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In Sakurai's quantum mechanics, the derivation of momentum operator and Hamlitonian operator is based on spatial translation and time translation as below,

  1. for spatial translation and momentum operator, we have $$\mathfrak{T}(d\pmb{x})|\alpha\rangle = \left(1 - \frac{i\pmb{p}\cdot d\pmb{x}}{\hbar}\right)|\alpha\rangle = \int{d\pmb{x}\ \mathfrak{T}(d\pmb{x})\ |\pmb{x}\rangle\langle\pmb{x}|\alpha\rangle} = \int{d\pmb{x}|\pmb{x}+d\pmb{x}\rangle\langle\pmb{x}|\alpha\rangle}$$ and if we change the variable $$\left(1 - \frac{i\pmb{p}\cdot d\pmb{x}}{\hbar}\right)|\alpha\rangle = \int{d\pmb{x}|\pmb{x}\rangle}\langle\pmb{x}-d\pmb{x}|\alpha\rangle = \int{d\pmb{x}|\pmb{x}\rangle}\left(\langle\pmb{x}|\alpha\rangle-d\pmb{x}\frac{\partial }{\partial \pmb{x}}\langle\pmb{x}|\alpha\rangle\right)$$ and then we get the momentum operator is, $$\pmb{p} = -i\hbar \frac{\partial}{\partial \pmb{x}}$$
  2. However, he derives Hamiltonian operator in a different manner. $$\mathfrak{U}(t + dt) - \mathfrak{U}(t)= \left(1 - \frac{iHdt}{\hbar}\right)\mathfrak{U}(t) - \mathfrak{U}(t) = -\frac{iHdt}{\hbar}\mathfrak{U}(t)$$ and then we get, $$\frac{\mathfrak{U}(t+dt)-\mathfrak{U}(t)}{dt} = \frac{d\mathfrak{U}(t)}{dt} = -i\frac{H}{\hbar}\mathfrak{U}(t)$$ and then we get Hamiltonian operator, $$H = i\hbar\frac{\partial}{\partial t}$$
  3. Here comes a problem, if we derive the momentum operator in similar manner as 2., we get $$\mathfrak{T}(\pmb{x} + d\pmb{x}) - \mathfrak{T}(\pmb{x})= \left(1 - \frac{i\pmb{p}\cdot d\pmb{x}}{\hbar}\right)\mathfrak{T}(\pmb{x}) - \mathfrak{T}(\pmb{x}) = -\frac{i\pmb{p}\cdot d\pmb{x}}{\hbar}\mathfrak{T}(\pmb{x})$$ and then we get, $$\frac{\mathfrak{T}(\pmb{x}+d\pmb{x})-\mathfrak{T}(\pmb{x})}{d\pmb{x}} = \frac{d\mathfrak{T}(\pmb{x})}{d\pmb{x}} = -i\frac{\pmb{p}}{\hbar}\mathfrak{T}(\pmb{x})$$ and then we get, $$\pmb{p} = i\hbar\frac{\partial}{\partial \pmb{x}}$$

There is a minus sign, and I don't know where is wrong , they should result in same result. Can someone tell me where is wrong?

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    $\begingroup$ Related: How does the momentum operator act on state kets? $\endgroup$
    – Qmechanic
    Sep 29, 2023 at 14:34
  • $\begingroup$ You are comparing apples and oranges. In 1, you are translating bras; whilst in 2 and 3 you are translating kets. 3 is highly nonstandard, in that 1 means $\hat p = \int dx ~ |x\rangle (-i\hbar\partial_x) \langle x|$, whilst 3 means $\hat p = \int dx ~ |x\rangle (i\hbar \overset{_\gets}\partial_x) \langle x|$ in your peculiar, unacknowledged conventions... When you integrate by parts, they align. $\endgroup$ Sep 29, 2023 at 15:31
  • $\begingroup$ @CosmasZachos Can you tell me why the derivations in 1. and 2. are different? Or is there a generalized method to derive these operator? Thanks! $\endgroup$ Sep 29, 2023 at 15:35
  • $\begingroup$ You have skipped the crucial steps, which has resulted in your symbol salad. $\mathfrak{U}$ is acting on kets, which you time-translate forward. But $\mathfrak{T}(dx)$ is acting on either bras or kets, as you found in 1. $|\alpha\rangle$ is a superfluous dummy ket and properly shouldn't be there. Recall the momentum as a derivative is a shared fiction/convention of the coordinate representation. You must specify what you mean by it. $\endgroup$ Sep 29, 2023 at 15:44

1 Answer 1

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The confused operators you wrote are undefined and sloppy. If only you wrote them in clean Dirac notation, there would be no ambiguity. Work in one dimension, w.l.o.g., so

  1. Acting on bras, you get the conventional sign, which defines the coordinate representation realization. Acting on kets, unconventional, you get the opposite sign. $$\mathfrak{T}(\delta x )= \left(1 - \frac{i \hat p \delta x }{\hbar}\right) = \int d x ~\mathfrak{T}( \delta x )\ | x \rangle\langle x| = \int dx~| x +\delta x \rangle\langle x | \\ = \int d x~ | x \rangle \langle x-\delta x | = \int d x ~\ | x \rangle\langle x| \mathfrak{T}( \delta x )~~ \implies \\ \hat p =\int d x~ | x \rangle ( -i\hbar\partial_x )\langle x |=\int d x~ | x \rangle ( i\hbar \overset{_\gets}\partial_x )\langle x | .$$

  2. Note the evolution operator $\mathfrak{U}(t)$ is defined on kets, not bras, $$\hat H = i\hbar\frac{\partial}{\partial t},$$ so it corresponds to the above, "unconventional", choice.

  3. So the option you believe is opposite in sign, is really the same as in 1., except you have integrated by parts to be acting on kets.

  4. After your additional comment. These time and space derivatives are meant to act on the Schroedinger wavefunction in the coordinate representation, $\phi(t,x)= \langle x| \phi(t)\rangle $, so $$i\hbar \partial_t \phi(t,x)= \langle x|\hat H |\phi(t)\rangle, \\ -i\hbar \partial_x \phi(t,x)= \langle x|\hat p|\phi(t)\rangle. $$

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  • $\begingroup$ I see the difference, $\hat{H}|\phi\rangle = i\hbar\frac{\partial}{\partial t}|\phi\rangle$ and $\langle\pmb{x}|\hat{\pmb{p}} = -i\hbar\frac{\partial}{\partial \pmb{x}}\langle\pmb{x}|$. However, another question arises, I don't know why these two operator can be combine into $p^{\mu} = i\partial^{\mu}$, where $E = i\frac{\partial}{\partial x^0}$ and $\pmb{p} = -i\nabla$. It seems to me that $E$ and $\pmb{p}$ acting on ket and bra respectively. So why they can be combined like $p^{\mu}$? $\endgroup$ Sep 30, 2023 at 6:44
  • $\begingroup$ I saw in Peskin's an introduction to quantum field theory page.xx in Notations and Conventions section, there is just operator form but no ket nor bra. $\endgroup$ Sep 30, 2023 at 12:21

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