# Tag Info

6

This issue is a bit confused in textbooks, however the statement of the professor is physically wrong (mathematically all the procedure can be rigorously justified using the theory of distributions). The point is that the claimed position operator is not the position operator because it is not even self-adjoint (nor Hermitian) in the relevant Hilbert space ...

4

Hint: When an eigenvalue for an operator $P$ is degenerate, there are more than one way to chose a set of eigenvectors. If the other commuting operator $H$ lifts that degeneracy, there will be a preferred choice of common eigenvectors. More generally, a set of diagonalizable operators commutes if and only if the set is simultaneously diagonalizable.$^1$ ...

4

Call $u_1, u_2, u_3, u_4$ the eigenvectors described by you, respectively. Your claims are all right, but realize that both $u_1$ and $u_2$ share the same eigenvalue, that is $1$, i.e., $Pu_1=u_1$ and $Pu_2=u_2$. Hence, any linear combination of $u_1$ and $u_2$ will also be eigenvectors with the same eigenvalue $1$. Try to find eigenvectors of $H$ of the ...

4

It seems OP's question (v4) is related to the proper handling of derivatives of Dirac delta distributions. Reductions are performed with the help of (the appropriate 3D generalizations of) the following formulas: $$\tag{A} \{\partial_x+\partial_y\}\delta (x-y)~=~ 0,$$ $$\tag{B} \{f(x)-f(y)\}~\delta (x-y)~=~ 0,$$ $$\tag{C} \{f(y)-f(x)\}~\partial_x \delta ... 3 I think your plotting routine is expecting angles in radians, not degrees. The starting angle of both graphs is wrong. The first is not 45^\circ, and the second is nowhere near 30^\circ. \tan(30)=-6.4 (if 30 is in radians) and that would explain the negative start. Also: did you square the cosine in the denominator? The formulas above the graph ... 3 i do not understand the part of clamping the center of the coil or cutting it off The solution method appeals to symmetry. Essentially, the system is symmetric (or even about the center of the spring) and this symmetry is exploited to arrive at an answer. If you have taken a first semester in electrostatics, you may be aware of the method of images. ... 3 As suggested by Frederic, you should integrate the$$ \int d^n x \: \alpha^a(x) \cdot \nabla_a \delta \Psi $$term (with whatever \alpha should be) by parts using the Ostrogradsky-Gauss theorem and the boundary condition \delta \Psi (\infty) \rightarrow 0. But Euler and Lagrange have already done it for you, so you can just use the formula for an ... 2 Your maths is almost correct. I think you should treat the lever as two separate rods which each have a pivot at one end. If the mass per unit length is \sigma = M/L, then the following applies to the two rods of length r_1 and r_2 (with r_2=L-r_1), where I split the rod up into infinitesimal chunks of length dr, mass dm=\sigma dr and moment of ... 2 For the eye, draw a picture. You have an isosceles triangle with the small angle one arc minute and base 10 AU. What is the altitude? Bisect the small angle and you have two right triangles. For the second, you need to compute the small angle of the same triangle from the diffraction limit on a diameter of 2.4 meters. Can you find that equation in your ... 2 The phrase "the function is spherically simmetrical" means that, if G is an orthogonal transformation (that sends spheres into themselves), then$$f(G\mathbf r, G\mathbf p,t)=f(\mathbf r , \mathbf p, t).$$If you know \mathbf r^2, \mathbf p^2, \mathbf r \cdot \mathbf p you can calculate f by taking an orthogonal transformation which maps \mathbf ... 2 Here is the answer (I will not consider the constants on the denominator of your Fourier transform for simplicity, however they are there ;-) ). When you write the operator \hat{\phi} you have to be careful. I will drop the hats, because it will be clearer I think (maybe here the hat stands for an operator and not for the fourier transform). Your operator ... 2 You can start by reading the wikipedia article on the method of characteristics. You will see that in our case the tangent of the characteristic is (1,-\partial H/\partial q,\partial H/ \partial p) where the components are in order t,p,q. When you formulate the equation of the characteristic, you will actually find out you get equations of motion of a ... 2 Though I agree with the logic, I find I cannot reproduce this quantitatively. I get an electron number density of 1.2 \times 10^{41} m^{-3} (is it just a unit thing?) for a Fermi energy of 30MeV. In a carbon white dwarf with 2 mass units per electron, the Fermi energy of the electrons reaches 30MeV at densities of 4\times 10^{14} kg/m^{3} - i.e. at ... 1 The inherent idea is, from that equation$$I = \exp (\frac{eU}{k_B T})$$if you plot (\ln I) versus U, that would be a straight line (of the form y=mx), with a slope$$\alpha = \frac{e}{k_B T}$$That's all you have to do in the experiment, use least square fitting to find an accurate value of \alpha and then find the Boltzmann constant using the ... 1 Lie algebras are not a group w.r.t. to the commutator (the Lie bracket). The first reason is that the commutator is not associative. Another is that they almost always lack an identity element, since the identity matrix is, for example, not in \mathfrak{su}(2), and Schur's lemma would, in the fundamental representation, guarantee that only multiples of ... 1 Your Hilbert space is finite-dimensional (specifically, 4-dimensional). It means that you need no fancy way of deriving the spectrum, just proceed with the standard approach: Write your Hamiltonian in the matrix form and solve the characteristic equation  \det ( H - \lambda \cdot 1_{4 \times 4} ) = 0  with respect to \lambda, which would give you ... 1 Start with the lower expression:$$ (c/2)\eta^{bc}\eta^{ae}\partial_{a}\left(g_{be,c} + g_{ce,b} - g_{bc,e}\right) - (c/2)\eta^{ae}\eta^{bc}\partial_{c}\left(g_{be,a} + g_{ae,b} - g_{ba,e}\right).\\ = \frac{c}{2}\eta^{bc}\eta^{ae}\left(g_{be,ca} + g_{ce,ba} - g_{bc,ea}-g_{be,ac} - g_{ae,bc} + g_{ba,ec}\right)\\ =\frac{c}{2}\eta^{bc}\eta^{ae}\left(g_{ce,ba} - ...

1

Just for the sake of having an answer: In $\phi \nabla ^2 \phi$, consider the term $\phi \frac{\partial ^2 \phi}{\partial x^2}$. By the product rule, this is equal to $\frac{\partial}{\partial x} \left(\phi \frac{\partial \phi}{\partial x}\right) - \left(\frac{\partial \phi}{\partial x}\right)^2$. Combining all three terms, we get $\phi \nabla^2 \phi = ... 1 The equation of motion for one mass$m$attached to a spring with spring constant$k$(with the other end fixed) on Earth is $$\ddot{x} + \omega^2 x = g$$ where$2 \pi f = \omega = \sqrt{k / m}$. If it's not obvious that the$g$term on the right hand side doesn't affect the frequency, notice that if we changed our origin so that$z = 0$when$x = ...

1

As was said in the comments, my attempt to a solution is to decompose $\hat{x}$ as a linear combination $\alpha \hat{A} + \beta \hat{B}$. This seems to be general, because if $\hat{p}$ is a linear combination of $\hat{A}$ and $\hat{B}$, then the position will be too, since it's the inverse Fourier transform of $\hat{p}$ and such transformation is linear. I'd ...

1

The answer depends entirely on the composition of the air currently in the tube, and the temperature. I think we can assume that you have saturated the volume of air - which is why you have condensation. How much water is in the air is a sensitive function of temperature - I create a plot below using data from ...

1

You need some kind of a force, either friction, or gravity (if the road is slanted for instance) to account for the "forces" side of the 2nd Newton law: $$\sum\vec{F}=m\vec{a}$$ Whereas the Right Hand Side (RHS) should contain the centripetal acceleration, i.e.: $$m\vec{a}=-m\frac{v^2}{r}\hat{r}$$ And the minus means it is directed towards the center of ...

1

Use $d=v_i t +\frac{1}{2}a t^2$ and then substitute from the first equation that $v_i=-a t$. You get $$d= -at^2+\frac{1}{2}at^2=-\frac{1}{2}at^2$$ from which you can infer $$a=-\frac{2d}{t^2}$$ Once you have the acceleration you get the initial velocity from $v_i=-a t$.

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You can always write any density matrix in the form $\rho = e^{-H_A}$ in a limiting sense, as long as you allow some of the parameters entering $H_A$ to become infinite. This follows from the identity $$\rho = e^{\ln \rho},$$ but notice that if $\rho$ has some eigenvalues equal to zero, then the eigenvalues of the exponent become infinite (or are not ...

1

All celestial bodies loose atmosphere due to a portion of the gas "near space" exceeding escape velocity. The velocity distribution of an ideal gas can be found using the Maxwell-Boltzmann distribution. So an easy approximation for this problem is to say we only want $10^{-6}$ of the molecules to have escape velocity. Using oxygen at 300K, results in an ...

1

It seems that that you're converting from a volume ($R_0^3$) (is the minus sign a typo?) to a mass($M_0$). The constant of proportionality is a density, presumably the $\rho_0$ that you're given. In more general terms, the constants that are hidden by the $\propto$ symbol can have units too.

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As Lelesquiz points out, that looks like a somewhat standard matrix differential equation to me. The Wikipedia link does give a solution method for the matrix, but I think that it might be easier to do some remapping: \begin{align} \rho_{00}&\to v_1\\ \rho_{01}&\to v_2\\ \rho_{10}&\to v_3\\ \rho_{11}&\to v_4 \end{align} and write it as ...

1

Since you have the expression of $\Omega$ your work is almost done. First remember that the entropy for a micro-canonical (fixed energy) system at thermal equilibrium is given the very famous Boltzmann's formula : $$S=k_B\,ln(\Omega)$$ Then, simply use the Stirling's approximation to evaluate $ln(N!)\approx Nln(N)-N$ (because $N>>1$, i.e. very long ...

1

B will move faster, the reason is that the acceleration, $a$, of A is smaller for two reasons (remember that $F_{applied}-F_{drag}=ma$) : 1) the same force forward is applied so the contribution to the acceleration on the smaller ball will be larger 2)the drag force on the larger ball will be larger (see Rennie's comment) on A because the cross sectional ...

1

It would be highly unreasonable to obtain the uncertainty of the average of the measurements to be greater than the uncertainty of any one measurement (√5/2 > 1). After all, what's the point in taking averages if it just makes your readings more uncertain? I believe the ETS people used the argument that the harmonic sum of the individual variances should ...

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