Quantum Mechanics: Expectation values (Griffiths)

[WWCY]

Homework Statement

A few questions:

Q1) How does 1.29 flow to 1.30 and 1.31? How was the integral-by-parts done?
Q2) The author states that <v> = d<x>/dt represents the expectation value of velocity. What does this actually mean? I tried to rationalise that d<x>/dt represented the velocity at which the average position of the particle was moving about as time progressed. Does it make any sense?
Q3) If my understanding in 2 is right, would <p> then be the momentum associated with the motion of the average position <x>?

Apologies if I'm not making much sense, I'm just starting to properly learn some QM.

Thanks in advance!

Homework Equations

The Attempt at a Solution

 

[NFuller]

I am not sure which part of your post is homework and which part is your attempt at solving the problem. Please correctly fill out the homework template by stating the full problem in section 1, any relevant equations in section 2, and your attempts and reasoning in section 3.

 

[WWCY]

I am not sure which part of your post is homework and which part is your attempt at solving the problem. Please correctly fill out the homework template by stating the full problem in section 1, any relevant equations in section 2, and your attempts and reasoning in section 3.

Apologies, this isn't really homework. It was more of case of me not understanding a passage in a text. I felt that this was low-level stuff and decided to post under the coursework forum so as to prevent cluttering on other threads (it is related to my coursework too). Should I shift it?

 

[WWCY]

Homework Statement

Correction to original post:

I was reading through Griffith's text and was stuck at the section shown above regarding expectation values of position, velocity and momentum. There are a few questions I'd like to ask.

Q1) How does 1.29 flow to 1.30 and 1.31?
Q2) The author states that <v> = d<x>/dt represents the expectation value of velocity. What does this actually mean?
Q3) If my understanding in 2 is right, would <p> then be the momentum associated with the motion of the average position <x>?

Homework Equations

The Attempt at a Solution

1) I tried integrating (equation 1.29) the part with the partial derivative and differentiating x while also trying it the other way round. I can't seem to arrive at expression 1.30. How do I approach this?
2) I tried to rationalise that d<x>/dt represented the velocity at which the average position of the particle was moving about as time progressed. Does it make any sense?

Thanks in advance!

 

[Ray Vickson]

Homework Statement

View attachment 210131
Correction to original post:

I was reading through Griffith's text and was stuck at the section shown above regarding expectation values of position, velocity and momentum. There are a few questions I'd like to ask.

Q1) How does 1.29 flow to 1.30 and 1.31?
Q2) The author states that <v> = d<x>/dt represents the expectation value of velocity. What does this actually mean?
Q3) If my understanding in 2 is right, would <p> then be the momentum associated with the motion of the average position <x>?

Homework Equations

The Attempt at a Solution

1) I tried integrating (equation 1.29) the part with the partial derivative and differentiating x while also trying it the other way round. I can't seem to arrive at expression 1.30. How do I approach this?
2) I tried to rationalise that d<x>/dt represented the velocity at which the average position of the particle was moving about as time progressed. Does it make any sense?

Thanks in advance!

For Q1: just standard integration by parts--yes, really! If you are not seeing it, post the details of your attempt, so we can see where you went wrong.

For Q2: Yes, it makes sense. The expected value of velocity is
$$ \lim_{\Delta \to 0} \frac{\langle x \rangle (t+\Delta t) - \langle x \rangle (t) } { \Delta t},$$
and this evaluates to what was given by Griffith.

 

[WWCY]

For Q1: just standard integration by parts--yes, really! If you are not seeing it, post the details of your attempt, so we can see where you went wrong.

1) Treating the 1.29 as ∫f(x)g'(x) = f(x)g(x) - ∫f'(x)g(x)dx

I tried to integrate the part with the partial derivative (g'(x)) whilst keeping x (f(x)) constant in the first part of the RHS. This gives the expression in 1.30 in the second part (- ∫f'(x)g(x)) of the RHS. I can't however, make the part f(x)g(x) go to zero as it is in the form (x)g(x) with limits of ∞ from -∞. What am I doing wrong?

3) Since 2) made sense, could I similarly describe <p> as the momentum associated with the motion of the average position <x>?

Thank you for your patience.

 

[Ray Vickson]

1) Treating the 1.29 as ∫f(x)g'(x) = f(x)g(x) - ∫f'(x)g(x)dx

I tried to integrate the part with the partial derivative (g'(x)) whilst keeping x (f(x)) constant in the first part of the RHS. This gives the expression in 1.30 in the second part (- ∫f'(x)g(x)) of the RHS. I can't however, make the part f(x)g(x) go to zero as it is in the form (x)g(x) with limits of ∞ from -∞. What am I doing wrong?

3) Since 2) made sense, could I similarly describe <p> as the momentum associated with the motion of the average position <x>?

Thank you for your patience.

In order to have convergence (that is, the finiteness of $\int_{-\infty}^{\infty} |\psi(x)|^2 \, dx$) we absolutely NEED $|\psi(x)| \to 0$ as $x \to \pm \infty.$

 

[WWCY]

In order to have convergence (that is, the finiteness of $\int_{-\infty}^{\infty} |\psi(x)|^2 \, dx$) we absolutely NEED $|\psi(x)| \to 0$ as $x \to \pm \infty.$

bringing the limits into f(x)g(x) means ∞*0 - (-∞)*0, does it not? How can i resolve these indeterminate forms?

Also, I don't mean to sound pushy but do you mind answering this? :/

3) Since 2) made sense, could I similarly describe <p> as the momentum associated with the motion of the average position <x>?

Thanks!