Ground state of harmonic oscillator

[unscientific]

Shouldn't the integrating factor be ##exp(\frac{m\omega x}{\hbar})##?
[tex]\frac{\partial <x|0>}{\partial x} + \frac{m\omega x}{\hbar} <x|0> = 0 [/tex]

This is in the form:

[tex]\frac{\partial y}{\partial x} + P_{(x)} y = Q_{(x)} [/tex]

Where I.F. is ##exp (\int (P_{(x)} dx)##

 

[Bill_K]

And P(x) = mωx/ħ, and ∫P(x) dx = mωx²/2ħ, and the integrating factor is exp(mωx²/2ħ).

 

[vanhees71]

You can as well solve the linear first-order ODE directly, i.e., with [itex]u_0(x)=\langle x|0 \rangle[/itex] given by the book to read
[tex]u_0'=-\frac{x}{2l^2} u_0.[/tex]
This you can write as
[tex]\frac{u_0'}{u_0}=-\frac{x}{2l^2}.[/tex]
This can be integrated
[tex]\ln \left (\frac{u_0(x)}{u_0(0)} \right )=-\frac{x^2}{4l^2} \; \Rightarrow \; u_0(x)=u_0(0) \exp \left (-\frac{x^2}{4l^2} \right ).[/tex]
Further the wave function should be normalized to 1, i.e.,
[tex]\int_{\mathbb{R}} \mathrm{d}x \; |u_0(x)|^2=|u(0)|^2 \int_{\mathbb{R}} \mathrm{d} x exp \left (-\frac{x^2}{2l^2} \right )=\sqrt{2 \pi l^2} |u_0(0)|^2 \stackrel{!}{=}1.[/tex]
From this we find, up to an irrelevant phase factor,
[tex]u_0(0)=\left (\frac{1}{2 \pi l^2} \right )^{1/4}.[/tex]
This completes the derivation for the ground state of the harmonic oscillator in position representation.