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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]

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# Homework Help: Expected Values in a Harmonic Oscillator

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