MHB Momentum operator in the position representation

[Fantini]

Hi! :) I'm trying to understand the following calculation. The book Quantum Mechanics by Nouredine Zettili wants to determine the form of the momentum operator $\widehat{\vec{P}}$ in the position representation. To do so he calculates as follows: $$\begin{aligned}
\langle \vec{r} | \widehat{\vec{P}} | \psi \rangle & = \int \langle \vec{r} | \widehat{\vec{P}} | \psi \rangle \langle \vec{p} | \psi \rangle \, d^3 p \\
& = \int \vec{p} \langle \vec{r} | \vec{p} \rangle \langle \vec{p} | \psi \rangle \, d^3 p \\
& = \frac{1}{(2 \pi \hbar)^{3/2}} \int \vec{p} \exp \left( \frac{i \vec{p} \cdot \vec{r} }{\hbar} \right) \Psi(\vec{p}) \, d^3 p.
\end{aligned}$$ The momentum operator is defined as $\widehat{\vec{P}} | \vec{p} \rangle = \vec{p} | \vec{p} \rangle$.

I don't understand he's doing from the first to the second line, after applying the definition of $\langle \vec{r} | \widehat{\vec{P}} | \psi \rangle$. I know that $$\langle \vec{r} | \vec{p} \rangle = \frac{1}{(2 \pi \hbar)^{3/2}} \exp \left( \frac{i \vec{p} \cdot \vec{r}}{\hbar} \right)$$ and $$\langle \vec{p} | \psi \rangle = \Psi(\vec{p}),$$ but I don't see how he gets to that. The book claims to use that $ | \vec{p} \rangle \langle \vec{p} | = \widehat{I}$, the identity operator, and $\langle \vec{r} | \vec{p} \rangle = \ldots$.

Here's what I thought: if we use that $ | \vec{p} \rangle \langle \vec{p} | = \widehat{I}$ then $$\begin{aligned} \langle \vec{r} | \widehat{\vec{P}} | \vec{p} \rangle \langle \vec{p} | \psi \rangle & = \langle \vec{r} | \widehat{\vec{P}} \widehat{I} | \psi \rangle \\ & = \langle \vec{r} | \widehat{\vec{P}} | \psi \rangle \\ & = \langle \vec{r} | \vec{p} | \vec{p} \rangle | \psi \rangle \\ & = \vec{p} \langle \vec{r} | \vec{p} \rangle | \psi \rangle. \end{aligned}$$ That doesn't seem to be what's happening. If we don't use that $ | \vec{p} \rangle \langle \vec{p} | = \widehat{I}$, then I find an extra $| \vec{p} \rangle$ lying around.

Where am I going wrong?

 

[topsquark]

Hi! :) I'm trying to understand the following calculation. The book Quantum Mechanics by Nouredine Zettili wants to determine the form of the momentum operator $\widehat{\vec{P}}$ in the position representation. To do so he calculates as follows: $$\begin{aligned}
\langle \vec{r} | \widehat{\vec{P}} | \psi \rangle & = \int \langle \vec{r} | \widehat{\vec{P}} | \psi \rangle \langle \vec{p} | \psi \rangle \, d^3 p \\
& = \int \vec{p} \langle \vec{r} | \vec{p} \rangle \langle \vec{p} | \psi \rangle \, d^3 p \\
& = \frac{1}{(2 \pi \hbar)^{3/2}} \int \vec{p} \exp \left( \frac{i \vec{p} \cdot \vec{r} }{\hbar} \right) \Psi(\vec{p}) \, d^3 p.
\end{aligned}$$ The momentum operator is defined as $\widehat{\vec{P}} | \vec{p} \rangle = \vec{p} | \vec{p} \rangle$.

I don't understand he's doing from the first to the second line, after applying the definition of $\langle \vec{r} | \widehat{\vec{P}} | \psi \rangle$. I know that $$\langle \vec{r} | \vec{p} \rangle = \frac{1}{(2 \pi \hbar)^{3/2}} \exp \left( \frac{i \vec{p} \cdot \vec{r}}{\hbar} \right)$$ and $$\langle \vec{p} | \psi \rangle = \Psi(\vec{p}),$$ but I don't see how he gets to that. The book claims to use that $ | \vec{p} \rangle \langle \vec{p} | = \widehat{I}$, the identity operator, and $\langle \vec{r} | \vec{p} \rangle = \ldots$.

Here's what I thought: if we use that $ | \vec{p} \rangle \langle \vec{p} | = \widehat{I}$ then $$\begin{aligned} \langle \vec{r} | \widehat{\vec{P}} | \vec{p} \rangle \langle \vec{p} | \psi \rangle & = \langle \vec{r} | \widehat{\vec{P}} \widehat{I} | \psi \rangle \\ & = \langle \vec{r} | \widehat{\vec{P}} | \psi \rangle \\ & = \langle \vec{r} | \vec{p} | \vec{p} \rangle | \psi \rangle \\ & = \vec{p} \langle \vec{r} | \vec{p} \rangle | \psi \rangle. \end{aligned}$$ That doesn't seem to be what's happening. If we don't use that $ | \vec{p} \rangle \langle \vec{p} | = \widehat{I}$, then I find an extra $| \vec{p} \rangle$ lying around.

Where am I going wrong?

Sorry it's taken me so long to respond. A visit to my parents, two week long flus, and a 3 day anxiety attack pretty much fills up a month.

Anyway, where did you get that first line from? Maybe it's right but to me it doesn't even look wrong. Also [math]I = \int | \vec{p} > < \vec{p} |~d^3 \vec{p}[/math]

Let's take it step by step.
[math]< \vec{r} | \vec{P} | \psi > ~ = ~ < \vec{r} | \vec{P} I | \psi >[/math]

[math]= < \vec{r} | \vec{P} \int | \vec{p} > < \vec{p} | d^3 \vec{p} ~ | \psi >[/math]

[math]= \int < \vec{r} | \vec{P} | \vec{p} > < \vec{p} | \psi > ~d^3 \vec{p}[/math]

[math]= \int \vec{p} < \vec{r} | \vec{p} > < \vec{p} | \psi > ~d^3 \vec{p}[/math]

[math]= \frac{1}{(2 \pi \hbar)^{3/2}} \int \vec{p} \psi( \vec{p} ) e^{i p \cdot r/ \hbar} ~d^3 \vec{p}[/math]

-Dan

 

[Fantini]

Thank you for answering, topsquark!

I think I understand everything. You write $\vec{P} = \vec{P}\cdot \widehat{I}$ and uses that $$\widehat{I} = \int |\vec{p} \rangle \langle \vec{p} | \, d^3 p.$$ Then you pass the ket $| \psi \rangle$ inside the integral and pair it up with the bra $\langle p |$, forming $\psi(\vec{p})$. Next you do the same with $\langle \vec{r} | \vec{P}$ and apply the definition of $\vec{P} | \vec{p} \rangle = \vec{p} | \vec{p} \rangle$, pair up $\langle \vec{r} | \vec{p} \rangle$ and the rest follows through.

One question: the relation $\vec{P} | \vec{p} \rangle = \vec{p} | \vec{p} \rangle$ makes no sense for me at all. I read this as if he's saying the vector $\vec{p}$ is an eigenvalue associated to the vector $| \vec{p} \rangle$, but eigenvalues are numbers, not vectors! Can you shed some light on this?

 

[topsquark]

Thank you for answering, topsquark!

I think I understand everything. You write $\vec{P} = \vec{P}\cdot \widehat{I}$ and uses that $$\widehat{I} = \int |\vec{p} \rangle \langle \vec{p} | \, d^3 p.$$ Then you pass the ket $| \psi \rangle$ inside the integral and pair it up with the bra $\langle p |$, forming $\psi(\vec{p})$. Next you do the same with $\langle \vec{r} | \vec{P}$ and apply the definition of $\vec{P} | \vec{p} \rangle = \vec{p} | \vec{p} \rangle$, pair up $\langle \vec{r} | \vec{p} \rangle$ and the rest follows through.

One question: the relation $\vec{P} | \vec{p} \rangle = \vec{p} | \vec{p} \rangle$ makes no sense for me at all. I read this as if he's saying the vector $\vec{p}$ is an eigenvalue associated to the vector $| \vec{p} \rangle$, but eigenvalues are numbers, not vectors! Can you shed some light on this?

Good notation is not what Quantum Mechanics is famous for.
[math]\vec{P} | \vec{p} > = \vec{p} | \vec{p} >[/math]

[math]| \vec{p} > [/math] is the eigenstate (eigenvector, eigenket, whatever) that "carries" the eigenvalue [math]\vec{p}[/math].

[math]\vec{P}[/math] is the (Hermitian) operator that returns the momentum of the eigenvector [math]| \vec{p} >[/math]. [math]\widehat{P}[/math] is another common symbol for this operator.

So letting the operator [math]\vec{P}[/math] act on [math]| \vec{p} > [/math] gives [math]\vec{P} | \vec{p} > = \vec{p} | \vec{p} > [/math]. And because [math]\vec{P}[/math] is Hermitian we also have [math]< \vec{p} | \vec{P} = < \vec{p} | \vec{p} [/math].

I'm using the symbols that I was taught. You need to make sure you know the terminology that your instructor is using. Please let me know if you have trouble understanding notation and I'll be happy to explain it out to you.

-Dan

 

[Fantini]

That seems standard to me. It's just that my mathematician-trained mind sees eigenvalues as numbers, not vectors. I'll get used to it, I guess. I'm self-studying this. I want to apply for a masters in mathematical physics in a physics institute. To get funding I need to do well in a test consisting of 10 questions, divided in 2 questions of classical mechanics, 2 questions of electromagnetism, 2 questions of quantum mechanics, 2 questions of modern physics and 2 questions of thermodynamics and statistical mechanics.

I'm self-teaching myself three (QM, TSM and MP), while I had one course in CM and EM each.

But I understand your reasoning perfectly. Thanks! :)

 

[topsquark]

That seems standard to me. It's just that my mathematician-trained mind sees eigenvalues as numbers, not vectors. I'll get used to it, I guess. I'm self-studying this. I want to apply for a masters in mathematical physics in a physics institute. To get funding I need to do well in a test consisting of 10 questions, divided in 2 questions of classical mechanics, 2 questions of electromagnetism, 2 questions of quantum mechanics, 2 questions of modern physics and 2 questions of thermodynamics and statistical mechanics.

I'm self-teaching myself three (QM, TSM and MP), while I had one course in CM and EM each.

But I understand your reasoning perfectly. Thanks! :)

If it makes you feel any better look at it this way: We have three operators [math]P_x \text{, } P_y \text{, and } P_z[/math]

Acting on a ket [math]| \vec{p} > [/math] we get [math]P_x = p_x | \vec{p} > [/math], etc. where [math]p_x[/math] is the x component of the momentum.

-Dan