# Tag Info

7

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 ...

6

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. ...

3

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 ...

3

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 ...

3

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 ...

3

$e^{i\theta} + e^{-i\theta}$ 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.

3

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 ...

2

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 ...

1

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 ...

1

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 ...

1

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 ...

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