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### Derivation of momentum operator [duplicate]

From a video lecture on quantum mechanics at MIT OCW I found that you didn't need to know Schrödinger's equation to know the momentum operator which is $\frac{\hbar}{i}\frac{\partial}{\partial x}$. ...
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### Why is quantum mechanical momentum the derivative of the wave function with respect to the position? [duplicate]

In classical mechanics the momentum is defined as mass times the time-derivative of position. In quantum mechanics, however, the time-derivative of the wave function is the hamiltonian, while the ...
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### Does the canonical commutation relation give a unique solution for the momentum operator? [duplicate]

So lets say we are in a 1d system and in the position basis just for simplicity. The CCR is: $$[x,p]=i$$ and the momentum operator is $-i\partial_x$. Is this solution unique or are there other ...
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### How to get the position operator in the momentum representation from knowing the momentum operator in the position representation?

I know that $$\tag{1}\hat{p}~=~-i\hbar \frac{\partial}{\partial x}~.$$ How can I get $$\tag{2}\hat{x}~=~i\hbar \frac{\partial}{\partial p}~?$$ I think this simple and I'm just over thinking it, ...
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### Is the Momentum Operator a Postulate?

I've been studying the postulates of QM and seeing how to derive important ideas from them. One thing that I haven't been able to derive from them, however, is the identity of the momentum operator. ...
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### Derivation of position operator in QM

For the definition of the momentum operator $$\hat{P } = -i \hbar \nabla$$ in quantum mechanics, as I understand you can derive this by either considering a more general definition of momentum, i.e. '...
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### Momentum operator ambiguous?

In nonrelativistic quantum mechanics, are different operators possible as a candidate for the momentum operator, given that one has fixed one position operator and a hilbert space that this position ...
Starting from my previous question Commutators in quantum mechanics and considering that the commutator $$\left[i\hbar\frac{\partial}{\partial x},x\right]=i\hbar, \tag{1}$$ the associated linear ...