Ground state of harmonic oscillator

Knot Head
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Homework Statement


Verify that the ground state (n=0) wavefunction is an eigenstate of the harmonic
oscillator Hamiltonian. Using the explicit wavefunction of the ground state to calculate
the average potential energy <0|\hat{v}|0> and average kinetic energy <0|\hat{T}| 0>


Homework Equations



\int^{\infty}_{0}(x^{2n} e^{-ax^{2}})dx=\frac{1x3x5x...x(2n-1)}{2^(n+1)a^n}\sqrt{\frac{\pi}{a}}


The Attempt at a Solution


I did the ground state harmonic oscillation standard alteration with "a" and got \\Psi_{0}(x)=1/(\pi^{1/4}\sqrt{x_{0}})*e^-x^2/2x^{2}_{0}
 
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You're going to have to make a bit more of an effort to work the problem on your own before you'll get any help here.
 
ill just go ask this question another website, thanx
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
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