Linked Questions

0
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1answer
1k views

How to prove that the position operator in momentum is $i\hbar \partial/\partial p$ - One Missing Sign [duplicate]

I am trying to prove that the position operator in momentum space is $i\hbar \partial/\partial p$ but my derivation is missing one sign. Can someone spot the error? Start with $$<\hat x> = \...
36
votes
7answers
6k views

What is Quantum Mechanics really about? [duplicate]

This question might sound very silly, so I'm sorry if that's the case. I'll try my best to make my point clear here. Before explaining, just to make clear, I'm not confused because of the Math ...
26
votes
6answers
37k views

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, ...
15
votes
6answers
2k views

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. ...
16
votes
3answers
4k views

How does one determine ladder operators systematically?

In textbooks, the ladder operators are always defined," and shown to 'raise' the state of a system, but they are never actually derived. Does one find them simply by trial and error? Or is there a ...
20
votes
2answers
3k views

What does the Canonical Commutation Relation (CCR) tell me about the overlap between Position and Momentum bases?

I'm curious whether I can find the overlap $\langle q | p \rangle$ knowing only the following: $|q\rangle$ is an eigenvector of an operator $Q$ with eigenvalue $q$. $|p\rangle$ is an eigenvector of ...
17
votes
6answers
1k views

Is the harmonic oscillator potential unique in having equally spaced discrete energy levels?

I was wondering if the good old quadratic potential was the only potential with equally spaced eigenvalues. Obviously you can construct others, such as a potential that is infinite in some places and ...
11
votes
2answers
2k views

Are all scattering states un-normalizable?

I am an undergraduate studying quantum physics with the book of Griffiths. in 1-D problems, it said a free particle has un-normalizable states but normalizable states can be obtained by sum up the ...
3
votes
4answers
7k views

Interesting relationship between diffraction and Heisenberg's Uncertainty Principle?

I recently came across an interesting explanation of diffraction through an aperture which does not use Huygens' Construction but instead relies on Heisenberg's Uncertainty Principle: The ...
6
votes
1answer
2k views

What physical significance has the Heisenberg Group?

I read that the canonical commutation relation between momentum and position can be seen as the Lie Algebra of the Heisenberg group. While I get why the commutation relations of momentum and momentum, ...
2
votes
1answer
534 views

Heisenberg relation

Given that $A(k)=\frac{N}{k^2+\alpha^2}$, show that $\Delta k \Delta x >1$. Considering the above example, according to my textbook, it is written that I must square the above function and ...
1
vote
1answer
479 views

In quantum mechanics, why position and momentum are related by Fourier Transformation (only)? [duplicate]

We know that if we take Fourier transform of momentum we go to position space. But why Fourier transform only.
1
vote
4answers
263 views

Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis

I am trying to calculate $\langle x\ |\ \hat{x}\ |\ p\rangle$. I can work in the $x$-basis like so: $$\langle x\ |\ \hat{x}\ |\ p\rangle=\int dx'\langle x\ |\ \hat{x}\ |\ x'\rangle\langle x'\ |\ p\...
1
vote
2answers
238 views

Fourier transformation [closed]

I have recently studied Fourier and Laplace transformation in maths. I wanted to understand the utility in physics with some examples that requires this change in dimension and the reason why.
4
votes
1answer
86 views

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 ...