"Quantum Cheshire Cat" observed - A particle separated in space from one of its properties - Polidicks

[QUOTE=Tamschi;45544261]Not quite. The integral goes from -infinity to infinity and the vectors are expressed as functions: [img]http://latex.codecogs.com/gif.latex?\left<\psi\middle|A\middle|\phi\right>=\int_{-\infty}^{\infty}\psi^*(x)A\phi(x)dx[/img] The order of the multiplication is important, but if A is a simple scalar (which can be complex), it can be exchanged with other elements. Otherwise you have to add the commutator of the exchanged items, which is [code][noparse][A,B] = AB - BA[/noparse][/code] (if I didn't mix it up).[/QUOTE] Bear in mind it's not always -∞ to ∞ (same goes to [B]JohnnyMo[/B], too). I mean, technically any limits you pick otherwise are going to fall between -infinity and infinity, but that's not always very helpful for calculations. If you try to use the boundaries -∞ and ∞ to solve the Schrödinger equation for the infinite square well you're going to have a bad time. [editline]31st July 2014[/editline] Your commutation relationship is correct, btw.

[QUOTE=sltungle;45544302]Bear in mind it's not always -∞ to ∞ (same goes to [B]JohnnyMo[/B], too). I mean, technically any limits you pick otherwise are going to fall between -infinity and infinity, but that's not always very helpful for calculations. If you try to use the boundaries -∞ and ∞ to solve the Schrödinger equation for the infinite square well you're going to have a bad time.[/QUOTE] strictly speaking though, that's because the particle in a box is unphysical

[QUOTE=Tamschi;45544197]Yes! These logical states are in fact what quantum computing exploits. The advantage over classical computers is that the information density can be vastly higher for self-interfering wave functions. I didn't check all of it, but you got the final result of F|x[B]>[/B] right. There's a mistake where you calculate F'|-> and F|[B]-[/B]> though.[/QUOTE] Ha! Thanks! How the heck I got harder calculation right and simpler calculation wrong though... gotta redo it.

wait no I was thinking the wave function was undefined outside the box but it's just zero. It's still unphysical though because the derivative of the wave function is discontinuous. but you can still integrate over all space. You're just right that it might be difficult. Point is I'm technically correct and that is the best kind of correct.

[QUOTE=JohnnyMo1;45544368]wait no I was thinking the wave function was undefined outside the box but it's just zero. It's still unphysical though because the derivative of the wave function is discontinuous. but you can still integrate over all space. You're just right that it might be difficult. Point is I'm technically correct and that is the best kind of correct.[/QUOTE] [I]Difficult?[/I] Evaluating sin(∞) is slightly more annoying than difficult :v: Even the finite square well (which is continuous if I'm not misremembering ) has points involved that aren't infinity (thankfully the points which [I]are[/I] integrated to infinity are decaying exponentials, though!).

[QUOTE=Fourier;45544338]Ha! Thanks! How the heck I got harder calculation right and simpler calculation wrong though... gotta redo it.[/QUOTE] It's a small "sign flip" mistake... except the sign is a variable name here :v:

[QUOTE=sltungle;45544418]Even the finite square well (which is continuous if I'm not mistaken) has points involved that aren't infinity (thankfully the points which [I]are[/I] integrated to infinity are decaying exponentials, though!).[/QUOTE] Well no, you're still integrating a function from -infinity to +infinity. It's just that the rules of integration allow us to break it up. [editline]30th July 2014[/editline] "That's correct but tough to do in practice!" Sounds like [I]experimentalist[/I] talk to me. [editline]30th July 2014[/editline] [QUOTE=sltungle;45544418][I]Difficult?[/I] Evaluating sin(∞) is slightly more annoying than difficult :v:[/QUOTE] Looking back on this, I'm a little confused where you're getting sin(infinity) from.

[QUOTE=JohnnyMo1;45544449]Well no, you're still integrating a function from -infinity to +infinity. It's just that the rules of integration allow us to break it up.[/QUOTE] Yeah, I guess. I suppose it's just semantics. I'd still consider the middle part of the integral separate and you're definitely not evaluating it to infinity, but yeah, it's definitely part of a larger picture to infinity. [editline]31st July 2014[/editline] [QUOTE=JohnnyMo1;45544449]Looking back on this, I'm a little confused where you're getting sin(infinity) from.[/QUOTE] Well if you mess up and try to normalise the infinite square well between -∞ and ∞ you're going to get sin(∞) terms.

[QUOTE=sltungle;45544214]In quantum mechanics |a> would normally just be your wavefunction. You also meant 3y, I presume (typo?); you did move the 3y after all. [B]<a|a>[/B] would be (using the percentage sign as the integral): [B]% a*a dx[/B] (or dy, dz, dr, whatever). Limits of your integral are given by your boundary conditions (-L and L in an infinite square well for example). <a|a> is normalised to be equal to 1 in quantum mechanics, though (the integral of a*a gives you the probability function associated with your wavefunction and the probability of finding your particle [I]somewhere[/I] within your limits has to be 100% (or 1) obviously).[/QUOTE] Yes, I meant 3y not 3x.. sorry! Oh, I got it kind of wrong then, but I see now. In short, <a|a> is integral with limits [0,1] and it returns probability function of finding particle, which has max somewhere (100% chance of finding particle)? And just one thing, if there is 1 chance of finding particle somewhere (100%), is this <a|a> function flat everywhere else (0%) or is it negative? (Because I think there can't be 100% chance of finding something at position 1. and 50% chance somewhere of finding something at position 2. It is just not logical.) Also, don't get me wrong, why is this important? So you know where to find particle? What do you do with him then.. read his state? It is kind of mindblowing stuff, really.

[QUOTE=sltungle;45544418][I]Difficult?[/I] Evaluating sin(∞) is slightly more annoying than difficult :v: Even the finite square well (which is continuous if I'm not misremembering ) has points involved that aren't infinity (thankfully the points which [I]are[/I] integrated to infinity are decaying exponentials, though!).[/QUOTE] Physical wave functions are square integrable and normalised though, which forces them to zero at infinity. This means you never actually get sin(infinity) as term you have to calculate, outside of a convergent limit.

[QUOTE=Fourier;45544508]Yes, I meant 3y not 3x.. sorry! Oh, I got it kind of wrong then, but I see now. In short, <a|a> is integral with limits [0,1] and it returns probability function of finding particle, which has max somewhere (100% chance of finding particle)? And just one thing, if there is 1 chance of finding particle somewhere (100%), is this <a|a> function flat everywhere else (0%) or is it negative? (Because I think there can't be 100% chance of finding something at position 1. and 50% chance somewhere of finding something at position 2. It is just not logical.) Also, don't get me wrong, why is this important? So you know where to find particle? What do you do with him then.. read his state? It is kind of mindblowing stuff, really.[/QUOTE] Read over the last few posts. Technically (in any physically realisable situation) the limits are (-∞,∞). This is one of the fundamental points of quantum mechanics. Particles aren't really truly particles as we understand them in the classical limit - you can't truly know the absolute position with a particle; they have a wave-like nature to them and there's a probability associated with finding them over a certain region of space, and over the entire universe (-∞,∞) you have to find it SOMEWHERE (probability of 1). About the peak of the wave function you might have like an 80% chance of finding the particle, and then it might drop off and taper towards zero very rapidly, but there isn't any point (or region) of the wavefunction (other than the entire thing taken to infinity) where you'll be 100% likely to find the particle.

[QUOTE=sltungle;45544485]Yeah, I guess. I suppose it's just semantics. I'd still consider the middle part of the integral separate and you're definitely not evaluating it to infinity, but yeah, it's definitely part of a larger picture to infinity. [editline]31st July 2014[/editline] Well if you mess up and try to normalise the infinite square well between -∞ and ∞ you're going to get sin(∞) terms.[/QUOTE] Well if you'll permit me to be purposely pedantic for another moment, normalizing the infinite square well over all space is still what you're doing when you just integrate over the well, it just so happens the part of the function outside the well contributes nothing to the integral, so we might as well have left it out. It seems like what you mean is that you'll get sin(infty) terms if you incorrectly assume that the value of the wave function inside the well is the same as over all space.

[QUOTE=Tamschi;45544512]Physical wave functions are square integrable and normalised though, which forces them to zero at infinity. This means you never actually get sin(infinity) as term you have to calculate, outside of a convergent limit.[/QUOTE] Exactly. But that was my point - you can't evaluate the [I]INFINITE[/I] square well between -∞ and ∞. But the infinite square well isn't a truly physically realisable case (but it's a bloody good approximation in some cases). It would be a mistake to try to normalise the infinite square well between -∞ and ∞; you have to integrate between the potential barriers. [editline]31st July 2014[/editline] [QUOTE=JohnnyMo1;45544564]Well if you'll permit me to be purposely pedantic for another moment, normalizing the infinite square well over all space is still what you're doing when you just integrate over the well, it just so happens the part of the function outside the well contributes nothing to the integral, so we might as well have left it out. It seems like what you mean is that you'll get sin(infty) terms if you incorrectly assume that the value of the wave function inside the well is the same as over all space.[/QUOTE] Yeah, that's about right actually. I guess you are still integrating over infinity you just get nothing outside of the well. I suppose that's misplaced limits to try and put the infinities around the sin terms.

[QUOTE=Fourier;45544508]Yes, I meant 3y not 3x.. sorry! Oh, I got it kind of wrong then, but I see now. In short, <a|a> is integral with limits [0,1] and it returns probability function of finding particle, which has max somewhere (100% chance of finding particle)? And just one thing, if there is 1 chance of finding particle somewhere (100%), is this <a|a> function flat everywhere else (0%) or is it negative? (Because I think there can't be 100% chance of finding something at position 1. and 50% chance somewhere of finding something at position 2. It is just not logical.) Also, don't get me wrong, why is this important? So you know where to find particle? What do you do with him then.. read his state? It is kind of mindblowing stuff, really.[/QUOTE] Yes, that's correct. However, the infinite square well is not physical. A physical one (that is: one with finite depth), like the area an electron in a (very simple) atom is in, always has some stray probability outside of it. This stray probability is what causes quantum tunneling, the "passing through" insurmountable potential barriers (with a very low likelihood).

[QUOTE=Tamschi;45544261]Not quite. The integral goes from -infinity to infinity and the vectors are expressed as functions: [url]http://latex.codecogs.com/gif.latex?\left<\psi\middle|A\middle|\phi\right>=\int_{-\infty}^{\infty}\psi^*(x)A\phi(x)dx[/url] The x that's integrated over here would be for example the position. The bra (<psi|) part could be the shape of the measurement, so for example a sharp area with value 1 around the origin, zero everywhere else. The ket (A|phi>) part describes what is measured, and assigns a probability to every position. Now when you calculate the integral, all that's left is the probability of A|phi> giving a result that is in the test area <psi|. You can chain these together as much as you want by emitting a state on the left again (by prepending a ket), and as long as the system doesn't decohere your quantum computer behaves like it should, with terms in each integral correctly. However, if the system decoheres the integral breaks in two and you get the sum of two integrals instead of the integral over a sum. (Not the same with complex numbers!) Avoiding this is very difficult in practise, which is why we don't have good quantum computers yet. The order of the multiplication is important, but if A is a simple scalar (which can be complex), it can be exchanged with other elements. Otherwise you have to add the commutator of the exchanged items, which is [code][noparse][A,B] = AB - BA[/noparse][/code] (if I didn't mix it up).[/QUOTE] [code] <ψ|A|Φ> [/code] I kind of get it now. It ain't so complex at all. (just this formula, not all quantum physics) So, bra (<ψ|) is like pulse function sweeping over ket (|Φ>), we integrate everything and get result of position, (where <ψ|A|Φ> = 1); x being the position? [editline]30th July 2014[/editline] Guys, sorry for not answering to you all, everytime I answer to something, like 3 news answers pop up with new information, and it is kind of hard to process everythig. [editline]30th July 2014[/editline] [QUOTE=sltungle;45544551]Read over the last few posts. Technically (in any physically realisable situation) the limits are (-∞,∞). This is one of the fundamental points of quantum mechanics. Particles aren't really truly particles as we understand them in the classical limit - you can't truly know the absolute position with a particle; they have a wave-like nature to them and there's a probability associated with finding them over a certain region of space, and over the entire universe (-∞,∞) you have to find it SOMEWHERE (probability of 1). About the peak of the wave function you might have like an 80% chance of finding the particle, and then it might drop off and taper towards zero very rapidly, but there isn't any point (or region) of the wavefunction (other than the entire thing taken to infinity) where you'll be 100% likely to find the particle.[/QUOTE] Damn waves. This looks more and more like topic of digital signal processing than anything else :D Also yes, I kind got of that there can't be 100% of finding particle, it just seemed too simple. Cool stuff man. Just one more thing and I will stop asking. I am programmer. Is this stuff I just learned about (<+| and |->, <ψ|A|Φ>, etc ) useful in designing quantum algorithms? Or is this really another topic? Just wondering, might be useful in future.

[QUOTE=Fourier;45544613][code] <ψ|A|Φ> [/code] I kind of get it now. It ain't so complex at all. (just this formula, not all quantum physics) So, bra (<ψ|) is like pulse function sweeping over ket (|Φ>), we integrate everything and get result of position, (where <ψ|A|Φ> = 1); x being the position? [...][/QUOTE] The result you get here is a probability for the measurement to be positive, the <ψ| is a mask selecting a certain area. If you want to get the expected position, you have to determine the expectation value of the position operator on let's say the particle |ψ> with the wave funtion ψ(x): [code](This is in position space! There are other representations like momentum wave functions where you need to apply a Fourier transform first.) <x> = <ψ|X|ψ> = ∫ψ*(x)xψ(x)dx = ∫x||ψ(x)||²dx[/code] <ψ(x)> is the square magnitude of the wave function at x, which is the probability to find the particle exactly there. ([del]I'm not too sure whether to use <> or || here for the norm, no idea which is more commonly used in QM.[/del] [del]I think <> is the squared magnitude and || is for the root of that.[/del] No, that wasn't it either :v:) If you know ψ(x) you can (usually) calculate this precisely, but there is also uncertainty for that result which I don't know the formula for right now. [editline]30th July 2014[/editline] [QUOTE=Fourier;45544613][...] Guys, sorry for not answering to you all, everytime I answer to something, like 3 news answers pop up with new information, and it is kind of hard to process everythig. [...][/QUOTE] Take your time, it's normal to have difficulties wrapping one's head around this at first. (I know I did :v:) [editline]30th July 2014[/editline] [QUOTE=Fourier;45544613][...] Damn waves. This looks more and more like topic of digital signal processing than anything else :D Also yes, I kind got of that there can't be 100% of finding particle, it just seemed too simple. Cool stuff man. Just one more thing and I will stop asking. I am programmer. Is this stuff I just learned about (<+| and |->, <ψ|A|Φ>, etc ) useful in designing quantum algorithms? Or is this really another topic? Just wondering, might be useful in future.[/QUOTE] You can use this kind of notation to describe quantum algorithms, yes. It just gets a bit unwieldy at a certain size which is why these circuit diagrams exist: [thumb]http://royalsocietypublishing.org/content/roypta/370/1979/5270/F4.large.jpg[/thumb] (At least I think that's one, it came up in the Google search and looks somewhat like the ones I saw. The boxes are operators, the lines distinct states.) QM is more like analog signal processing I think. You can use the same kind of maths to construct audio filters for example. DSP comes in when you have square wells and the like I suppose. Also: If you have further questions, please continue to ask! I'm glad if I can help and explaining it also makes it easier and faster for me to apply this stuff.

[QUOTE=Tamschi;45544758]The result you get here is a probability for the measurement to be positive, the <ψ| is a mask selecting a certain area. If you want to get the expected position, you have to determine the expectation value of the position operator on let's say the particle |ψ> with the wave funtion ψ(x): [code](This is in position space! There are other representations like momentum wave functions where you need to apply a Fourier transformation first.) <x> = <ψ|X|ψ> = ∫ψ*(x)xψ(x)dx = ∫x|ψ(x)|dx[/code] |ψ(x)| is the square magnitude of the wave function at x, which is the probability to find the particle exactly there. (I'm not too sure whether to use <> or || here for the norm, no idea which is more commonly used in QM.) If you know ψ(x) you can (usually) calculate this precisely, but there is also uncertainty for that result which I don't know the formula for right now. [editline]30th July 2014[/editline] Take your time, it's normal to have difficulties wrapping one's head around this at first. (I know I did :v:) [editline]30th July 2014[/editline] You can use this kind of notation to describe quantum algorithms, yes. It just gets a bit unwieldy at a certain size which is why these circuit diagrams exist: [thumb]http://royalsocietypublishing.org/content/roypta/370/1979/5270/F4.large.jpg[/thumb] (At least I think that's one, it came up in the Google search and looks somewhat like the ones I saw. The boxes are operators, the lines distinct states.) QM is more like analog signal processing I think. You can use the same kind of maths to construct audio filters for example. DSP comes in when you have square wells and the like I suppose. Also: If you have further questions, please continue to ask! I'm glad if I can help and explaining it also makes it easier and faster for me to apply this stuff.[/QUOTE] I am kind of familiar with Fourier transformation So in essence, position space is time space, where as momentum wave functions space is frequency space/domain? Also rest of your post, sorry, my head became kind of slow (lame excuse lol) and I need work to do, I will check later this and comment on it... Haha, saw block diagram when I answered, I suppose you were editing while I was answering.

[QUOTE=Fourier;45544989]I am kind of familiar with Fourier transformation So in essence, position space is time space, where as momentum wave functions space is frequency space/domain? Also rest of your post, sorry, my head became kind of slow (lame excuse lol) and I need work to do, I will check later this and comment on it... Haha, saw block diagram when I answered, I suppose you were editing while I was answering.[/QUOTE] I also made one more edit afterwards to that formula, didn't get the notation quite right. You're absolutely correct about the relation between position and momentum space, but this also applies in reverse since the transform is (almost) self-inverse. This is also the exact reason why position and momentum can't be measured exactly at the same time: If you have an exact value in one domain (represented by the Dirac delta), you get a wavelet in the other one. [editline]30th July 2014[/editline] Seems like I can't edit it any more because of that raised 2... It should say: ||ψ(x)||^2 is the square magnitude of the wave function at x, [...] The reason I could use [code]ψ*(x)xψ(x) = xψ*(x)ψ(x) = x||ψ(x)||^2[/code] there is because the x inside the integral is a scalar and as such can be exchanged with anything.

We're still on topic?

[QUOTE=nerdster409;45545171]We're still on topic?[/QUOTE] I don't even know what they're talking about to tell if it is or not.

Found an article that explains, what we were talking about: [url]http://www.lecture-notes.co.uk/susskind/quantum-entanglements/lecture-2/quantum-states/[/url] Don't know, if it has connection to the OP, but I pressume there is big connection.

[QUOTE=DowntownTiger;45539318]so... what are the practical effects of this?[/QUOTE] There's always someone asking this in every single fucking thread about scientific findings. I'm sure there's a specific circle in hell for you.

[QUOTE=paul simon;45549150]There's always someone asking this in every single fucking thread about scientific findings. I'm sure there's a specific circle in hell for you.[/QUOTE] I think that is pretty legitimate question, I too am interested, what does this finding enables / what doors opens for us?

read about this in NewScientist, shits pretty cash, but i'm not entirely too sure on what doors it opens up for us research wise

Here's a reddit post about how this is much less interesting than the article says, just to crush some dreams: [url]http://www.reddit.com/r/Physics/comments/2c2k1o/quantum_cheshire_cat_observed/cjbju7y[/url]

[QUOTE=Fourier;45551523]I think that is pretty legitimate question, I too am interested, what does this finding enables / what doors opens for us?[/QUOTE] A bit above your post on the first page, in reply to someone else asking the same question: [QUOTE=JohnnyMo1;45539422][...] woah lookie who didn't read the article [quote]Their delicate apparatus could have useful applications in high-precision metrology, the researchers say. "For example, one could imagine a situation in which the magnetic moment of a particle overshadows another of the particle's properties which one wants to measure very precisely. "The Cheshire Cat effect might lead to a technology which allows one to separate the unwanted magnetic moment to a region where it causes no disturbance to the high-precision measurement of the other property."[/quote][/QUOTE] [editline]31st July 2014[/editline] [QUOTE=JohnnyMo1;45552747]Here's a reddit post about how this is much less interesting than the article says, just to crush some dreams: [url]http://www.reddit.com/r/Physics/comments/2c2k1o/quantum_cheshire_cat_observed/cjbju7y[/url][/QUOTE] So to verify the validity someone has to actually do the weak measurements at the same time. I really wonder why they didn't do that, is there some prohibitively high cost do the set-up to be replicated, or is the machine too large and the separation too small to be feasible?