Two-fold question about wave function

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When an atom travels from point A to B and interacts with other particles, its wave function collapses, indicating it behaves like a particle in our classical world. However, quantum objects inherently possess both wave-like and particle-like characteristics, with the observed behavior depending on the interaction context. Wave function collapse does not eliminate the wave nature; it merely alters the wave function in accordance with the measurement. The uncertainty principle applies to all objects, regardless of size, because everything is fundamentally quantum. To grasp these concepts fully, a solid understanding of quantum mechanics mathematics, including integrals and vectors, is essential.
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So let's say I shoot an atom from A to B. If it interacts with other atoms, electrons or photon's, the wave function collapses right? And the particle exists in our 'normal' world, no quantum tunneling etc. And it looks as if it was a normal particle all along right?

But let's say it goes from A to B. It would go through the slit in the double slit experiment. And it interacts with a photon. Will it spread out in a wave function again if it does not interact with anything between the interaction and the screen?

And what causes a particle or photon to behave as a wave? What conditions have to be present? I know what makes a wave function collapse, but what conditions have to be met to create a probability wave function in an electron/atom etc.

TIA
 
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jimmylegss said:
So let's say I shoot an atom from A to B. If it interacts with other atoms, electrons or photon's, the wave function collapses right? And the particle exists in our 'normal' world, no quantum tunneling etc. And it looks as if it was a normal particle all along right?

That's not really right, although you'll see it described that way in books that are trying to spare you from the math.

A quantum object always has some wave-like characteristics and some particle-like characteristics; which one you see more of depends on the details of the interaction you're observing. (Note that I am not saying that the particle is both a wave and a particle - it is neither, it is a quantum object with some characteristics of both. You wouldn't say that a sheep is both a table and pillow and discuss sheep in terms of "table-pillow duality" because a sheep has four legs like a table and is fuzzy like a pillow, so you shouldn't say that a quantum object is both a wave and a particle and talk about "wave-particle duality" either).

Wave function collapse doesn't eliminate the waviness, even for a moment. It just changes the wave function in a particular way, one that is consistent with the measurement that caused the collapse. The wave function continues evolving from there according to Schrodinger's equation.

(Also, you should be aware that although wave function collapse is a common way of describing what the mathematical formalism of QM is telling us, it is not actually part of that formalism. It is possible to do QM without ever thinking in terms of wave function collapse, and indeed that's the easiest way of attacking some QM problems).
 
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I supose I should really start learning all the math behind it then.

But as you say, when something is measured, it basicly still is a wave function? Except we call it a particle because it is a very narrow wave? Is it still the same wave?

Is that the reason that the uncertainty principle still counts for large objects?

And if i want to understand this properly I probably have to understand the whole integral vector thing really well? And trigonometry as well?
 
jimmylegss said:
But as you say, when something is measured, it basicly still is a wave function?

Everything is a wave function because everything is quantum - but its better to call it a quantum state rather than wavefunction - the distinction will become clearer when you learn the detail..

jimmylegss said:
Except we call it a particle because it is a very narrow wave? Is it still the same wave?

If it wasn't destroyed by the observation, the observation is what is known as a filtering type observation which is a state preparation procedure. You have prepared the particle in a different state. If it was a position measurement then you have prepared the particle is a state of definite position etc. These days state and state preparation procedure are pretty much synonymous. Whether such a view resolved anything is another mater - but it certainly is a lot clearer.

jimmylegss said:
Is that the reason that the uncertainty principle still counts for large objects?

It counts for all objects, large or small, it doesn't matter, because everything is quantum. We simply don't notice it here in the macro world.

jimmylegss said:
And if i want to understand this properly I probably have to understand the whole integral vector thing really well? And trigonometry as well?

I suggest 3 books:
https://www.amazon.com/dp/0471827223/?tag=pfamazon01-20
https://www.amazon.com/dp/0465075681/?tag=pfamazon01-20
https://www.amazon.com/dp/0465036678/?tag=pfamazon01-20

There are also associated video lectures:
http://theoreticalminimum.com/

Thanks
Bill
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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