I was reading the book Introduction to Quantum Mechanics by Daniel Griffith, and also following Brant Carlson's videos. He basically makes videos about parts of the book. The book was discussing $\frac{\mathrm d\langle x\rangle}{\mathrm dt}$, and this spiraled into getting the expectation value of momentum. We are then introduced to the operator used for momentum:

$$\langle p\rangle =\int \psi^*\left(-~\mathrm i\hbar\left(\frac{\partial}{\partial x}\right)\right)\psi~\mathrm dx$$ But in the Brant Carlson video on the topic, he states:

$$\langle \hat p\rangle =\int \psi^*\left(-~\mathrm i\hbar\left(\frac{\partial}{\partial x}\right)\right)\psi ~\mathrm dx$$

My question is whether this means that $\langle p\rangle =\langle \hat p\rangle\,.$ If this statement is true then the expectation value of $p$ is the same as the expectation value of the momentum operator.

This is the link to the video.

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    $\begingroup$ Yes. $\langle p\rangle$ and $\langle \hat{p}\rangle$ are synonymous, as the formulas show. $\endgroup$ – Gert Nov 16 '16 at 3:52
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    $\begingroup$ If $|p\rangle$ is a base state, then $\hat p|p\rangle ~=~ p|p\rangle\,.$ $\endgroup$ – user36790 Nov 16 '16 at 4:03

They are simply using different notations for the same mathematical object (the momentum operator). Some authors use the "hat" notation for operators and write the momentum operator as $\hat p$ and its eigenvalues as $p$, other authors (like Griffiths and Sakurai for example) write both as $p$.

Notice that if $p$ was an eigenvalue of the operator $p$ we would trivially have

$$\langle p \rangle = p$$

In fact, for every complex number $c$, $\langle c \rangle = c$.

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