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My textbook states that the prefactor $(2 \pi \hbar)^{-\frac{3}{2}}$ is not required for the following superpositioned wave function, but should be included for practical reasons without stating what the practical reason is.

$$\Psi(\vec{r},t) = (2 \pi \hbar)^{-\frac{3}{2}} \int \phi(\vec{p}) \ e^{i(\vec{p}\cdot \vec{r} - Et)/\hbar} d^3p$$

So what is the purpose of this prefactor? And where does it come from? I see that Schrödingers equation is derived while including this, so I assume it is "required" somewhow.

UPDATE: Is it somehow related to the definition of the momentum operator $\hat{\vec{p}} \equiv\frac{\hbar}{i}\nabla$?

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  • $\begingroup$ Have you reviewed the normalizations of probability densities in coordinate and momentum space, respectively? $\endgroup$
    – Cosmas Zachos
    Commented Nov 28, 2018 at 17:21
  • $\begingroup$ @CosmasZachos I have reviewed normalization of probability densities in general, however, are there special considerations for position-/momentum-space? $\endgroup$
    – Gjert
    Commented Nov 28, 2018 at 17:29
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    $\begingroup$ How do you arrange $\int d^3x ~|\Psi|^2=1=\int d^3p ~|\phi|^2 $? $\endgroup$
    – Cosmas Zachos
    Commented Nov 28, 2018 at 17:43
  • $\begingroup$ @CosmasZachos Ah, through the Fourier transform given in the answer below. Nice, I found the definition in the book now. Thank you :) $\endgroup$
    – Gjert
    Commented Nov 28, 2018 at 17:52
  • $\begingroup$ The better way to do this is to write the Fourier transform in terms of $k$ instead of $p$. $\endgroup$
    – DanielSank
    Commented Nov 28, 2018 at 18:56

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The $\hbar$ is for units, and the factors of $\sqrt{2\pi}$ are there to fit the unitary convention for the Fourier transform. In quantum mechanics language, we need $\langle x|p\rangle = \langle p|x\rangle^*$.

For why the factors of $2\pi$ are needed in the Fourier transform at all, it's because that is the period of sine and cosine in radians, leading to a formal definition of the delta function as $$\delta (x-y) = \int_{-\infty}^\infty \frac{\mathrm{e}^{ik(x-y)}}{2\pi}\operatorname{d}k.$$

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  • $\begingroup$ If $\hbar$ is for units, is there a special reason for $\hbar^{-3/2}$? I understand the $(2\pi)^{-3/2}$ from the def. of Fourier transforms, but not sure where $\hbar$ is coming from. $\endgroup$
    – Gjert
    Commented Nov 28, 2018 at 17:42
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    $\begingroup$ The power comes from the dimensionality of the problem. In general, in $d$ dimensions the power is $\frac{d}{2}$. Note that $\hbar$ also appears because in your case the integration variable is $p$ instead of $k$. $\endgroup$
    – eranreches
    Commented Nov 28, 2018 at 17:47
  • $\begingroup$ Great explanation, thank you! Also, it's a bit off topic, but could you lead me in the direction of how the momentum operator was motivated? I've tried looking for it in Classical Mechanics book, but just states it as a definition without reasoning. $\endgroup$
    – Gjert
    Commented Nov 28, 2018 at 17:54

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