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$$x^0_\;\mathrm{nov}=\frac{x^0-\frac{v}{c}x^1}{\sqrt{1-\(\frac{v}{c}\)^2}}\,,\quad x^1_\mathrm{nov}=\frac{-\frac{v}{c}x^0 x^1}{\sqrt{1-\(\frac{v}{c}\)^2}}\,,\quad x^2_\mathrm{nov}= x^2\,,\quad x^3_\mathrm{nov}= x^3$$ leaves the spacetime interval invariant. * Set $Δx=Δx=0$ in the definition of a spacetime interval. You should get
$$(\Delta s)^2 = -\(\Delta x^0\)^2 \(\Delta x^1\)^2$$
What is the region of the plane ( $Δx^{0},Δx¹;)$ where the spacetime interval ( $Δs)$ is positive? Where negative? What is the curve ( $Δs)=0?$
$$F_\;\mathrm{d} = m a_\mathrm{d} = \frac{\alpha}{r^2}\,,$$
where $ris$ the radius of a circle and $α$ is some constant. Then
Tip Youshould have encountered radial motion in your high-school education and also the relationships between displacement, velocity and acceleration. Use them and then the integration of action along the circumference of the circle with a constant $r$ shall become easier (constant quanties can be easily factored out of the integral). Don't forget that the path integral of „nothing“ is merely the length of the integrated path.
$$\mathbf{y}(t)&=\left(2t,t\right) \,,\\\mathbf{y}(t)&=\left(1-\cos{(\pi t)} \frac{1}{\pi}\sin{2\pi t}, t\right) \,,\\\mathbf{y}(t)&=\left(2t, \frac{\;\mathrm{e}^t-1 t^2(t-1)}{\mathrm{e}-1}\right) \,,$$
where e is the Euler number. Tip First find the derivative of $\textbf{y}(t)$, put it into the equation for action and integrate.
$$y(x,t)=\sin ({\alpha} x)\left [a\sin {({\omega} t)} b\cos {({\omega} t)}\right ]$$
a solution of the wave equation? Tip Subsitute into the equation for motion and use the boundary conditions.
$$X({\tau} , {\sigma} )=(c{\tau} , R\cos {{\sigma} }, R\sin {{\sigma} },0)$$
for $τ∈\langle0,2π\rangle$. Furthermore sketch the worldsheet of this string (forget about the last zero component) and how the line of a constant $τ$ and $σ$ look.
src=„https://latex.codecogs.com/gif.latex?\hat%20X“>$
and momentum $<img$ src=„https://latex.codecogs.com/gif.latex?\hat
%20P“>$ acts on the components of the state vector in $x-$
representation (wave function) and calculate their comutator, in other
words
<img src=„https://latex.codecogs.com/gif.latex?(\hat%20{X})_x%20\left((\hat%20
{P})_x%20{\psi}%20(x)\right)%20-%20(\hat%20{P})_x%20\left((\hat%20{X})_x%20
{\psi}%20(x)\right)%20“>
Tip Find out what happens when you take the derivative
of two functions multiplied together
for $V(x)=0$ has the
following form:
<img src=„https://latex.codecogs.com/gif.latex?-\frac%20{\hbar%20^2}
{2m}%20\dfrac{\partial^2%20{\psi}%20(x)}{\partial%20x^2}=%20E%20{\psi}%20
(x)\,.“>
( $x)=e^{αx}$ as the solution
and find out for what $α$ (a general complex number)
is $Epositive$ (only use such $α$ from now on).
(wavelength)?
(in the $x-representation)?$ If yes find the relation between
wavelength and momentum (in other words the respective eigenvalue) of the state.
particle in space.naší vlnové funkci podle vzorce uvedeného v textu. Pravděpodobnost, že se
částice vyskytuje v celém prostoru by měla být pro fyzikální hustotu pravděpodobnosti 1,
tj. <img src=„https://latex.codecogs.com/gif.latex?\int_\mathbb{R}%20\rho
(x)%20\mathrm{d}%20x=1.“> Show that our wave function can't be
$normalized$ (in other words multiply by some constant) so that its formal
density of probability according to the equation from the text was a real
physical density of probability.
uncertainity of a position of a particle is if the wave function it has is close
to ours (In other words it approaches it in all properties but it always has a
normalized probability density and thus is a physical state) Can we (using Heisenberg's relation of uncertainty) determine what is
the lowest possible imprecision while finding the momentum?
Tip Take care when dealing with complex numbers. For
example the square of a complex number is different than that of its magnitude.
electron in hydrogen using reduced action. Due to a random happenstance the
solution of the spectrum of the hamiltonian in a coulombic potential of a
proton would lead to thecompletely same energy,in other words
<img src=„https://latex.codecogs.com/gif.latex?E_n%20=%20-{\mathrm{Ry}}%20\frac
%20{1}{n^2}“>
where Ty = 13,6 eV is an ernergy constant that is known
as the Rydberg constant. An electron which falls from a random energy
level to $n=2$ shall emit energy in the form of a proton
and the magnitude of the energy shall be equal to the diference of the energies
of the two states. Which are the states that an electron can fall from so that
the light will be in the visible spectrum? What will the color of the spectral
lines be?
Tip Remember the photoelectric
effect and the relation between the frequency of light and its
wavelength.
Where can the strings end in the case of three parallel D2-branes?
$$\mathcal{P}_{\mu}^{\tau}$$
ot $$\mathcal{P}_{\mu}^{\sigma}$$ that was defined in the first part of the series and find its explicit
form (in other words a direct dependence on $$\dot{X}^{\mu}$$ and <img
src=„https://latex.codecogs.com/gif.latex?X'^{\mu}“>). Show that the conditions $$\vect{X}'\cdot \dot{\vect{X}}=0$$
and $$|\dot{\vect{X}}|^2=-|\vect{X}'|^2$$
$$\hat{H}=\frac{\hat{p}^2}{2m} \frac{1}{2}m\omega^2\hat{x}^2$$
The second expression is clearly the potential energy while the first gives after substituting in $$\hat{p}=m\hat{v}$$ kinetic energy. We define linear combination as
$$\hat{\alpha}=a\hat{x} \;\mathrm{i} b\hat{p}$$ . Find the real constants <img
src=„https://latex.codecogs.com/gif.latex?a“> a $b$ , such that the Hamiltonian will have the form of
<img src=„https://latex.codecogs.com/gif.latex?\hat{H}=\hbar \omega \left(\hat{\alpha} ^{\dagger}\hat{\alpha}+\frac{1}
{2}\right)\,,“> where $$\hat{\alpha} ^{\dagger}$$ is the complex conjugate <img
src=„https://latex.codecogs.com/gif.latex?\hat{\alpha}“>.
$$\hat{x}$$
and $$\hat{p}$$ that the following is true
<img src=„https://latex.codecogs.com/gif.latex?\left[\hat{\alpha},\hat{\alpha}\right]=0\,,\quad\left[\hat{\alpha} ^{\dagger},\hat{\alpha} ^
{\dagger}\right]=0\,,\quad\left[\hat{\alpha} ,\hat{\alpha} ^{\dagger}\right]=1\,.“>
amount of oscilating. Lets call it $$|0\rangle$$ . This state must fulfill <img
src=„https://latex.codecogs.com/gif.latex?\alpha |0\rangle =0“>. Show that its energy is equal to $$\hbar\omega/2$$ , ie. $$\hat{H}|0\rangle=\hbar\omega/2|0\rangle$$ . Furthermore prove that if $$\alpha |0\rangle \neq 0$$ then we have a contradiction with the fact that <img
src=„https://latex.codecogs.com/gif.latex?|0\rangle“> has a minimal energy ie. <img src=„https://latex.codecogs.com/gif.latex?\hat{H}\alpha |0\rangle=E\alpha
$$E<\hbar\omega/2$$ . All the eigenstates of the Hamiltonian can be described
as $$\left(\alpha^{\dagger}\right) ^n|0\rangle$$
for $$n=0,1,2,\dots$$ Find the energy of these states, in other words find such numbers <img
src=„https://latex.codecogs.com/gif.latex?E_n“> that <img src=„https://latex.codecogs.com/gif.latex?\hat{H}\left(\alpha^{\dagger}\right) ^n|0\rangle=E_n\left
(\alpha^{\dagger}\right)^n|0\rangle“>.
Tip Use the commutation relation for $$\hat{\alpha}^{\dagger}$$ a <img
src=„https://latex.codecogs.com/gif.latex?\hat{\alpha}“>.
$$V(\phi)=\frac{1}{3\alpha'}\frac{1}{2\phi _0}(\phi-\phi _0)^2\left (\phi \frac{1}{2}\phi _0\right )\,,$$
where $$\alpha'$$
$$\{A,B\}=AB BA$$
Find two such $$2\times 2$$
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