Expectation values of operators

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MIT 8.04 Quantum Physics I, Spring 2016
Instructor: Barton Zwiebach

License: Creative Commons BY-NC-SA

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When you integrate by parts, remember that the UV term will go away because you evaluate it at the limits. That held me up for a while.

zacharythatcher

15:14 I was confused why Prof. Zwiebach substituted the p for a momentum operator with an extra minus sign( - (hbar/i) (d/dx)).
But I just realized that he WASN'T using the momentum operator. He was just trying to find a more useful form of the integrand of integral p, with an equivalent result, which just turned out to look similar to the momentum operator in position space.
In fact, (as he mentioned) the two integrands above and beneath the underline are exactly the same.
Wrote this so no one else gets stuck over nothing for as long as I did ;-)

허지원-yv

Really really insightful! Couldn't find any more beautiful explanation elsewhere.

shivamsinghaswal

A great series of lecture ! I am gratefull to the MIT for sharing that. Just a remark : the computations simply much by using Plancherel's Theorem which states that, not only the L^2 norm is conserved, but also the inner product. Then, in particular, the Dirac mass is no more needed.

rougirelarnaud

Fantastic presentation, great prelude to Wigner's 1932 paper ! Only one quibble, the expectation is really of Q, not Qhat... Qhat is the operator, Q is the variate and it is an expectation of the variate that is of interest. The operator simply gets us there.

paulg

how does he suddenly jump the generalization that just because it works for the momentum operator it works for energy? doesn't this require the wavefunctions fouriered over energy space in terms of coordinates to cancel nicely? or generalise further for any operator?

nicktohzyu

At 14:25, Prof. Zwiebach used the momentum operator as a derivative, but there is an extra negative sign there (the i is in the denom.)

The problem is that: following the derivation (without the negative), the extra negative we pick from integration by parts will leave me with negative the momentum operator at the end :/

faroukmokhtar

My physical chem book doesn't tell me this.
This was an amazing lecture.

hyunwoopark

so it only works for k.e. operator. but not for Hamiltonian operator, which includes potential energy?

SergeyPopach

7:30 Could you help me why we could similarly have(? or find?) the probability density for the momentum space?

Leowlion

Interesting that given a particle with the expected value of the momentum = 0, the expected value of the kinetic energy operator is the variance of the momentum

rajinfootonchuriquen

at 18.35 (-h/id/dx) is an operator, how can you do integration by parts in that way? Integral uv' = a term that vanishes + u'v

gpolix

11:57 I need those blackboards at my University

israeltulmo

@ 2:24 The expectation value of throwing 1 dice is 3.5. In reality you will NEVER see this outcome!!! Did you 'expect' to measure that???

jacobvandijk