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## Hot answers tagged hilbert-space

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 ...

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 $\Psi$: $$... 2 Consider a N-dimensional vector space V and let \{\chi_1, \cdots, \chi_N\} be basis of V. Next focus attention on the anti symmetric space (V\otimes\cdots \otimes V)_A where V occurs M\leq N times. A basis of (V\otimes\cdots \otimes V)_A can be constructed out of \{\chi_1, \cdots, \chi_N\} making use of the projector$$A: V\otimes\cdots ...

1

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 ...

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 ...

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 ...

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