hi, not an issue of quantum mechanics but a Bayesian statistics, I know the two envelopes problem, https://en.wikipedia.org/wiki/Two_envelopes_problem

Why don' you try integration by ##\theta## at first. The order should not matter for this exercise at least.

Change of variable ##t=\cos\theta## would make \int_{-1}^1 \frac{dt}{\sqrt{A-Bt}} Does it make sense ?

[m3/m3] has no physical dimension but [molecules/cm3] has physical dimension of L^-3. I am afraid some more details of the exercise are required to help you.

Supplement to my post #2 Say three points ##(a_1,a_2),(b_1,b_2),(c_1,c_2)## is on the circle which is caratterized by (l,m, n) \begin{pmatrix} a_1& a_2 & 1 \\ b_1& b_2 & 1 \\ c_1& c_2 & 1 \\ \end{pmatrix} \begin{pmatrix} l \\ m \\ n \\ \end{pmatrix} =- \begin{pmatrix} a_1^2+a_2^2 \\...

Contd. from my previous post The equation of line which connects two commom points of the cirlces i and j, is (l_i-l_j)x+(m_i-m_j)y+n_i-n_j=0 The system of such six lines meet at a point (x,y) are recuced to three equations. \begin{pmatrix} l_1-l_2 & m_1-m_2 & n_1-n_2 \\ l_2-l_3 & m_2-m_3 &...

I interpeted which is which as I draw on your sketch below. Does it make sense in the story of the textbook ? What is the title of the textbook ?

Interesting exercise. I take Cartesian coordinates to look at as attached figure. We can write the equation of circle touching three points as x^2+y^2+lx+my+n=0 Giving three (x,y)s of touching points, we get a set (l,m,n) . We get four (l,m,n) sets for the four circles. The system of four...

I am afraid that the potential of deutron is a finite well potential as for radius r as you have used in your calculation rather than 1/r.

So if your coefficent to the formula in post #26 is ##-4\pi A^2 V_0##, we get the same result. I agree with you that the question has something wrong.

I tried =-2\pi V_0 A^2 \int_0^R (1- \cos 2k_1r)r^2dr =-\frac{\pi A^2}{4k_1^3} V_0 \int_0^{2k_1R} (1- \cos y)y^2dy =-\frac{\pi A^2}{4k_1^3} V_0 \{ \frac{u^3}{3}-(u^2-2)\sin u -2u \cos u \} where u=2k_1R Though I am careless in calculation, do you share it with me ? Does it have anything good to...

I do not go into the mathematics but I am convinced that velocity of fluid at particle surface has no radial component. In fact \mathbf{\hat{r}} \cdot \textbf{v}_s(\hat{\textbf{r}}) = \sum {2\over{{n(n+1)}}} B_n (\mathbf{\hat{r}}\cdot (\hat{\textbf{e}}\cdot \hat{\textbf{r}}) \hat{\textbf{r}} -...

You are right but not an error because e^{2n\pi i}=1

The answer seems to make a_x=-\alpha_k (g+a)<0 negative. You can make it positve one with your setting of ##-0.5a_xt^2##.

Polar form of complex number, -1+i= \sqrt{2}\ e^{3/4\ \pi\ i}, is easy to handle power 1/3. Dividing phase by 3 and multiplying cubic roots of 1, (-1+i)^{1/3}= \sqrt[6]{2}\ e^{1/4\ \pi i + 2/3\ n\pi i} where n=-1,0,1.

Do you see continuing from post #4, ...=-\frac{\hbar^{2}}{2m} \int_{0}^{\infty} dr \psi^* \frac{\partial}{\partial r} (4\pi r^2 \frac{\partial \psi}{\partial r})=-\frac{\hbar^{2}}{2m} \int_{0}^{\infty} 4\pi r^2dr \frac{1}{r^2} \psi^*\frac{\partial}{\partial r} ( r^2 \frac{\partial \psi}{\partial...

Last year I read the news article https://scitechdaily.com/physicists-prove-that-the-imaginary-part-of-quantum-mechanics-really-exists/ which says i is necessary for QM. You may be interested in it also.

I agree with your minus result.