Translation operator. Infinite potential well.

[LagrangeEuler]

For potential well problem for well with potential which is zero in the interval $[0,a]$ and infinite outside we get $\psi_n(x)=\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}$. If I want to get this result for well with potential which is zero in the interval $[-\frac{a}{2},\frac{a}{2}]$ and infinite outside we must do some translation. May I say that the wave function for this problem is
\psi_n(x-\frac{a}{2})?

 

[The_Duck]

Probably you want $\psi_n(x + a/2)$.

In general, you can probably convince yourself that if $\psi(x)$ satisfies

$-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi(x) + V(x) \psi(x) = E \psi(x)$

then

$\phi(x) = \psi(x + a)$

satisfies

$-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \phi(x) + V(x+a) \phi(x) = E \phi(x)$

 

[LagrangeEuler]

I'm not sure why. I need that wave function be zero in $-\frac{a}{2}$ and $\frac{a}{2}$. In solved problem its zero in $0$ and $a$. Isnt then wave function for potential well with potential which is zero in the interval $[−\frac{a}{2},\frac{a}{2}]$ and infinite outside we
\psi_n=\sqrt{\frac{2}{a}}\sin \frac{n\pi (x-\frac{a}{2})}{a}?

 

[jtbell]

Hint: consider the trig identity for $\sin(\alpha \pm \beta)$.

 

[LagrangeEuler]

I know the trig identy I just asked is transformation OK?

 

[Khashishi]

The Duck already answered your question. To translate a function to the "left", you need to add to x, not subtract. (More specifically, to translate a function to the "right" by b, you replace x by x-b. Moving to the "left" is opposite of that.)

 

[jtbell]

If it's still not clear whether you want a + or - sign, draw a diagram of the potential well, and label the the sides of the well with the coordinates (x) in the untranslated system (origin at left side of the well) and the coordinates (x') in the translated system (origin at the center of the well).

 

[LagrangeEuler]

The Duck already answered your question. To translate a function to the "left", you need to add to x, not subtract. (More specifically, to translate a function to the "right" by b, you replace x by x-b. Moving to the "left" is opposite of that.)

That confused me. Why?

 

[LagrangeEuler]

If it's still not clear whether you want a + or - sign, draw a diagram of the potential well, and label the the sides of the well with the coordinates (x) in the untranslated system (origin at left side of the well) and the coordinates (x') in the translated system (origin at the center of the well).

If I understand you origin in translated system has coordinate $x'=0$ and in untranslated $x=\frac{a}{2}$. So $x'=x-\frac{a}{2}$.

 

[jtbell]

So $x'=x-\frac{a}{2}$.

Correct. But you're beginning with an equation expressed in terms of x, and you want to re-write it in terms of x'. So you need to substitute x = ...