Expected value of momentum P in terms of k

Click For Summary
The discussion confirms that when given a wave function in terms of momentum, the expected values of momentum can be expressed as ##\langle p \rangle = \hbar \langle k \rangle## and ##\langle p^2 \rangle = \hbar^2 \langle k^2 \rangle##. This relationship is justified based on the definitions of momentum and wave number in quantum mechanics. The equations provided illustrate how to compute these expected values from the wave function. The clarification sought pertains to the validity of these expressions in quantum mechanics. The conclusion affirms that the expressions are indeed correct.
Kyuubi
Messages
18
Reaction score
8
Homework Statement
Just need to clarify this point.
Relevant Equations
$$ \langle p \rangle = \int_{-\infty}^{\infty}\bar{\phi}(p,t) p \phi(p,t)dp $$
Now if I'm given a ##\phi(k)##, and I'm asked to find ##\langle p \rangle##, ##\langle p^2 \rangle##, etc. Am I justified to say that ##\langle p \rangle = \hbar \langle k \rangle## and that ##\langle p^2 \rangle = \hbar^2 \langle k^2 \rangle## ?
 
Physics news on Phys.org
Kyuubi said:
Homework Statement: Just need to clarify this point.
Relevant Equations: $$ \langle p \rangle = \int_{-\infty}^{\infty}\bar{\phi}(p,t) p \phi(p,t)dp $$

Now if I'm given a ##\phi(k)##, and I'm asked to find ##\langle p \rangle##, ##\langle p^2 \rangle##, etc. Am I justified to say that ##\langle p \rangle = \hbar \langle k \rangle## and that ##\langle p^2 \rangle = \hbar^2 \langle k^2 \rangle## ?
Yes.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 2 ·
Replies
2
Views
818
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
748
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
763
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K