Linked Questions

3
votes
1answer
207 views

Where does the expression $\langle q \mid p\rangle=\frac{1}{(2\pi\hbar)^{n/2}}e^{i q\cdot p/\hbar}$ come from? [duplicate]

I have been given the following complete systems of eigenvectors $$\mathbf{Q}\mid\mathbf{q} \rangle=\mathbf{q}\mid\mathbf{q} \rangle, \quad \mathbf{P}\mid\mathbf{p} \rangle=\mathbf{p}\mid\mathbf{p} \...
0
votes
0answers
394 views

How to prove $\hat{p}|x\rangle=i\hbar\frac{\partial}{\partial x}|x\rangle$,using $[\hat{x},\hat{p}]=i\hbar$? [duplicate]

How to prove $$\hat{p}|x\rangle=i\hbar\frac{\partial}{\partial x}|x\rangle,$$ using $$[\hat{x},\hat{p}]=i\hbar~?$$ The question seems to be uncomplete because for any $f(x)$ $$[\hat{x},\hat{p}+f(x)]=i\...
2
votes
0answers
98 views

Is my expansion of the state $| x \rangle$ correct? [duplicate]

In my quantum mechanics textbook it says that the relation between the basis $|x\rangle$ and $|p\rangle$ is given by: $\langle p | x \rangle = \Large \frac{e^{-ip x/ \hbar}}{\sqrt{2\pi \hbar}} \, .$ ...
2
votes
0answers
92 views

How can I derive $\langle p | x \rangle=e^\frac{ipx}{\hbar}$ without resorting to wave mechanics? [duplicate]

$\langle p | x \rangle=e^\frac{ipx}{\hbar}$ can pretty easily be derived from wave mechanics. But how can it be derived without resorting to it? I've seen it be derived from the translation operator $\...
28
votes
6answers
39k 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, ...
11
votes
5answers
6k views

How does the momentum operator act on state kets?

I have been going through some problems in Sakurai's Modern QM and at one point have to calculate $\langle \alpha|\hat{p}|\alpha\rangle$ where all we know about the state $|\alpha\rangle$ is that $$\...
6
votes
2answers
2k views

Proving that $i\hbar\frac{\partial}{\partial \mathbf{p}}$ is the operator of $\mathbf{x}$ in momentum space

How can I prove that $i\hbar\frac{\partial}{\partial \mathbf{p}}$ is the operator of $\mathbf{x}$ in momentum space?
6
votes
3answers
2k views

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. '...
13
votes
1answer
2k views

Normalizing Propagators (Path Integrals)

In the context of quantum mechanics via path integrals the normalization of the propagator as $$\left | \int d x K(x,t;x_0,t_0) \right |^2 ~=~ 1$$ is incorrect. But why? It gives the correct pre-...
9
votes
1answer
3k views

Why path integral approach may suffer from operator ordering problem?

In Assa Auerbach's book (Ref. 1), he gave an argument saying that in the normal process of path integral, we lose information about ordering of operators by ignoring the discontinuous path. What did ...
1
vote
2answers
6k views

From position space to momentum space

Lets say I have a state vector $\left|\Psi(t)\right\rangle$ in a position space with an orthonormal position basis. If I now use an operator $\hat{p}$ on this basis I will get basis which corresponds ...
5
votes
2answers
2k views

Derivation of canonical position-momentum commutator relation

We know that the position-momentum commutator is fundamental in quantum mechanics, but would it be possible to derive it starting from a different set of first principles, more specifically starting (...
5
votes
2answers
1k views

What is the most general expression for the coordinate representation of momentum operator?

I have a question about deriving the coordinate representation of momentum operator from the commutation relation, $[x,p]= i$. One derivation (ref W. Greiner's Quantum Mechanics: An Introduction, 4th ...
4
votes
1answer
744 views

What conservation law corresponds to this local $U(1)$ symmetry of the CCR?

It is known that canonical commutation relations do not fix the form of momentum operator. That means that if canonical commutation relations (CCR) are given by $$[\hat{x}^i,\hat{p}_j]~=~i\hbar~\...
0
votes
2answers
473 views

Derivation of plane wave from inner product of position ket and momentum ket

In textbooks it seems to be taken for granted that $$\langle \mathbf{r}|\mathbf{k}\rangle ~=~ \frac{1}{\sqrt{\Omega}}\exp(i\mathbf{k}\cdot\mathbf{r}).$$ I'm sure it's obvious but is there a ...

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