- #1
Domnu
- 178
- 0
Problem
Show that in the [tex]n[/tex]th state of the harmonic oscillator
[tex]\langle x^2 \rangle = (\Delta x)^2[/tex]
[tex]\langle p^2 \rangle = (\Delta p)^2[/tex]
Solution
This seems too simple... I'm not sure if it's correct...
It is obvious that [tex]\langle x \rangle = 0[/tex]... this is true because the parity of the square of the eigenfunction is [tex]1[/tex] (in other words, the probabiliity density is an even function). Now, we know that [tex](\Delta x)^2 = \langle x^2 \rangle - \langle x \rangle ^2 [/tex], but [tex]\langle x \rangle = 0[/tex], so by substitution, the desired result follows. A similar argument can be made for the momentum. [tex]\blacksquare[/tex]
Show that in the [tex]n[/tex]th state of the harmonic oscillator
[tex]\langle x^2 \rangle = (\Delta x)^2[/tex]
[tex]\langle p^2 \rangle = (\Delta p)^2[/tex]
Solution
This seems too simple... I'm not sure if it's correct...
It is obvious that [tex]\langle x \rangle = 0[/tex]... this is true because the parity of the square of the eigenfunction is [tex]1[/tex] (in other words, the probabiliity density is an even function). Now, we know that [tex](\Delta x)^2 = \langle x^2 \rangle - \langle x \rangle ^2 [/tex], but [tex]\langle x \rangle = 0[/tex], so by substitution, the desired result follows. A similar argument can be made for the momentum. [tex]\blacksquare[/tex]