B How does quantum superposition really work?

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Quantum superposition involves expressing quantum states as linear combinations of basis states, similar to how vectors can be decomposed into components. This concept is not magical; it is a mathematical representation that allows for the analysis of quantum systems. The choice of basis affects how states are expressed, but the underlying quantum state remains the same regardless of representation. For practical applications, such as quantum computing, selecting an appropriate basis simplifies calculations and predictions. Understanding superposition is crucial, but it is only one aspect of the broader complexities of quantum mechanics and quantum computing.
Curious Cat
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I understand how to put a quantum particle into a superposition, by
passing it thru 2 slits or a beam splitter, which U can say actually
splits the wave, function, but
How does putting it potentially inbetween 2 basis states, which U
will measure only later, on, do it!? That seems like magic, to me. And
yet I accept that it does. Somehow.
 
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Have you ever studied vectors? E.g. although gravity is a vertical force, we can decompose it into components in any basis. E.g. when analysing motion along an inclined plane we decompose it into components normal to and tangential to the plane.

That's superposition. Not magic!
 
PeroK said:
Have you ever studied vectors? E.g. although gravity is a vertical force, we can decompose it into components in any basis. E.g. when analysing motion along an inclined plane we decompose it into components normal to and tangential to the plane.

That's superposition. Not magic!
Thank you, for your reply:
Well, yes, of course.
Are you saying that that's all QSP is!?
Then how does that explain the quantum computers?
 
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Curious Cat said:
Thank you for your reply:
. Well, yes, of course.
. Are you saying that that's all QSP is!?
Then how does that explain the quantum computers?
That's all superposition is: quantum States are (abstract) vectors and may be expressed in any basis - which means an infinite choice of superposition for any state.

It takes more than the concept of superposition to explain QM and quantum computers.
 
Curious Cat said:
How does putting it potentially inbetween 2 basis states, which U
will measure only later, on, do it!?
As @PeroK says above, states add like vectors so any state can be written as the sum (more precisely, a linear combination) of other states. Typically when we start in on a problem we choose a set of vectors that we call the basis, and then we write everything else as a superposition of these basis vectors; but we could choose a different basis and then some of the states that were written as superpositions before might no longer be.

For example: Let's call the photon state that we informally describe as "polarized at a 45-degree angle clockwise" ##|45\rangle##; there's a state that is orthogonal to that one, we'll call it ##|-45\rangle##. If we choose to use these two vectors as our basis then we would write the state of a vertically polarized photon as the superposition ##\frac{\sqrt{2}}{2}(|45\rangle+|-45\rangle)##. However, we could also choose to use the states "polarized vertically" and "polarized horizontally" as our basis (let's call them ##|V\rangle## and ##|H\rangle##); now the state of a vertically polarized photon would be written as simply ##|V\rangle## - no superposition. But it's the same state either way, and there's no more magic involved than in noting that we can write "5" or "3+2" and it's the same number either way.

In practice we generally try to choose a basis that make the problem at hand easy. If we're planning to send a photon through a vertical polarizer at some stage of our experiment we'll probably want to use the ##|H\rangle##,##|V\rangle## basis just because we can read the probability amplitude for passing through the filter directly from the state. In this basis the state will look like ##\alpha|V\rangle+\beta|H\rangle## and the probability amplitude will be ##\alpha##. If instead we chose to use the ##|45\rangle##,##|-45\rangle## basis we'd have to do some extra algebra to get the same result.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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