# Tag Info

2

Griffiths' formulation makes it explicit that operators which commute are not restricted by the uncertainty principle. Your boxed expression obscures this physical and mathematical insight.

2

The most correct relation is the following general relation, that actually contains both terms. If you omit the inequality $(|z|^2\geq(Im(z))^2)$ from the derivation, the next steps toward the uncertainty relation would be: $$\sigma_A^2 \sigma_B^2\geq|\langle f\lvert g\rangle|^2$$ $$|\langle f\lvert g\rangle|^2=\langle f\lvert g\rangle\langle g\lvert ... 1 The fields satisfy the wave equation. We can therefore write \begin{split} \phi(x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ a({\bf p}) e^{i p \cdot x} + b^\dagger({\bf p} ) e^{- i p \cdot x} \right] \\ \phi^\dagger (x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ b({\bf p}) e^{i p \cdot x} + ... 2 We interpret OP's question (v4) as: How do we recover the phase ambiguity from the generator of translation method in Ref. 1? Recall that an eigenvector for an operator can be rescaled with a non-zero multiplicative factor. The main point is that the position eigenket | x \rangle, which satisfies$$\tag{A} \hat{x}| x \rangle~=~ x| x \rangle, $$... 0 "projection operators commute → they're the same" Are you sure he said this predicate ? or it is your own consequence? However, it is not true ! Consider two dimensional X-Projector And Y-Projector , they commute but they are not the same! 0 A complete set of eigenstates spans the whole space, not just the subspace the projection operators project on. In this set of eigenstates you also have a basis of the subspace belonging to the eigenvalue 0. 3 The easiest thing for this exercise is to use Levi-Civita symbol for the vector product:$$\vec{a} \times \vec{b} = a_i b_j e_k \varepsilon_{ijk},$$where I denote by e_i the canonical basis of \mathbb{R}^3. Using this notation, we have:$$[L_j,p_i]=[r_k p_l \varepsilon_{klj},p_i]= i \hbar p_l \varepsilon_{ilj}.$$and$$[L^2,\vec{p}]=e_i[L_j ...

1

Intuitively, if the potential energy is a function only of the position, if you measure the position precisely, you can just calculate the potential energy using that precise measurement. More formally, if $V(\hat x)$ is any function of $\hat x$, the position operator $$[V(\hat x), \hat x] = 0$$ which really is just that any operator commutes with itself.

3

Seeing as $$\langle k|k_1k_2\rangle = \langle 0| a(\mathbf{k}) a^{\dagger}(\mathbf{k_1}) a^{\dagger}(\mathbf{k_2}) |0\rangle$$ and $$a(\mathbf{k})a^{\dagger}(\mathbf{k_1}) = a^{\dagger}(\mathbf{k_1})a(\mathbf{k}) + f(\omega)\delta(\mathbf{k}-\mathbf{k_1})$$ you'll get one term that vanishes because you cannot destroy the vacuum and one term that simply ...

2

As you said if two operators commute they share eigenvectors. Physically this means that you can have a definite value for both. For example in the hydrogen atom the Hamiltonian $H$, which is the energy, and $J^2$, the magnitude of angular momentum, commute. A hydrogen atom can be in a state of definite energy and definite angular momentum. However, the ...

5

First, a note about the Hamiltonian and its time derivatives. I think that it is misleading to write that the Hamiltonian $$H = i\hbar\frac{d}{dt},$$ although the time-dependent Schrodinger equation is of course $$H\psi = i\hbar\frac{d}{dt} \psi.$$ To evaluate e.g. $\frac{d}{dt}H$ you should consider $H=H(p, q, t)$, rather than $H = i\hbar\frac{d}{dt}$. ...

1

The notation is short-hand for an expression utilizing the Backer Campbell Haussdorf formula. Let $X$ and $Y$ be operators, then $$e^{x}Ye^{-X} = Y + [X,Y] + \frac{1}{2!}[X,[X,Y]] + \frac{1}{3!}[X,[X,[X,Y]]] + ...$$ I assume $[X,Y]_{(n)}$ refers to the $n$th term in this expansion; it roughly counts how many times the commutators are nested in each other. ...

3

I think this is the Baker-Campbell-Haussdorff formula, and the notation means to iterate the commutator. That is, $$[L, M]_1 = [L, M]$$ And $$[L,M]_{n+1} = [L, [L,M]_{n}].$$

1

This is not standard notation, and one would typically expect any text that uses it to define it at its first occurrence. Since you understandably cannot provide us with a reference, your best bet is hunting for all occurrences of that notation, starting from there and going up through the text, until it explains what it means. Trust me, it will be there.

0

Commutation does become transitive, and thus an equivalence relation (reflexive and symmetric are trivial), when you impose an extra condition: nondegeneracy. If $A$, $B$, $C$ are Hermitian operators, and each of them has only unique eigenvalues, then $AB\! =\! BA\, \cap\, BC\! =\! CB$ implies $AC = CA$. Proof: for a nondegenerate operator, the eigenbasis ...

4

No, it does not! Let me give you a counterexample: Consider the Hermitian operators $\mathsf{1}$ (identity operator), $p$ (momentum) and $x$ (position) in 1D. Now, the trivial commutation relations $[\mathsf{1},x]=0$ and $[\mathsf{1},p]=0$ do not imply $[x,p]=0$ as the correct relation is $[x,p]=\mathrm i\hbar\neq 0$.

3

Commuting is not an equivalence relation. All components of angular momentum commute with $J^2$ but they don't commute with each other. How to find a complete set of mutually commuting observables is a difficult problem and I don't think you can give an algorithmic answer. It depends very much on the specific problem. An observable that commutes with the ...

0

Since the vector $\Lambda | \omega_i \rangle$ has the same eigenvalue as $| \omega_i \rangle$, it must be in the same invariant subspace as $| \omega_i \rangle$, which Shankar takes to be one dimensional.

1

When $\lambda_1$ is an eigenvalue of a matrix and $v_1$ and $v_2$ are the components of the corresponding eigenvector, then the following equation holds: $\begin{pmatrix} a-\lambda_1 & b \\ c &d-\lambda_1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ Now when you scale up the eigenvector (say by three) ...

2

Note that he explains above: "Consider first the case where at least one of the operators is nondegenerate, i.e. to a given eigenvalue, there is just one eigenvector, up to a scale." So he uses the assumption that the operator is nondegenerate and the definition of nondegeneracy (or a statement equivalent to the definition of nondegeneracy, if you use a ...

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