# Tag Info

1 vote

### How do operators on kets and wavefunctions correspond?

Is it by showing that this holds for position and momentum operators, and then have the result follow from any observable having to be a function of these? Yes, naturally. Your text or instructor ...
• 64.2k
1 vote
Accepted

### What is the mathematical support for the formula $f_n = n f_1$, used to calculate the frequency of a standing wave?

Let's start with a quick answer: the boundary conditions fix the frequencies of the harmonics. It's worth emphasizing that boundary conditions come from physics, not mathematics. Below, I'll consider ...
• 51.2k

### Negative kinetic energy on a step potential

Negative kinetic energy is absurd, right? What's wrong with this calculation? Kinetic energy value we assign to the particle, based on measurement or the psi function, cannot be negative. When you ...
• 39k
Accepted

### Negative kinetic energy on a step potential

I'm having trouble with the explanation of the kinetic energy on the classically forbbiden region on a step potential ($V=0$ for $x<0$, $V=V_0$ for $x>0$ and $E<V_0$). ... On the classically ...
• 21.9k

### Negative kinetic energy on a step potential

Expectation values obey classical rules (Ehrenfest's theorem) and to be in such a region where the potential energy is positive and higher than your total energy you would need to have a negative ...
• 978
1 vote

A simple answer which slightly expands on the other (excellent) answers here, coming at it from an information-theory perspective: The Kolmogorov complexity of Graham's number is significantly smaller ...
• 19

Bekenstein Bound The Numberphile mathematicians were clearly describing an explicit representation of the bits in Graham's number, rather than any symbolic one (obviously, just writing "G" ...
• 2,602
1 vote

Without watching the video and going with your headline, my first guess would have been to use some black hole thermodynamics. If we trust a well known internet black hole calculator, then a 1.6e-35m ...
• 3,075
Accepted

Simply put, what's stated in the video is misleading to an extent that it's fair to say it's wrong. Ignoring problems of how you define the word 'imagining' (this is physics stack exchange, after all),...
• 3,645
Accepted

### Why is the tensor product not a multilinear application?

The tensor product map $$\otimes : V_1\times \cdots\times V_n \to V_1\otimes \cdots \otimes V_n$$ is multi linear as the codomain is a linear space and the map is separately linear in each entry. ...
• 73.8k
Dicing with infinities is usually dangerous. Start with the idea that a small part of the object, size o$\times$o, produces an image of size i$\times$i. Now let o become smaller and smaller. What ...