New answers tagged commutator
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Unitary transformation of the momentum operator in quantum mechanics
Work in the coordinate representation,
$\hat O(\hat{\mathbf{r}}) |x\rangle=O(\mathbf{r})|x\rangle$,
so, then,
$$[\hat{\mathbf{p}}, \hat{U}_0] \mapsto -i\hbar {U}_0( {U}_0^\dagger \nabla_{\mathbf{r}} ...
- 65.2k
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Schrodinger basis representation of modes in 2D CFT (Polchinski section 2.8)
$\psi(x)$ are represented by $a= (ix+p)/\sqrt{2}$ and $a^\dagger = (ix-p)/\sqrt{2}$. Here $p= i\partial_x$ when acting on wave functions. In the Schr"odinger picture we thus have
\begin{align}
\...
- 3,136
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Schrodinger basis representation of modes in 2D CFT (Polchinski section 2.8)
Yes, the second term in eq. (2.8.25a) is needed in order for the commutation relations
$$\begin{align}[X_n,X_m]~=~&0, \cr [\frac{\partial}{\partial X_n},X_m]~=~&\delta^n_m, \cr [\frac{\partial}...
- 209k
6
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Are quantum measurements on two different particles always commutative?
Yes, they commute. However, it's not correct to talk about the commutator $[A,B]$ because these are operators defined on different Hilbert spaces.
When you have a system of two particles, the state of ...
- 81
3
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Are quantum measurements on two different particles always commutative?
I haven't seen it explicitly stated either, but it follows directly from definitions.
If $A$ is an observable associated with only the first particle, then by definition, it actually takes the form $...
- 104k
3
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Spacetime translation in QFT
Your premise is incorrect. In quantum mechanics, the correct formula is
$$
p | x \rangle = i \frac{d}{dx} | x \rangle . \tag{1}
$$
In QFT, $|x\rangle = \phi(x) |0\rangle$, so it follows that
$$
[ p , \...
- 311
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Proving quantum operator relationship used in derivation of Radial Schrodinger Equation
I have managed to find a solution but I would appreciate some verification if the logic of this solution is correct ...
So I can make it work if I should interpret any vector operator "square&...
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Proving quantum operator relationship used in derivation of Radial Schrodinger Equation
Your mistake happens in the last line. When you use the commutator. You get $r^2p^2=rppr+i\hbar rp$.
But you still have it on the right side of your equation. Push it to the left so you get the minus ...
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3
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Accepted
Is string theory a particular non-commutative field theory (whether the commutator of the position coordinates in string theory is non-zero)?
The relation between string theory and non-commutative geometry is very poorly understood. Non-commutativity is only seen when studying open strings. The positions of close $ N$ D-branes is described ...
- 76
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How to derive the commutation coefficient from coordinate basis (GR)?
An anholonomic basis is basically just the linear combination of holonomic bases where the expansion coefficients $f$ are functions of coordinate variables -
$$\begin{equation}\tag{1}\mathbf{e}_{i}=f^{...
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What is the mathematically precise definition of raising and lowering operators?
There is a simple way in which I understand this concept of raising and lowering operators. I'll try to explain it in two stages.
First, given the assumed relation
$$ [\hat{a},\hat{a}^{\dagger}] = \...
- 15.6k
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What is the mathematically precise definition of raising and lowering operators?
There are two main different ways in which one can generalize the notion of ladder operators, but neither is particularly useful in practice:
Naive ladder operators. The nice thing about linear ...
- 127k
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What is the mathematically precise definition of raising and lowering operators?
In general, a ladder operator $O_{\pm}$ is an operator defined for some states $|n\rangle$ as
$$O_{\pm}|n\rangle = C_{\pm}(n)|n\pm1\rangle \ \ \mathrm{and} \ \ O_+^\dagger
=O_- .$$
The term $C_\pm$ is ...
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