Expected value of momentum P in terms of k

AI Thread 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
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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## ?
 
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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.
 
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