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

[Bradyns]

[QUOTE=DowntownTiger;45540484]doesnt sound very practical[/QUOTE]

What even is pure research? :downs:

[pkhzor]

[QUOTE=Bradyns;45541165]What even is pure research? :downs:[/QUOTE]

Seems like his brain was separated from his body when he was making that post.

[Zeke129]

[QUOTE=Fourier;45539858][IMG]http://i.imgur.com/zb3x60y.png[/IMG][/QUOTE]

Scientists are just making shit up now to see if we fall for it

last I checked the absolute value of positive greater than less than positive minus the absolute value of negative greater than less than negative doesn't equal oz

[Cl0cK]

[QUOTE=Zeke129;45541232]Scientists are just making shit up now to see if we fall for it

last I checked the absolute value of positive greater than less than positive minus the absolute value of negative greater than less than negative doesn't equal oz[/QUOTE]

Oz is a tv series set in prison, those science nerds just spoiled the show with arrows and shit!

[LoganIsAwesome]

[B][I]you fools never look outside the box[/I][/B]

[B][I][U]we can transport things now [/U][/I][/B]

[Adius]

I'm just in awe by what humanity is discovering everyday. Better understanding the physical plane of existence we find ourselves in. It's like reverse engineering a super complicated game engine, stretching the limitations of the physics engine, digging deeper into the source code that defines the universe.

[Cheshire_cat]

With my help, science can explain a complex quantum phenomenon! Rejoice!

[Awesomecaek]

lets separate mass from atoms, then we can obviously send them elsewhere faster than speed of light

not sure what to do with them once they get there nor what to do with the mass left over, but I am sure somebody gonna figure that out

[Vitamine D]

if you could do this repeatedly on like a spaceship every few seconds or so, couldnt you effectively half (or fully reduce) the weight of it?

[Kondor]

[QUOTE=DowntownTiger;45539318]so... what are the practical effects of this?[/QUOTE]

i'm sure people were saying the same thing when they found out about atomic decay

[Cabbage]

if we can separate the grin from the cheshire cat

can we separate the ban from that cat

[Awesomecaek]

[QUOTE=Kondor;45542114]i'm sure people were saying the same thing when they found out about atomic decay[/QUOTE]

Eh, probably not, at first they started briefly losing their shit because it was finally possible to "make gold". Also, it was immediately obvious how massive amounts of energy are released through atomic decay.

But for example electricity was seen as at best a funny party trick for a quite a while.

[elix]

[QUOTE=Awesomecaek;45542152]But for example electricity was seen as at best a funny party trick for a quite a while.[/QUOTE]

[URL="http://en.wikipedia.org/wiki/Aeolipile"]The ancient Greeks invented the first steam engine[/URL], but it was just a fun toy to them. They didn't realize it could be harnessed to do work. Took humanity 17 centuries to work it out.

The railway had already been around for five hundred years but horses pulled the load.

[Awesomecaek]

[QUOTE=elixwhitetail;45542384][URL="http://en.wikipedia.org/wiki/Aeolipile"]The ancient Greeks invented the first steam engine[/URL], but it was just a fun toy to them. They didn't realize it could be harnessed to do work. Took humanity 17 centuries to work it out.

The railway had already been around for five hundred years but horses pulled the load.[/QUOTE]

To be fair, Greeks probably weren't capable of the metallurgy necessary to make a steam machine feasible. Even if they invented the actual piston steam engine (which is necessary to make steam power worthwhile, they only toyed with reactive propulsion), all the stuff would be horribly unreliable and high maintenance. They weren't capable of creating strong and light enough materials, they couldn't machine the high stress parts, they couldn't build a high pressure boiler.

[TheDecryptor]

[QUOTE=LoganIsAwesome;45541345][B][I]you fools never look outside the box[/I][/B]

[B][I][U]we can transport things now [/U][/I][/B][/QUOTE]

Yes, by putting said things back into the box, and then moving the box to a new location.

[Tamschi]

[QUOTE=LoneWolf_Recon;45539818]Fission with a selectable decay mode?[/QUOTE]

Separating something along a part of the path doesn't automatically mean you can meaningfully change it.

If I'm not mistaken weak interactions here are the same as weak measurements, so they don't disentangle but also can't force a select state.
Even strong measurements or direct manipulation wouldn't work here, the former can't choose the outcome and the latter just discards the entangled state and replaces it with a fully local one.

(I haven't read the paper though, going to do that in a bit.)

[editline]30th July 2014[/editline]

[QUOTE=Fourier;45539858][IMG]http://i.imgur.com/zb3x60y.png[/IMG]

Please, explain.

I want to know, what are those weird closures > and <

Also I literary have no idea, what really this means but I think it is good since there is possible deep manipulation of matter if we can disable properties...?[/QUOTE]

[QUOTE=JohnnyMo1;45540083]As LoneWolf_Recon said, the |+> is a quantum state. <+| is the dual vector representation of the same state. Because of the way the states are set up, sigma_z is is an operator (having state vectors face each other in that |><| configuration allows them to act on other state vectors).[/QUOTE]

I'm going to expand on this a bit because I think not everyone here has a firm grasp on linear algebra in Hilbert space :wink:
(Actually I don't really either, but I know enough of the basics to explain this stuff in a way that most here should understand.)

As in normal geometry, if you multiply a vector by a normalized one, you get the product of the individual lengths times the cosine of the angle between the vectors as scalar (plain number).
The only difference here is that the vectors all contain possibly complex numbers, and that as such you have to multiply the dual (complex conjugate = same vector but with inverted complex part) to get the same result.

I assume |+> and |-> are orthogonal states here, so the following equations apply:

(<x||y> is written as <x|y>, likely because the repeated | is unnecessary effort.)

<+|+> = 1
<-|-> = 1
<+|-> = 0
<-|+> = 0

Other states in the resulting vector space can be represented by a sum:

|x> := 2|+> + 3|->
<+|x> = <+|2|+> + <+|3|-> = 2<+|+> + 3<+|-> = 2 * 1 + 3 * 0 = 2
<-|x> = <-|2|+> + <-|3|-> = 2<-|+> + 3<-|-> = 2 * 0 + 3 * 1 = 3

(I'm aware that these would usually be fractions adding to 1, I'm just using 2 and 3 here for simplicity. The same applies to the operators, normally they'd not result in larger vectors than the input.)

With that you can make projection operators, which filter vectors into their components:

A := |+><+|
B := |-><-|

A|+> = |+><+|+> = |+>1 = |+>
A|-> = |+><+|-> = |+>0 = 0 // The last 0 here is the null vector, I could have written it as |0> if I assigned that beforehand but just writing a plain 0 is more common I think.

B|+> = |-><-|+> = |->0 = 0
B|-> = |-><-|-> = |->1 = |->

|x> := 2|+> + 3|->
A|x> = |+><+|x> = |+>2 = 2|+>
B|x> = |-><-|x> = |->3 = 3|->

You can add and subtract operators too:

C := A + B = |+><+| + |-><-|
D := A - B = |+><+| - |-><-| // This is sigma_z.

|x> := 2|+> + 3|->

C|+> = A|+> + B|+> = |+> + 0 = |+>
C|-> = A|-> + B|-> = 0 + |-> = |->
C|x> = A|x> + B|x> = 2|+> + 3|-> = |x>
// As you have probably seen by now, C is the identity operator here.
// The above is not a general proof of that fact though.

D|+> = A|+> - B|+> = |+> - 0 = |+>
D|-> = A|-> - B|-> = 0 - |-> = -|->
D|x> = A|x> - B|x> = 2|+> - 3|->
// D behaves a bit like C, but it flips the sign on the |-> component.

Each operator usually represents a device in the experiment or the time and space in-between.
Operators can also "rotate" one direction onto another one, like these do:

E := |+><-| + |-><+|
F := 1/2 (|+><+| - |-><-|) + |+><-|

As an exercise, you can calculate yourself what these do when applied to |+>, |-> and |x>.
(In part because you'll remember it better and in part because it's annoying to type them out :eng101:)

The examples here are fairly simple, since |+> and |-> are linearly independent.
If they weren't we would get additional fractions in the results, or, in the case of more complicated relations, really nasty terms that may or may not be impossible to simplify.

That's pretty much where the problems come in: Most vectors here have actually infinitely many dimensions representing the probability to get a certain value and particles (like an electron or a fridge) are described by wave packets that spread out over time.
Anything as heavy/large as a grain of sand takes an incredibly long time to spread out, but for elementary particles it takes just a fraction of a second to double their "width".

To get actually sensible expectation values, you have to integrate over all the probabilities, which can be fairly difficult depending on the equation you're looking at.
Many of them are in fact so complicated that it becomes impossible to solve them by hand, or even with a computer. You can still approximate the solutions though.

[B](Someone who knows more about this than me please read over this really quick. It's likely that there's at least some inaccuracy.)[/B]

[sltungle]

Double vertical lines aren't necessary in Bra-Ket notation as far as I've encountered. It probably IS a result of Dirac being a lazy physicist, but... yeah; not necessary.

You should probably also have mentioned in your second little math (or code) box that if the Bra doesn't operate on what's between it and the Ket, and what's inbetween the Bra and the Ket doesn't operate on the Ket, then you can treat it as a constant and drag it out the front. That's the reason why you can make <+|1|+> = 1, because:

<+|1|+> = 1 <+|+>

<a|a> = 1

Therefore 1 <+|+> = 1 * 1 = 1.

[Fourier]

[QUOTE=LoneWolf_Recon;45539884]Also, holy shit, does this mean we could bypass Heisenberg's principle?

[editline]29th July 2014[/editline]

State vectors of a particle(s), IIRC, the + and - are a declaration of magnetic moments. There's also up and down arrows which also represent spin states.[/QUOTE]

Thanks for explanation. I kind of know what you are talking about but still I am lost more than monkey on acid trip.

[editline]30th July 2014[/editline]

[QUOTE=JohnnyMo1;45540083]As LoneWolf_Recon said, the |+> is a quantum state. <+| is the dual vector representation of the same state. Because of the way the states are set up, sigma_z is is an operator (having state vectors face each other in that |><| configuration allows them to act on other state vectors).[/QUOTE]

Thank you too for explanation. I need to learn physics more...

[sltungle]

[QUOTE=Fourier;45543584]Thanks for explanation. I kind of know what you are talking about but still I am lost more than monkey on acid trip.

[editline]30th July 2014[/editline]

Thank you too for explanation. I need to learn physics more...[/QUOTE]

<a|a> is an integral over a*a (where a* is the Hermitian conjugate of a - the complex conjugate transposed; it's not a simple multiplication).

If you stick an operator in the middle and get, say, <a|b|a>, and if b can be treated as a constant in the integral, i.e. it's not acted on by whatever is left of it, and it doesn't act on whatever is right of it, then in Bra-Ket notation you can simply drag the b out the front and be left with b<a|a>.

In other words imagine a as being a function in x, a(x) and b is a function in y, b(y), then if you had the integral of

a*(x)b(y)a(x) dx

(where, again, * denotes the Hermitian conjugate) you could just drag b(y) out of the front of the integral because it's not dependent on x.

That's not a very mathematical explanation, but hopefully it gets the point across.

[LoneWolf_Recon]

[QUOTE=Tamschi;45542713]Separating something along a part of the path doesn't automatically mean you can meaningfully change it.

If I'm not mistaken weak interactions here are the same as weak measurements, so they don't disentangle but also can't force a select state.
Even strong measurements or direct manipulation wouldn't work here, the former can't choose the outcome and the latter just discards the entangled state and replaces it with a fully local one.

(I haven't read the paper though, going to do that in a bit.)

[...][/QUOTE]

Ah, I see so, in that case we could simply measure the potential energy of a certain atom, and then predict what it's decay mode, when-it-will-decay, etc more precisely? So this could help oodles with new isotopes, or new elements?

(Also thank you for the fantastic explanation, I'm actually taking linear algebra this semester at my Uni
. But I'm thankful for a more meaningful application/explanation early on because I have a semi-basic understanding apart from basic complex space)

[Fourier]

[QUOTE=Tamschi;45542713]Separating something along a part of the path doesn't automatically mean you can meaningfully change it.

If I'm not mistaken weak interactions here are the same as weak measurements, so they don't disentangle but also can't force a select state.
Even strong measurements or direct manipulation wouldn't work here, the former can't choose the outcome and the latter just discards the entangled state and replaces it with a fully local one.

(I haven't read the paper though, going to do that in a bit.)

[editline]30th July 2014[/editline]

I'm going to expand on this a bit because I think not everyone here has a firm grasp on linear algebra in Hilbert space :wink:
(Actually I don't really either, but I know enough of the basics to explain this stuff in a way that most here should understand.)

As in normal geometry, if you multiply a vector by a normalized one, you get the product of the individual lengths times the cosine of the angle between the vectors as scalar (plain number).
The only difference here is that the vectors all contain possibly complex numbers, and that as such you have to multiply the dual (complex conjugate = same vector but with inverted complex part) to get the same result.

I assume |+> and |-> are orthogonal states here, so the following equations apply:

(<x||y> is written as <x|y>, likely because the repeated | is unnecessary effort.)

<+|+> = 1
<-|-> = 1
<+|-> = 0
<-|+> = 0

Other states in the resulting vector space can be represented by a sum:

|x> := 2|+> + 3|->
<+|x> = <+|2|+> + <+|3|-> = 2<+|+> + 3<+|-> = 2 * 1 + 3 * 0 = 2
<-|x> = <-|2|+> + <-|3|-> = 2<-|+> + 3<-|-> = 2 * 0 + 3 * 1 = 3

(I'm aware that these would usually be fractions adding to 1, I'm just using 2 and 3 here for simplicity. The same applies to the operators, normally they'd not result in larger vectors than the input.)

With that you can make projection operators, which filter vectors into their components:

A := |+><+|
B := |-><-|

A|+> = |+><+|+> = |+>1 = |+>
A|-> = |+><+|-> = |+>0 = 0 // The last 0 here is the null vector, I could have written it as |0> if I assigned that beforehand but just writing a plain 0 is more common I think.

B|+> = |-><-|+> = |->0 = 0
B|-> = |-><-|-> = |->1 = |->

|x> := 2|+> + 3|->
A|x> = |+><+|x> = |+>2 = 2|+>
B|x> = |-><-|x> = |->3 = 3|->

You can add and subtract operators too:

C := A + B = |+><+| + |-><-|
D := A - B = |+><+| - |-><-| // This is sigma_z.

|x> := 2|+> + 3|->

C|+> = A|+> + B|+> = |+> + 0 = |+>
C|-> = A|-> + B|-> = 0 + |-> = |->
C|x> = A|x> + B|x> = 2|+> + 3|-> = |x>
// As you have probably seen by now, C is the identity operator here.
// The above is not a general proof of that fact though.

D|+> = A|+> - B|+> = |+> - 0 = |+>
D|-> = A|-> - B|-> = 0 - |-> = -|->
D|x> = A|x> - B|x> = 2|+> - 3|->
// D behaves a bit like C, but it flips the sign on the |-> component.

Each operator usually represents a device in the experiment or the time and space in-between.
Operators can also "rotate" one direction onto another one, like these do:

E := |+><-| + |-><+|
F := 1/2 (|+><+| - |-><-|) + |+><-|

As an exercise, you can calculate yourself what these do when applied to |+>, |-> and |x>.
(In part because you'll remember it better and in part because it's annoying to type them out :eng101:)

The examples here are fairly simple, since |+> and |-> are linearly independent.
If they weren't we would get additional fractions in the results, or, in the case of more complicated relations, really nasty terms that may or may not be impossible to simplify.

That's pretty much where the problems come in: Most vectors here have actually infinitely many dimensions representing the probability to get a certain value and particles (like an electron or a fridge) are described by wave packets that spread out over time.
Anything as heavy/large as a grain of sand takes an incredibly long time to spread out, but for elementary particles it takes just a fraction of a second to double their "width".

To get actually sensible expectation values, you have to integrate over all the probabilities, which can be fairly difficult depending on the equation you're looking at.
Many of them are in fact so complicated that it becomes impossible to solve them by hand, or even with a computer. You can still approximate the solutions though.

[B](Someone who knows more about this than me please read over this really quick. It's likely that there's at least some inaccuracy.)[/B][/QUOTE]

Actually, this is quite good explanation!

Looks like some sort of boolean operators to me which carry value with it's logical state.. :)

Also lil excercise just to confirm my understanding :)

E := |+><-| + |-><+|

E|+> = |+><-|+> + |-><+|+> = |+> * 0 + |-> * 1 =  |->
E|->  = |+><-|-> + |-><+|-> = |+> * 1 + |-> * 0 = |+>

|x> := 2|+> + 3|->
E|x> = |+><-|x> + |-><+|x> = (|+><-|3|->) + (|-><+|2|+>) = 3|+> + 2|->

I converted here F to F' for easier calculation .. might have made more complications than necessary..

F := 1/2 (|+><+| - |-><-|) + |+><-|
F' = 2F = (|+><+| - |-><-|) + 2(|+><-|)

F'|+> = (|+><+| - |-><-|) +> +  2|+><-|+> = (|+><+|+> - |-><-|+>) + 2|+><-|+> = (|+>) + 0 = |+>
F|+> = (F'|+>) / 2 = (1/2) |+>

F'|-> = (|+><+| - |-><-|) |-> + 2(|+><-|) |-> = (|->)  + 2 (|->) = 3|->
F|+> = (F'|->) / 2 = (3/2) |->


F'|x> = (|+><+| - |-><-|) |x> + 2(|+><-|) |x> = (|+><+|2|+> - |-><-|3|->) + 2(|+><-|) 3|-> = (2|+> - 3|->) + 6|+> = 8|+> - 3|-> 

F|x>= 4|+> - (3/2)|->

Damn, I hope I didn't messed anything up because I really took myself 30 mins to complete those.. :D It isn't nearly hard as it looks like..!

[Tasm]

fuckin nerds

[highlight](User was banned for this post ("Shitpost" - AshMan55))[/highlight]

[Fourier]

[QUOTE=sltungle;45543651]<a|a> is an integral over a*a (where a* is the Hermitian conjugate of a - the complex conjugate transposed; it's not a simple multiplication).

If you stick an operator in the middle and get, say, <a|b|a>, and if b can be treated as a constant in the integral, i.e. it's not acted on by whatever is left of it, and it doesn't act on whatever is right of it, then in Bra-Ket notation you can simply drag the b out the front and be left with b<a|a>.

In other words imagine a as being a function in x, a(x) and b is a function in y, b(y), then if you had the integral of

a*(x)b(y)a(x) dx

(where, again, * denotes the Hermitian conjugate) you could just drag b(y) out of the front of the integral because it's not dependent on x.

That's not a very mathematical explanation, but hopefully it gets the point across.[/QUOTE]

So lets make this clear.. <a| is a* (Hermitian conjugate) where as |a> is just normal a (no conjugate)?

I know from previous post that <a||a> = <a|a> that is why I must ask about conjugates.

About integrals.. yes I kind of get it. Really I do. It is basically same way in integrals like this:
I will use symbol % for 'integral of'

//we move 2 out because it doesn't change 
% 2x dx = 2% x dx

//we move 3x and 2 out because it doesn't change
% 2x 3y dx = 6y % x dx 

So, are you saying notation <a|a> is same as (from a*) % (to a) x dx? Integral from a* to a?

[Tamschi]

[QUOTE=sltungle;45543651]<a|a> is an integral over a*a (where a* is the Hermitian conjugate of a - the complex conjugate transposed; it's not a simple multiplication).

If you stick an operator in the middle and get, say, <a|b|a>, and if b can be treated as a constant in the integral, i.e. it's not acted on by whatever is left of it, and it doesn't act on whatever is right of it, then in Bra-Ket notation you can simply drag the b out the front and be left with b<a|a>.

In other words imagine a as being a function in x, a(x) and b is a function in y, b(y), then if you had the integral of

a*(x)b(y)a(x) dx

(where, again, * denotes the Hermitian conjugate) you could just drag b(y) out of the front of the integral because it's not dependent on x.

That's not a very mathematical explanation, but hopefully it gets the point across.[/QUOTE]

Right, that's a more accessible explanation than just saying they are infinitely dimensional vectors.

As an aside: This infinite vector space I mentioned is in fact just the vector space of these wave functions, since functions can be seen as vectors in certain spaces. There are further limits to the actually physically relevant ones though, like being square-integrable.

[DrLuke]

[QUOTE=Awesomecaek;45541507]lets separate mass from atoms, then we can obviously send them elsewhere faster than speed of light

not sure what to do with them once they get there nor what to do with the mass left over, but I am sure somebody gonna figure that out[/QUOTE]

If you remove mass from things, that would directly mean that said things have no energy whatsoever anymore. I'm pretty sure that's a rather bad thing.

[Tamschi]

[QUOTE=Fourier;45543978]Actually, this is quite good explanation!

Looks like some sort of boolean operators to me which carry value with it's logical state.. :)

Also lil excercise just to confirm my understanding :)
[maths]

Damn, I hope I didn't messed anything up because I really took myself 30 mins to complete those.. :D It isn't nearly hard as it looks like..![/QUOTE]

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.

[sltungle]

[QUOTE=Fourier;45544102]So lets make this clear.. <a| is a* (Hermitian conjugate) where as |a> is just normal a (no conjugate)?

I know from previous post that <a||a> = <a|a> that is why I must ask about conjugates.

About integrals.. yes I kind of get it. Really I do. It is basically same way in integrals like this:
I will use symbol % for 'integral of'

//we move 2 out because it doesn't change 
% 2x dx = 2% x dx

//we move [B]3x[/B] and 2 out because it doesn't change
% 2x 3y dx = 6y % x dx 

So, are you saying notation <a|a> is same as (from a*) % (to a) x dx? Integral from a* to a?[/QUOTE]

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).

[JohnnyMo]

[QUOTE=Fourier;45544102]So lets make this clear.. <a| is a* (Hermitian conjugate) where as |a> is just normal a (no conjugate)?

I know from previous post that <a||a> = <a|a> that is why I must ask about conjugates.

About integrals.. yes I kind of get it. Really I do. It is basically same way in integrals like this:
I will use symbol % for 'integral of'

//we move 2 out because it doesn't change 
% 2x dx = 2% x dx

//we move 3x and 2 out because it doesn't change
% 2x 3y dx = 6y % x dx 

So, are you saying notation <a|a> is same as (from a*) % (to a) x dx? Integral from a* to a?[/QUOTE]

not quite, it's the integral from -infinity to infinity of a*(x)a(x)dx where a is a function, probably complex-valued.

[Awesomecaek]

8=======D

hehehehe

[Tamschi]

[QUOTE=Fourier;45544102]So lets make this clear.. <a| is a* (Hermitian conjugate) where as |a> is just normal a (no conjugate)?

I know from previous post that <a||a> = <a|a> that is why I must ask about conjugates.

About integrals.. yes I kind of get it. Really I do. It is basically same way in integrals like this:
I will use symbol % for 'integral of'

//we move 2 out because it doesn't change 
% 2x dx = 2% x dx

//we move 3x and 2 out because it doesn't change
% 2x 3y dx = 6y % x dx 

So, are you saying notation <a|a> is same as (from a*) % (to a) x dx? Integral from a* to a?[/QUOTE]

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

[noparse][A,B] = AB - BA[/noparse]

(if I didn't mix it up).