# Tag Info

3

The modern version of Pauli's principle requires completely antisymmetry of a state of $n$ fermions. Instead, the argument discussed in the body of the question only implies that a state of $n$ fermions has to be either symmetric or antisymmetric under interchange of a pair of particle. It is impossible, following this way, to prove that the full state is ...

6

This argument just replaces one axiom by another. It assumes that if a quantum system consists of identical particles, then the state of the system should not change (it get's multiplied by a phase) under exchange of quantum numbers. Although this is (perhaps) a more intuitive way of thinking about states of identical particles, it's still a strong ...

2

One way to understand it is to recognize that for the spherical harmonic $|l,m\rangle$ with $l=0$ (and obviously $m=0$), we have $\hat L_i|0,0\rangle=0$, where $\hat L_i$ is the angular momentum operator in the direction $i=x,y,z$. It is obvious for $\hat L_z$, which eigenvalue is $m=0$, and can be verified for the other two. Then, the rotation operator ...

1

Suppose that there existed a spherically symmetrical wavefunction $\psi({\bf r})=f(r)$ for which $l\neq0$. This cannot be, for if we calculate $\langle \psi | L^2 | \psi \rangle$ we will always get zero, as each term in $L^2$ has derivatives with respect to $\theta$ and $\phi$. Conceptually speaking, a spherically symmetric state gives the electron the ...

0

I'm not quite sure what your goal is. If the user specifies both $p$ and $q$, you end up with one value for the energy, rather than a plot. For a proper plot you'd have to allow for a range of values $p,q\in\mathbb{Z}$. This will of course result in a 3d graph. You may cut that down to 2d, like in the plot you posted, but you'd have to give some relation ...

1

In one sense you are right: the only free space "perfectly collimated" optical field is the plane wave in the sense that these are the only eigenfields of Maxwell's equations, being fields which conserve their form under propagation and only undergo scaling by an eigenvalue in such propagation. Since Maxwell's equations conserve energy in free space, ...

3

Two cents from an experimentalist. It is always good to keep in mind that a wavefunction for a real particle in the lab is a solution of Schrodinger's equation with specific boundary conditions given by the experimental setup that makes the measurement. Every measurement changes the boundary conditions for the solution that describes the particle's ...

1

The electron doesn't get destroyed when you measure it (though photons usually do), but its wavefunction doesn't go back to how it was before. Instead it gets a new wavefunction, different from the old one. If you measured the position of the electron, this new wavefunction will be a delta function (a single infinitely sharp spike) centred at the position ...

11

Assuming wave-function collapse is correct (which can be a relatively hefty philosophical claim in some circles), then think of measurement this way: When you measure an observable on a system, you collapse the wave-function of the system into a Dirac delta function in the eigenbasis for that observable. If you measure position, you get a delta function in ...

1

I might add a few commas to that Wikipedia sentence, as "A perfectly collimated beam*,* with no divergence*,* cannot..." to show informative rather than additional parameters. To answer your question about "collimated" vs. "plane wave" , consider two point sources at th plane of focus of a lens. Each point source gives off spherical waves; the lens ...

3

No, it is not possible in general, because the particles are almost surely to be entangled (since they are interacting, as mentioned). The reason is that each $\rho_i$ is nothing but the reduced density operator of the $i$ th particle. Indeed, in the form you have written, \rho_i(x_i,x_i') = \langle x_i | \rho_i | x_i' \rangle = \langle x_i | ...

2

So now I'm wondering the following: is this the definition of a wavefunction or is there a wavefunction for every observable, being this wavefuntion just a map from the possible values of the observable to their probability amplitudes? There is indeed a wavefunction for every observable. The state $|\psi\rangle$ is a vector in a complex Hilbert space. ...

3

Really, there is one wavefunction that runs over all compatible observables, and it's time evolution is governed by the Schrodinger equation with some Hamiltonian. But often times certain observables don't interact much, so you can just treat them as a system in isolation. You can imagine the Hamiltonian as being a matrix acting on some vector space, and if ...

9

Actually, what your professor isn't telling you - what we always gloss over in intro quantum mechanics for simplicity, but I might as well give it away now because everything will make so much more sense once you get this - is that kets aren't wavefunctions at all. Forget that you ever learned about wavefunctions for a minute. Kets are a form of notation ...

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There is no significance in the choice between upper- and lower-case $\psi$ (or $\Psi$) to denote a system's wavefunction. The two are used interchangeably and it is the author's discretion to use either symbol. (On the other hand, of course, one shouldn't use the two symbols interchangeably within the same text; if both are used they would refer to ...

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The most probable position would be such as where the global maximum of the distribution is located. This is different to the expectation value of a distribution, but it happens that for a Gaussian function the mean and the most probable value are the same.

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Let $I \sim \sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R} \int_{V_{UC}} d^3r \Psi^*_{n\vec{k}}\left(\vec r\right) \Psi_{n'\vec{k'}}\left(\vec r\right)$ The term $\sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R}$ gives you a $\sim \delta(\vec{k} - \vec{k'})$ term. Now, you have : \$\Psi^*_{n\vec{k}}\left(\vec r\right) ...

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