Expected Values in a Harmonic Oscillator

[Domnu]

Problem
Show that in the nth state of the harmonic oscillator

\langle x^2 \rangle = (\Delta x)^2
\langle p^2 \rangle = (\Delta p)^2​

Solution
This seems too simple... I'm not sure if it's correct...

It is obvious that \langle x \rangle = 0... this is true because the parity of the square of the eigenfunction is 1 (in other words, the probabiliity density is an even function). Now, we know that (\Delta x)^2 = \langle x^2 \rangle - \langle x \rangle ^2, but \langle x \rangle = 0, so by substitution, the desired result follows. A similar argument can be made for the momentum. \blacksquare

 

[G01]

You can always explicitly show that <x> = 0 in the nth state if you feel you need to show more work.

To get started try representing x in terms of the raising and lowering operators in the following line, letting the operators act on any kets to their right, and simplifying:

<n|x|n> = ... |n> is the eigenket for the nth eigenstate.You should end up with delta functions that have to be zero.