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You don't. Sinusoidal wavefunctions like these, a.k.a. plane waves, are non-normalizable, because the integral which defines the norm does not converge. $$\langle\psi|\psi\rangle = \int_{-\infty}^{\infty} |A e^{ikx} + B e^{-ikx}|^2\mathrm{d}x = \infty$$ (EDIT: just thought I should mention that $\ldots = \infty$ doesn't mean the integral literally equals ...

It depends on the domain of $p$. If we take the domain of $p$ to be the Schwartz space on $\mathbb{R}$, then, by symmetry of $p$, $$ \langle p^2\rangle =\langle p\psi |p\psi \rangle =\left\| p\psi \right\| ^2 $$ This is $0$ iff $p\psi =0$ iff $\psi$ is constant. However, the only constant Schwartz function is $0$. Hence, $p^2$ is positive-definite. ...

Jerry Schirmer's answer applies in an infinite space. If you put the system in a box there is no problem: the normalized wavefunction is $\mathrm{e}^{ikx}/\sqrt{V}$. This is the usual theoretical device to make everything nice and well behaved. Then we take the limit $V\rightarrow\infty$ at the end of the day and, if we've done our job correctly, all of the ...

You can't, really. In the continuum, the "free particle wavefunction" is an abstraction to make the math easier. In a real context, you'd talk about a Wave envelope $\phi(x) = \int dk A(k)e^{ikx}$, where $A(k)$ is a bounded function that guarantees that $\int\phi^{*}\phi$ is finite and normalizable. Note that this is the generalization to the continuum of ...

You test a wave function for normalizability by integrating its square magnitude. If you get a finite result then it is normalizable. To spare you complicated integrations you can also take a simpler wave function that you know is normalizable and compare it using the usual arguments. An operator is not only defined by the mathematical operation it ...

$e^{i\theta} + e^{-i\theta}$ is just $2\cos \theta$. The superposed wavefunction is $$\Psi(x,t) = 2N\cos(ax) e^{i(f(x) + \omega t)}$$ Then $$\Psi^*\Psi = 4N^2\cos^2(ax)$$ The average height is $2N^2$ if $x_0a = n\pi/2$, in which case $N = \frac{1}{2}\sqrt{1/x_0}$. Otherwise you can do this integral.

If I understand correctly, your question basically comes down to identifying a basis for the space of square-integrable functions, $L^2(\mathbb{R})$, since any physical state $|\Psi\rangle$ can be constructed by performing the integral you listed in your question with a function $\Psi_x(x)\in L^2(\mathbb{R})$. $L^2$ is known to be a vector space, so a basis ...

The eigenfunctions of a self adjoint operator lie outside the Hilbert space of square integrable functions on the line. One solution is to work with a basis of eigenfunctions of a non-self adjoint operator such as $x+ip$. Of course these are the coherent states. For the coherent states, one has an ovecomplete basis and a partition of unity, thus it is not ...

I know in electrical engineering, particularly in power transmission fields, sometimes people use the so-called per-unit system, where quantities are normalized to the corresponding base value. Sometimes, subscript like $U_{\text{p.u.}}=\frac{U}{U_0}$ ($U_0=U_{\text{base}}$) is used. I don't think there are any proper notations for normalized scalars. You ...

There is not, to my knowledge, a uniform standard on this subject. I have seen normalized quantities expressed by adding a twiddle (tilde) over the character (as in $\tilde{A}=A/A_0$), but I this notation is often reserved to indicate a time-varying quantity, instead. Unless someone knows of a strong standard in this regard, I'd say your best bet is to ...

This is mostly a comment but I feel it's important enough that it warrants some space as an answer. Note carefully that the wavefunction $\psi=\psi_0 \sin(kx-\omega t)$ is not a solution to the free-particle Schrödinger equation. This is easy to see mathematically. The kinetic term, $-\frac{\partial^2}{\partial x^2}$, will return a sine term, while the ...