The kinetic energy operator: $$ \hat{T}=\frac{{}\hat{p^{2}}}{2m}=\frac{-\frac{\partial^2 }{\partial x^2}\hbar^{2}}{2m} $$
I think that the integral you're looking for is the following: $$ E(T)=-\frac{A^{2}\hbar^{2}}{2m}\int_{-\infty}^{\infty}e^{-\frac{x^{2}}{a^{2}}}(\frac{x^{2}}{a^{4}}-k^{2}-\frac{2xik}{a^{2}}-\frac{1}{a^{2}})dx $$
$$KE = frac{p^{2}}{2m}$$. $p^{2}$ $frac{1}{2m}$
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ReplyDelete\pi
ReplyDelete$\pi$
ReplyDelete$$\pi$$
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ReplyDelete$$
ReplyDelete\frac{-A^2 \hbar^2}{2m} \int_{-\infty}^{+\infty} e^{-x^2/a^2} (\frac{x^2}{a^4} - \frac{2ikx}{a^2} - k^2 - \frac{1}{a^2}) dx
$$
For 5.1(c), the momentum operator would be
ReplyDelete$$ \hat{p} =-i\hbar \frac{\partial}{\partial x} $$
For the kinetic energy operator, it would be
$$ \hat{KE} = \frac{hbar^{2}}{2m}\frac{\partial^{2}}{\partialx^{2}} $$
The kinetic energy operator: $$
ReplyDelete\hat{T}=\frac{{}\hat{p^{2}}}{2m}=\frac{-\frac{\partial^2 }{\partial x^2}\hbar^{2}}{2m}
$$
I think that the integral you're looking for is the following: $$
E(T)=-\frac{A^{2}\hbar^{2}}{2m}\int_{-\infty}^{\infty}e^{-\frac{x^{2}}{a^{2}}}(\frac{x^{2}}{a^{4}}-k^{2}-\frac{2xik}{a^{2}}-\frac{1}{a^{2}})dx
$$
I think he means this one:
ReplyDelete$$ n \approx kTD_(c)e^(-\frac(E_(c)-E_(f))(kT)$$
I think he means this one:
ReplyDelete$$ n \approx kTD_(c) e^(- \frac(E_(c)-E_(f))(kT)$$
I think he means this one:
ReplyDelete$$ n \approx kTD_{c}e^(-\frac(E_(c)-E_(f))(kT)$$
I think he means this one:
ReplyDelete$$ n \approx kTD_{c}e^{-\frac{E_{c}-E_{f}}{kT}}$$
Actually, this may be a good approximation:
ReplyDelete$$ E_{f} = frac{E_{c} + E_{v} + kTln(\frac{D_{v}}{D_{c}}}{2}
Actually, this may be a good approximation:
ReplyDelete$$ E_{f} = frac{E_{c} + E_{v} + kTln(\frac{D_{v}}{D_{c}}}{2} $$
Actually, this may be a good approximation:
ReplyDelete$$ E_{f} = \frac{E_{c} + E_{v} + kTln(\frac{D_{v}}{D_{c}}}){2}
Actually, this may be a good approximation:
ReplyDelete$$ E_{f} = \frac{E_{c} + E_{v} + kTln(\frac{D_{v}}{D_{c}}}){2}$$
Actually, this may be a good approximation:
ReplyDelete$$ E_{f} = \frac{E_{c} + E_{v} + kTln(\frac{D_{v}}{D_{c}})}{2}
ReplyDelete$$ E_{f} = \frac{E_{c} + E_{v} + kTln(\frac{D_{v}}{D_{c}})}{2}$$
Actually, this may be a good approximation:
ReplyDelete$$ E_{f} = frac{E_{c} + E_{v} + kTln(\frac{D_{v}}{D_{c}})}{2} $$
Actually, this may be a good approximation:
ReplyDelete$$ E_{f} = \frac{E_{c} + E_{v} + kTln(\frac{D_{v}}{D_{c}})}{2} $$
\(int_{E_c}^{E_f}D(E)EdE = E_{total}\)
ReplyDelete\(\int_{E_c}^{E_f}D(E)EdE = E_{total}\)
ReplyDeleteI think if you re-write \(n_\uparrow^2+n_\downarrow^2\) as \((n_\uparrow+n_\downarrow)^2 + (n_\uparrow-n_\downarrow)^2\) it comes out nicely.
ReplyDelete
ReplyDelete$$Omega = 2^{n-1} $$
$$ \dbinom{9}{3} $$
ReplyDelete$$\dbinom{n}{n_{down}}$$
ReplyDelete$$\dbinom{n}{n_{down}} = frac{n!}{n_{down}!(n-n_{down}) = frac{n!}{n_{down}n_{up}}$$
ReplyDelete$$\dbinom{n}{n_{down}} = \frac{n!}{n_{down}!(n-n_{down}) = \frac{n!}{n_{down}n_{up}}$$
ReplyDelete$$\dbinom{n}{n_{down}} = \frac{n!}{n_{down}!(n-n_{down)!} = \frac{n!}{n_{down}!n_{up}!}$$
ReplyDelete$$\dbinom{n}{n_{down}} = \frac{n!}{n_{down}!(n-n_{down}!} = \frac{n!}{n_{down}!n_{up}!}$$
ReplyDelete$$\dbinom{n}{n_{down}} = \frac{n!}{n_{down}!(n-n_{down}!)} = \frac{n!}{n_{down}!n_{up}!}$$
ReplyDelete$$\sum_{k=0}^{n}\binom{n}{k}$$
ReplyDelete$$\Omega = \sum_{k=0}^{n}\binom{n}{k}$$
ReplyDelete