Schrodinger using a Hermite Polynomial

[QuantumMech]

Can some1 help me solve a first energy level Schrodinger (\psi_{1})with a the Hermite polynomial and also show that it equals to \frac{3}{2}\hbar \omega?

I got as far as
<br /> \newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }\frac{\hbar^2}{2 m} \ \pd{\Psi}{x}{2} + V \Psi = \frac{\hbar}{2m} (2N_{1}\frac{x}{\alpha}e^\frac{-1}{2\alpha^2}x^2(\frac{1}{\alpha^2}+\frac{1}{\alpha^4}x^2)<br />

Thanks.

 

[dextercioby]

Is it for the simple 1D-HO...?That one has Hermite polynomials as eigenfunctions...

If so,how about checking any book on QM (any,all treat 1D-HO) for the famous algebraic method...?

U'll then get the \psi_{1}(x) by applying the creation operator on the ground state...

Daniel.

 

[QuantumMech]

The problem says: confirm that the 1st excited state wavefunction of a 1D HO given by the Hermitian equation

H_{1}(y)= 2y y = \frac{x}{\alpha}, \alpha = (\frac{\hbar^2}{mk})^\frac{1}{4}<br />

is a solution of the Schrodinger equation and that the energy is \frac{3}{2}\hbar\omega.


Thanks.

 

[dextercioby]

\psi_{1}(x)=:\langle x|1\rangle


\hat{H}=\hbar\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hat{1}\right)

\langle x|\hat{H}|1\rangle =\hbar\omega \left(\langle x|\hat{a}^{\dagger}\hat{a}|1\rangle +\frac{1}{2} \langle x|\hat{1}|1 \rangle \right) = \hbar\omega \left\langle x\left|\left(1+\frac{1}{2}\right)\right|1 \right\rangle=\frac{3}{2} \hbar\omega \langle x|1\rangle

which means

\hat{H}\psi_{1}(x)=\left(\frac{3}{2}\hbar\omega\right) \psi_{1}(x)

Q.e.d.


NOTE:I made use of

\left \{\begin{array}{c} \hat{a}|n\rangle =\sqrt{n}|n-1\rangle ,\ \mbox{for} \ n=1 \\ \hat{a}^{\dagger}|n\rangle =\sqrt{n+1}|n+1\rangle ,\ \mbox{for} \ n=0 \end{array} \right

which are typical for the operators which form the famous Heisenberg algebra of the 1D-HO.

Daniel.

 

[Data]

Yes... in order to show that a solution satisfies the equation, you don't actually have to solve the equation! Just substitute the proposed solution in and see if the resulting statement is true.

 

[QuantumMech]

I don't know bracket notation or what a means, but I put it in the HW I turning in.

 

[dextercioby]

It's the only elegant way to do it,really.

Daniel.