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Any one of operators Lx, Ly, or Lz can be called quantized. There exists a set of states which are eigenstates of Lz; the matrix Lz is diagonal but Ly and Lx are not. There exists a set of states which are eigenstates of Ly; the matrix Ly is diagonal but Lx and Lz are not. There exists a set of states which are eigenstates of Lx; the matrix Lx is diagonal ...


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Consider a $N$-dimensional vector space $V$ and let $\{\chi_1, \cdots, \chi_N\}$ be basis of $V$. Next focus attention on the anti symmetric space $(V\otimes\cdots \otimes V)_A$ where $V$ occurs $M\leq N$ times. A basis of $(V\otimes\cdots \otimes V)_A$ can be constructed out of $\{\chi_1, \cdots, \chi_N\}$ making use of the projector $$A: V\otimes\cdots ...


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For an antilinear operator, as the antiunitaries and the complex conjugation, the definition of adjoint is changed: $$\langle U^{a*}\psi,\phi\rangle=\overline{\langle \psi,U\phi\rangle}$$ where $a*$ stands for anti-adjoint. It is therefore easy to see that the anti-adjoint of $K$ is $K$ itself (and in general the anti-adjoint of an anti-unitary is ...


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Maybe it helps if we can place this in a context. You can have spaces like $A$ and $B$ and then you can make product spaces like $A \otimes B$. You can make linear operators like $S:A\rightarrow C$ and $T:B\rightarrow D$ and then since an arbitrary thing in $A \otimes B$ is spanned by things like $a \otimes b$ (with $a\in A$ and $b\in B$) you can clearly ...


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I) Let us here phrase the problem in the context of some position operator $\hat{q}$ of QM for simplicity. The generalization to QFT can formally be achieved by replacing the position operator $\hat{q}$ with a quantum field $\hat{\psi}({\bf x})$. We know that the overlap with Minkowski (M) signature is given as a path integral ...


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Here I constructed perturbation-like approximants converging to the vacuum in $\phi^4_2g(x)$ (technically an interacting QFT, although not translation invariant, so Haag's theorem does not apply). There are no "infinities" in this case.


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No, it has not discrete spectrum (on $L^2(\mathbb{R}^d)$). In fact $a+a^*$ is proportional to the position operator (or the momentum one, depends on your definition of $a$ and $a^*$; by the usual one the position operator $x$ is proportional to the real part $a+a^*$ and the momentum $p$ to the imaginary part $\frac{1}{i}(a-a^*)$). Both position and momentum ...


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But how can we guarantee that two solutions $\boldsymbol {\psi_1}$ and $\boldsymbol {\psi_2}$ to the time-dependent equation don't have $\boldsymbol {\psi_1(x,0)} = \boldsymbol {\psi_2(x,0)}$. If we can't guarantee this, then how do we know that the solution found by Griffith's method is unique? I interpret that your question basically asks how do we ...


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So I see your whole question as this: How can we guarantee that two solutions $\boldsymbol {\psi_1}$ and $\boldsymbol {\psi_2}$ to the time-dependent equation don't have $\boldsymbol {\psi_1(x,0)} = \boldsymbol {\psi_2(x,0)}$. If we can't guarantee this, then how do we know that the solution found by Griffith's method is unique? This is a really basic ...


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The form of the solution shown by Griffiths is not unique. That means that there exist cases where a basis $\{\psi_n(x)\}$ will reproduce $\Psi$ as $$ \Psi(x,t)=\sum_{n=1}^\infty c_n\psi_n(x) e^{-iE_nt/\hbar}, $$ but there exists a second, different basis $\{\varphi_n(x)\}$ which (with different coefficients) also reconstructs $\Psi$: $$ ...



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