When the ##\psi(2s)## particle decays also the following two transitions are observed
$$\psi(2S) \rightarrow \gamma + \eta({1}^S_0)$$
$$\psi(2S) \rightarrow \gamma + \chi_{c0}({3}^P_0)$$
The branching fraction for the first decay is about ##3.4*10^{-3}## while it's ##9.7*10^{-2}## for the...
I have insertet the equations for H and P in the relation for the commutator which gives
$$[H,P] = [\sum_{n=1}^N \frac{p_n^2}{2m_n} +\frac{1}{2}\sum_{n,n'}^N V(|x_n-x_n'|),\sum_{n=1}^N p_n]
\\ = [\sum_{n=1}^N \frac{p_n^2}{2m_n},\sum_{n=1}^N p_n]+\frac{1}{2}[\sum_{n,n'}^N...
The radiaton length for air is about $$X_0 = 30420cm$$.
This is the length at which the electron has decreased to 1/e of it´ s initially value.
I also know that the maximal value of interactions for a specific energy is given by $$ n_{max} = \frac{ln(\frac{E_0}{E_c})}{ln(2)} $$, where E_c is...
No, everything should be correct.
After substituting x=sin(Theta) one gets
$$ p(x) = \frac{1}{a} \int_{-a}^{a} \sqrt{y-V_0*sin^2(\Theta)}* cos(\theta)dx $$
that can be substituted by $$u = sin(\Theta)$$
which gets one
$$ p(x) = \frac{1}{a} \int_{-a}^{a} \sqrt{y-V_0*u^2}du $$
finally one can...
Not yet. I´ m also not sure which one to use.
I´ ve now written the equation as
$$ p(x) = \frac{1}{a} \int_{-a}^{a} \sqrt{V_0(a^2-x^2)-E} dx $$
but have now idea how to go further.
Not quite, there´ s missing one bracket. It should be:
$$p(x)=\frac{1}{\hbar} \int_{-a}^{a}\sqrt{2m(V_0(1-(\frac{x}{a})^2)-E)}dx$$
with the potential $$ V(x) = \begin{cases} V_0(1-(\frac{x}{a})^2) & \text{for -a $\leq x \leq $a} \\0 & \text{otherwise} \end{cases}$$
Homework Statement: The Task is to calculate the Transmission coefficient with the WKB Approximation of following potential: V(x) = V_0(1-(x/a)²) |x|<a ; V(x) = 0 otherwise
Homework Equations: ln|T|² = -2 ∫ p(x) dx
I have inserted the potential in the equation for p(x) and recieved
p(x) =...
Thanks for the hint, I now tried to solve it and got following result:
$$\tilde{\phi}(x) = \begin{cases} 0 & x < -L \\ \frac{A_0}{\sqrt{2pi}k} \cdot 2sin(kL)\ & -L \leq x \leq L \\ 0 & x > L\end{cases}$$
I have integrated from -L to L for the second interval. Is it correct ?
And how can I...
\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{ikx} \phi(x)\,dx
is the Fouriertransformation of \phi(x). It changes the dependence of the wavefunction from position x to momentum p.
Hello Physics Forum,
I am not sure what to to in this task, because the wavefunction is only given as A_0. Maybe someone can explain it to me.
Thanks in Advance,
B4ckflip