Expectation value

1. May 19, 2014

kashokjayaram

Why, in a Gaussian wave function the position and momentum expectation value coincide to be zero?
Does it have any physical interpretation?

I had an idea that expectation value is the average value over time on that state. But, for Gaussian it tells that it vanishes. Can you please explain.?

2. May 19, 2014

Simon Bridge

The gaussian wavefunction does not have to have position and momentum expectation values zero.
That is only for a special case where the wavefunction is centered on the origin.

The expectation values are not the average over time, but the average over space. $$<x>=\int_\infty\psi^\star x \psi\;dx$$... for instance.

The expectation value is just the average value - so <x>=0 just means that the particle's average position is at the origin. This is what you'd expect for, say, a simple harmonic oscillator.

While the particle spends most time near the middle, the momentum there is either positive or negative.
The average of +p and -p is zero ... so it is equally likely to be found going left-to-right as right-to-left.

Also see:
http://en.wikipedia.org/wiki/Wave_packet#Gaussian_wavepackets_in_quantum_mechanics

3. May 19, 2014

BruceW

yeah, the momentum expectation value can be nonzero, you just need to put the peak of the distribution (in momentum-space) at somewhere which is non-zero. you can do this by translating the distribution along the momentum-axis. Then, when you take Fourier transform, you get a Gaussian, multiplied by a plane wave. So if you take the expectation value of momentum in the position basis, you will also get the same non-zero expectation value. Also, if you want to create a wavepacket that has a non-zero expectation for space and momentum, then (I think) you also need to multiply the wavepacket by some complex constant with modulus 1. And then after that, if you want to add more complexity, you can start thinking about bringing time into the formula.