3

Given a momentum operator $$ \def\bra #1{\langle#1|} \def\ket #1{|#1\rangle} \def\braket #1{\langle#1\rangle} $$

$$ P \equiv - i \hbar \frac{\partial}{\partial x}, $$ I want to calculate $\bra{a} P \ket{a}$. Here are the steps provided by my professor.

$$\begin{aligned} \bra{a} P \ket{a} &= \int dx \int dx' \braket{a | x} \bra{x} P \ket{x'} \braket{x' | a} && \left( \text{insert two sets of bases} \int dx \ket{x} \bra{x} = I \right) \\ &= \int dx \int dx' \braket{a | x} \bra{x} \left(- i \hbar \frac{\partial}{\partial x'}\right) \ket{x'} \braket{x' | a} && \left( \text{pull in } P = - i \hbar \frac{\partial}{\partial x'} \right) \\ &= \int dx \int dx' - i \hbar \braket{a | x} \braket{x | x'} \left(\frac{\partial}{\partial x'}\right) \braket{x' | a} && \left( \text{move } \ket{x'} \text{ to the left} \right) \\ &= \int dx \int dx' - i \hbar \braket{a | x} \delta(x - x') \left(\frac{\partial}{\partial x'}\right) \braket{x' | a} \\ &= \int dx - i \hbar \braket{a | x} \left(\frac{\partial}{\partial x}\right) \braket{x | a} && \left( \int f(x') \delta(x - x')\, dx' = f(x)\right) \\ &= -i \hbar \int dx \Psi_a^*(x) \left(\frac{\partial}{\partial x}\right) \Psi_a(x) && \left( \braket{x | a} = \Psi_a(x) \right) \\ \end{aligned}$$

I have a few questions.

  1. Is $\ket{x'}$ a function? Can I treat it as $f(x')$.
  2. If $x$ and $x'$ mean "position", why they can become a series of the basis? $\int dx\, \ket{x} \bra{x} = I$
  3. Is the equation below correct?

$$ \left(- i \hbar \frac{\partial}{\partial x'}\right) \ket{x'} \braket{x' | a} = - i \hbar \left( \frac{\partial}{\partial x'} \ket{x'} \braket{x' | a} \right) $$

  1. Why $\ket{x'}$ can be moved to the left at the third step?
IvanaGyro
  • 267

1 Answers1

5

Your basic definition is simply wrong/ambiguous. Use @Emilio’s informal rule above, or the correct definition of Sakurai and Napolitano, (1.248), namely, $$ \def\bra #1{\langle#1|} \def\ket #1{|#1\rangle} \def\braket #1{\langle#1\rangle} $$ $$ P \equiv - i \hbar\int \!\! dx~\ket{x} \frac{\partial}{\partial x}\bra{x}, $$ sometimes encoded as $$ P_x \equiv - i \hbar \frac{\partial}{\partial x}, $$ as long as you know what you are doing, which your instructor muffed, to your detriment.

From the correct definition, note $$ \bra{y} P\ket{z}= \int\! dx \braket {y|x} (-i\hbar \partial_x) \braket{x|z}= -i\hbar \partial_y \delta(y-z), $$ and the rest will follow correctly. $$ \bra{a} P \ket{a}= \int\! dy d z ~\braket{a | y} \bra{ y} P \ket{ z} \braket{ z | a}\\ =\int\! dxdy d z ~\braket{a | y} \braket {y|x} (-i\hbar \partial_x) \braket{x|z} \braket{ z | a}\\ =\int\! dx ~ \braket {a|x} (-i\hbar \partial_x) \braket{ x | a}\\ =-i\hbar \int\!\! dx ~~ \Psi_a^*(x) \partial_x \Psi_a(x)~. $$
You can, of course, go from the first to the penultimate line directly. I left them in to humor your previous wrong derivation.

  1. Yes.
  2. No.

2&4. are meaningless.

Cosmas Zachos
  • 62,595
  • How about $P|x\rangle$? Does $P$ still have $\int dx~|{x}\rangle$ part? – IvanaGyro Dec 26 '22 at 08:17
  • You may easily compute it from the right formula given, but you should never have to use it! $P|x\rangle=-i\hbar \int dy ~|y\rangle \partial_y \delta(y-x)= i\hbar\partial_x |x\rangle$. Never! – Cosmas Zachos Dec 26 '22 at 08:43
  • Why the minus sign is disappeared? How to make it consistent with $\langle x | P | x \rangle$? – IvanaGyro Dec 30 '22 at 20:56
  • Do the math. You first integrate by parts to flip the sign, and only then you collapse the δ function! Remember, as @Emilio's answer emphasizes, you get a - when acting on the bra, and a + when acting on the ket! It is consistent! This is exactly how this matrix element is computed! – Cosmas Zachos Dec 30 '22 at 21:12