Translation operator. Infinite potential well.

  • #1
660
16
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
[tex]\psi_n(x-\frac{a}{2})?[/tex]
 
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Answers and Replies

  • #2
1,006
105
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)##
 
  • #3
660
16
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
[tex]\psi_n=\sqrt{\frac{2}{a}}\sin \frac{n\pi (x-\frac{a}{2})}{a}?[/tex]
 
  • #4
jtbell
Mentor
15,763
4,001
Hint: consider the trig identity for ##\sin(\alpha \pm \beta)##.
 
  • #5
660
16
I know the trig identy I just asked is transformation OK?
 
  • #6
Khashishi
Science Advisor
2,815
493
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.)
 
  • #7
jtbell
Mentor
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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).
 
  • #8
660
16
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?
 
  • #9
660
16
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}##.
 
  • #10
jtbell
Mentor
15,763
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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 = ...
 

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