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this is a description of an interaction between the electron and photons, which would collapse the wavefunction (right?). No this isn't right. As long as the system stays isolated, the interaction simply means that there are cross terms in the relevant Hamiltonian and that you have a two-particle quantum system, whose state space is the tensor product ...


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Let's look at where the electromagnetic interaction comes from in hydrogen. At first quantization you have a multiparticle system so the wavefunction is defined as $\psi=\psi(x_1,y_1,z_1,x_2,y_2,z_2,t)$ and the point is to write the Hamiltonian. And the Hamiltonian comes from the Lagrangian. For a single particle of charge $q$ in an external ...


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You are imagining the particle in the well as a classical system i.e. a point particle moving to and fro in the well. However this is not a good description of the system. A quantum particle does not have a position. By this I mean that it is meaningless to ask what the position of the particle is because position, in the sense we normally use the term, is ...


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$$\int_{-\frac{d}{2}}^{\frac{d}{2}}\psi^*(x)\psi(x)dx$$ is the probability of finding particle between $-\frac{d}{2}$ and $\frac{d}{2}$. Expectation value is : $$\langle x\rangle=\langle \psi|\hat{x}|\psi\rangle$$ $|\psi\rangle$ is the summation of probability amplitude times given basis kets $$|\psi\rangle = \sum_i c_i |{e_i}\rangle$$ ...


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The definition of the expectation value of an operator A is \begin{equation} \langle A\rangle=\int{\psi^* (x) A(x) \psi (x) dx} \end{equation} (because it represents "the value of the variable" $A(x)$ times "the probability of being in that configuration" $P(x)=\psi^* (x) \psi (x)$) and for the particular case of the expectation value of the position ...


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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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I think your error is in assuming that $E_{n+1} - E_{n}$ is proportional to $n$. At least, I assume you assumed it; it's the only way I can see that you could go from the statement $$ E_{n+1} - E_n \propto E_n^{-1/2} $$ to the statement $$ E_n \propto n^{-2}. $$ Really, what the first proportionality above implies is that $$ \frac{dE}{dn} \propto ...


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With this transformation $\rho = kr$ you can't possibly change the form of the equation, because it's a scale transformation that does nothing special to the derivatives, so I think that you have computed the derivatives wrong.



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