Quadratic reciprocity with the prime three givesThe latter symbol is $+1$ when $p\equiv 1(\mathrm{mod}3)$ and $-1$ when $p\equiv 2(\mathrm{mod}3)$. A prime greater than three occupies one of the classes $1,5,7,11$ modulo twelve. Combining its residues modulo three and four gives signs $+1,-1,-1,+1$ respectively. Hence

Over ${\mathbb{F}}_5$,The quadratic discriminant is $9-16\equiv 3$, while the squares modulo five are $0,1,4$. Thus that factor is irreducible. Adjoining either quadratic root creates a degree-two extension and supplies the other root by the sum-of-roots relation. Consequently the splitting field is ${\mathbb{F}}_{25}$ and $\boxed{[K:{\mathbb{F}}_5]=2}$.

Put $\alpha =\sqrt[3]{2}$ and $\zeta =e^{2\pi i/3}$. The three roots are $\alpha ,\alpha \zeta ,\alpha {\zeta }^2$, so the splitting field is $K=\mathbb{Q}(\alpha ,\zeta )$. Eisenstein's criterion at two gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$. This intermediate field is real, while $\zeta$ is not real, and $\zeta$ satisfies $x^2+x+1=0$. Its degree over the intermediate field is therefore exactly two. By the tower law, $\boxed{[K:\mathbb{Q}]=6}$.

The roots are $\pm \sqrt{-5}$, so one root already supplies both. The polynomial has no rational root and is irreducible of degree two. Thus its splitting field is $\mathbb{Q}(\sqrt{-5})$, with $\boxed{[K:\mathbb{Q}]=2}$.

The five specified masses sum to one, so there are no other possible outcomes. Symmetry gives $\mathbb{E}X=0$, while $\mathbb{E}Y=1/4$. The two terms with $Y=1$ have opposite $X$ values and equal masses, soNevertheless $\mathbb{P}(X=0,Y=1)=0$, whereas $\mathbb{P}(X=0)\mathbb{P}(Y=1)=(1/2)(1/4)=1/8$. Therefore $X$ and $Y$ are not independent. They are uncorrelated random variables, illustrating that zero covariance alone does not imply independence.

Independence means that for all subsets $A,B\subseteq \mathbb{Z}$,For discrete random variables, this is equivalent to the pointwise factorizationOne direction takes singleton sets; the other sums the pointwise identity over $A\times B$.

For integer-valued random variables, the joint probability mass function isThe marginal probability mass functions sum out the other variable:All masses are nonnegative, and their sums over the corresponding sample values equal one.

Put $s=1+it$. Logarithmic differentiation givesThereforeso the appropriate potential energy is $\boxed{V(x)=0}$. The wavefunction never vanishes, so no other time-independent potential gives the same equation for this precise phase choice.

The squared modulus isIt changes width and height with time, so this is not a stationary state. Its integral is the constant $\sqrt{\pi \hslash /m}$; multiplying the wavefunction by $(m/(\pi \hslash ){)}^{1/4}$ normalizes it. The normalized Gaussian wave packet stays centred at zero but spreads, with position variance $\hslash (1+t^2)/(2m)$. It represents a localized free particle with a spread of momenta, rather than an energy eigenstate. Free Gaussian wave packet evolution broadens the packet even with zero mean momentum.

The Time-dependent Schrödinger equation isA stationary state of energy $E$ has a wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, whereIts time dependence is only a global phase, and its probability density is independent of time.