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partyday

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- Homework Statement
- Write down the time independent one dimensional Schrodinger

equation for a particle of mass m in a potential V (x). Show that if V (x) = V (−x) and real, then the

solutions ψn(x) have the property that ψn(x) = ±ψn(−x).

- Relevant Equations
- V(x) = V(-x)

##C \psi''(x) + V(x)\psi(x)=E\psi(x)##

C is just the constant by ##\psi''##

My initial attempt was to write out the schrodinger equation in the case that x>0 and x<0, so that

$$ \frac {\psi'' (x)} {\psi (x)} = C(E-V(x))$$

and

$$ \frac {\psi'' (-x)} {\psi (-x)} = C(E-V(-x))$$

And since V(-x) = V(x) I equated them and rearranged them so

$$\frac {\psi'' (x)} {\psi (x)} = \frac {\psi '' (-x)} {\psi(-x)} $$ I feel like this is where I'm supposed to be in the problem, but I'm struggling to connect this result to ψn(x) = ±ψn(−x). Any suggestions?

My initial attempt was to write out the schrodinger equation in the case that x>0 and x<0, so that

$$ \frac {\psi'' (x)} {\psi (x)} = C(E-V(x))$$

and

$$ \frac {\psi'' (-x)} {\psi (-x)} = C(E-V(-x))$$

And since V(-x) = V(x) I equated them and rearranged them so

$$\frac {\psi'' (x)} {\psi (x)} = \frac {\psi '' (-x)} {\psi(-x)} $$ I feel like this is where I'm supposed to be in the problem, but I'm struggling to connect this result to ψn(x) = ±ψn(−x). Any suggestions?