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## New answers tagged hilbert-space

0

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 $$A: V\otimes\cdots ... 1 For an antilinear operator, as the antiunitaries and the complex conjugation, the definition of adjoint is changed:$$\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 ... 0 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 ... 1 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 ... 0 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. 1 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 ... 5 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 ... 0 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 ... 3 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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