Quick question: Momentum operator in QM

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SUMMARY

The discussion focuses on the two representations of the momentum operator in quantum mechanics: p = (-i hbar d/dx) and p = (hbar / i)d/dx. Participants clarify that the confusion arises from the properties of the imaginary unit, specifically that (-i)^2 equals -1, not 1. The mathematical manipulation involving the imaginary unit is essential for understanding the equivalence of the two forms of the momentum operator. The discussion emphasizes the importance of correctly applying the properties of complex numbers in quantum mechanics.

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Libohove90
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Homework Statement



There are two ways to write the momentum operator, p = (-i hbar d/dx) and p = (hbar / i)d/dx. How do you go from one to the other?

Homework Equations



The two I gave above.

The Attempt at a Solution



I tried to see if -ih = h/i by squaring both sides, but one came out positive and the other negative. Thanks for the help!
 
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-i^2 = 1, not -1. This makes (-i hbar)^2 not equivalent to (hbar / i)^2
 
Last edited:
Since when did (-i)^2 = 1?
 
Libohove90 said:
-i^2 = 1, not -1. This makes (-i hbar)^2 not equivalent to (hbar / i)^2

(-i)^2 = (-1)^2(i)^2
 
i^{-1} =- i
Why?
<br /> i^{-1} = e^{-ln(i)}=e^{-ln(e^{i\pi /2})}=e^{-i\pi /2}=-i<br />
since e^{\pm i\pi /2}= cos(\pm\pi /2) + isin(\pm\pi /2)=\pm i
 
Or you just multiply 1/i by i/i to get i/i^2 = i/(-1) = -i.
 

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