# Schrodinger using a Hermite Polynomial

1. Apr 7, 2005

### 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
$$\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }\frac{\hbar^2}{2 m} \ \pd{\Psi}{x}{2} + V \Psi = \frac{\hbar}{2m} (2N_{1}\frac{x}{\alpha}e^\frac{-1}{2\alpha^2}x^2(\frac{1}{\alpha^2}+\frac{1}{\alpha^4}x^2)$$

Thanks.

2. Apr 7, 2005

### 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.

3. Apr 7, 2005

### QuantumMech

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

$$H_{1}(y)= 2y y = \frac{x}{\alpha}, \alpha = (\frac{\hbar^2}{mk})^\frac{1}{4}$$

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

Thanks.

Last edited: Apr 7, 2005
4. Apr 7, 2005

### 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.

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

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

Daniel.

5. Apr 7, 2005

### 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.

6. Apr 7, 2005

### QuantumMech

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

7. Apr 7, 2005

### dextercioby

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

Daniel.