Operator inequality

[neelakash]

I got stuck to this problem:

To prove that (x p)^2 is not equal to (x)^2 (p)^2

where x and p are position and mometum operator in QM.

I approached this way:

Two operators A and B are equal iff Af=Bf for all f

So,here {(x p)^2 - (x)^2 (p)^2}f=0

Since this is valid for all f, we must have {(x p)^2 - (x)^2 (p)^2}=0

In other words, [(x p)^2 , (x)^2 (p)^2 ]=0

But I could not disprove this commutator.Calculations are big and after some steps I doubt whether this might at all be the correct way.

Can anyone please give some hint?

[BenTheMan]

Is f an eigenfunction of position, momentum, or both?

[neelakash]

Hi Ben,
I do not think that is necessary...it is the definition of equality of two operators...

However, this can be done as follows:
(xp)^2=xpxp
(x^2 p^2)=xxpp

If they are equal then xp=px which is wrong.

[Tom2]

It looks like you're trying to prove it by contradiction, but you can do it straightforwardly by using the commutator \[x,p\]=i\hbar

(xp)^2=xpxp=x(xp-i\hbar)p

Use the distributive property to conclude from there.

[temur]

Nice.

ps: I'm not trying to increase my post count.

[Frud11]

Is it called a "non-Abelian transform"? (Just guessing)

[neelakash]

OK...the matrices are non-commuting---in that sense you may call it non-Abelian.But I do not know a great deal about this.So,amyone more knowledgeable is welcome.

[Tom2]

The only times I ever see the terms "non-Abelian" and "transformation" used in the same breath are when "non-Abelian gauge transformations" are mentioned, and this isn't one of those.

[AlphaNumeric]

Is it called a "non-Abelian transform"? (Just guessing)I don't think it really has a name. x and p are conjugate operators and so their commutation relations is non-trivial.
OK...the matrices are non-commuting---in that sense you may call it non-Abelian.You can't even really consider them matrices because they don't obey standard identities for matrices

xp-px = [x,p] = ih'

Take the trace of both sides

LHS : Tr(xp-px) = Tr(xp)-Tr(px) = Tr(xp)-Tr(xp) = 0 (Using the identity Tr(AB...CD) = Tr(DAB...C), it's cyclic)

RHS : Tr(ih') = nih' where n is the dimension of the matrix. For finite dimensional systems this is trivially false. For infinite dimensional systems it's a little more up in the air but even so...

Things like a and a^{\dag} can be thought of as matrices, as can stuff like J_{\pm}. You can explicitly construct their matric expresentation via A_{ab} = \langle a|\hat{A}|b\rangle but that doesn't work when you've got quantised conjugate variables.

[Frud11]

The only times I ever see the terms "non-Abelian" and "transformation" used in the same breath are when "non-Abelian gauge transformations" are mentioned, and this isn't one of those.OK, but a transform is just an operation.
Turning a cube over is a transform, and turning it back is the inverse transform. It's to do with commutativity. Also automorphisms, or morphisms (iso- hetero- homo-, etc). These are just sets of transforms (operations), or permutations of them, that give a different state, or a rotation, say.

Vector algebra is non-Abelian.

[QuarkHead]

Vector algebra is non-Abelian.you mean non-commutative right? The term abelian (note the lower case "a") refers to algebraic sets, not operations upon them; operations are either commutative or not.

But in fact, in general, vector algebra is most definitely a commutative algebra (there are some wacky exceptions, I believe) - it is part of the definition of a vector space.

[przyk]

For finite dimensional systems this is trivially false.
I don't think \hat{x} and \hat{p} can have any common (non-trivial) finite-dimensional invariant subspaces.

Anyway in a finite-dimensional space you've got a serious problem if you calculate:
\tr \left[ \hat{A}, \, \hat{B} \right] \equiv \sum_{n} \left\langle \varphi_{n} \middle| \left[ \hat{A}, \, \hat{B} \right] \middle| \varphi_{n} \right\rangle

in two different ways and get two different answers. As you just demonstrated, the trace of the commutator of two self-adjoint operators is zero (if it's finite, otherwise it's undefined).

[przyk]

But in fact, in general, vector algebra is most definitely a commutative algebra (there are some wacky exceptions, I believe) - it is part of the definition of a vector space.
On a side note, it doesn't even need to be. You can derive the commutativity of vector addition from the other axioms that define a vector space.

[Tom2]

How's that?

[przyk]

Referring to the axioms on Wikipedia's page on vector spaces (http://en.wikipedia.org/wiki/Vector_space#Formal_definition), first you show that we can add \vec{0} and subtract just as well from the left as from the right (as in axioms (3) and (4)):

\begin{eqnarray}
(-\vec{v}) \, + \, \vec{v} & = & (-\vec{v}) \, + \, \vec{v} \, + \, \vec{0} \\
& = & (-\vec{v}) \, + \, \vec{v} \, + \, (-\vec{v}) \, + \, (--\vec{v}) \\
& = & (-\vec{v}) \, + \, \vec{0} \, + \, (--\vec{v}) \\
& = & (-\vec{v}) \, + \, (--\vec{v}) \\
& = & \vec{0}
\end{eqnarray}

(where --\vec{v} (= \vec{v}) is the additive inverse of -\vec{v})

\begin{eqnarray}
\vec{0} \, + \, \vec{v} & = & \vec{v} \, + \, (-\vec{v}) \, + \, \vec{v} \\
& = & \vec{v} \, + \, \vec{0} \\
& = & \vec{v}
\end{eqnarray}

Then with that out of the way you expand the expression (1 + 1)( \vec{v} \, + \, \vec{w}) in two different ways and compare the results:

(1 + 1)\vec{v} \, + \, (1 + 1)\vec{w} = \vec{v} \, + \, \vec{v} \, + \, \vec{w} \, + \, \vec{w}

1( \vec{v} \, + \, \vec{w}) \: + \: 1( \vec{v} \: + \: \vec{w}) = \vec{v} \, + \, \vec{w} \, + \, \vec{v} \, + \, \vec{w}

...which you equate and simplify to obtain \vec{v} \, + \, \vec{w} = \vec{w} \, + \, \vec{v}. Apparently this was only discovered quite a while after the vector space was first defined.

I originally only had vector addition in mind when I made that last post, but while we're at it:

\begin{eqnarray}
(\lambda + \mu) \vec{v} & = & \lambda \vec{v} \, + \, \mu \vec{v} \\
& = & \mu \vec{v} \, + \, \lambda \vec{v} \\
& = & (\mu + \lambda) \vec{v} \\ \\ \\ \\ \\
\Rightarrow \lambda \, + \, \mu & = & \mu \, + \, \lambda
\end{eqnarray}

Scalar multiplication doesn't have to be commutative though - in my first year linear algebra course, I saw vector spaces defined on division rings in general, and not just specifically fields.

[Frud11]

Who would like to clarify what is meant by "operation", "set", and "transform"?
Is there such a thing as a set of transforms? Or a commutative transform? Does a transform do something transitive?

Does an operation, like addition, transform anything?

Vector algebra is pseudo-commutative. It works in a pseudo-mathematical space, and you can find a determinant (in that space). The key is the connection to the real world (the "other half", of vector algebra).

If there's a set of transform operations defined in a particular space, any list (ordered subset) of them will produce a particular state, for some geometric object, say, or you can define some object in terms of a list of operations, like a graph of some kind.

[AlphaNumeric]

Who would like to clarify what is meant by "operation", "set", and "transform"?
Is there such a thing as a set of transforms? Or a commutative transform? Does a transform do something transitive?An operation is the effect an operator has. An operator does something to the thing it operates on.

For instance, the operator A defined by "A : x -> 2x" doubles whatever it's applied to. The differential operator "B : f(x) -> f'(x)" takes the derivative of the function it's applied to.

Strikly speaking you have to define the space upon which the operator works. For instance B is only defined on the space of functions which have derivatives. If f'(x) doesn't exist then B is not well defined.

A set is just a collection of things. {Bob, Fred, Sam} is a set of men. {1,2,3,4} is a set of numbers. Some sets have particular properties or can be given particular properties in a nice way. For instance, Z, the set of all integers (both positive and negative) is such that for any a and b in the set a+b is too.

A transform is somewhat of another name for an operator. It does something to the thing it's applied to. Generally it's applied to sets/spaces where the transformed object is in the set too. ie if T is a transform, T(x) is in the same set as x.

A commutative set of transforms means that it doesn't matter the order you do the transforms. For instance, if my transform is defined on the set of integers as A : x->x+1 and B : x->x+2 then applying A(B(x)) = B(A(x)), doesn't matter the order. Most transforms aren't commutative. A simple example would be if C : x->2x. A(C(x)) = A(2x) = 2x+1. C(A(x)) = C(x+1) = 2x+2. Not the same. When you get into the weird and wonderful world of quantum mechanics non-commuting transforms are common place. Infact, they are common place in most physics. Even something as simple as rotations in 3 dimensions.
Does an operation, like addition, transform anything?Depends on what your space is defined as. For the integers, then yes. A:x->x+1 will have that A(x) != x. Though if you define your space as x mod 1 then A(x)=x.
Does a transform do something transitive?Doesn't have to be. A transative group action means that if you've two elements in your group, x and y, there's always another element, z, such that z(x)=y. ie you can always map x to y using some group action.

This isn't always true.
Vector algebra is pseudo-commutative. It works in a pseudo-mathematical space, and you can find a determinant (in that space). The key is the connection to the real world (the "other half", of vector algebra).'Pseudo-mathematical' is meaningless. It's always mathematical, it's a mathematical system. Just because your usual notions of how things work don't apply doesn't make it any less mathematical. Quantum mechanics makes use of Grassman variables where a*b+b*a=0, something completely different to the usual properties of numbers, but the applications to the real world are huge (they represent fermions).

Mathematics is entirely a conceptual thing. Wether it has applications to the real world is irrelevent to it's own self consistency.
If there's a set of transform operations defined in a particular space, any list (ordered subset) of them will produce a particular state, for some geometric object, say, or you can define some object in terms of a list of operations, like a graph of some kind.I don't follow.

I'd recommend you read about simple groups, like the discrete permutation group. That's got plenty of interesting properties like non-commutating actions.

[Frud11]

What I mean by "pseudo-mathematical space", is the space that pseudo vectors are in.

You can define a square as a list of operations (productions of lines, or vectors), and so on for any geometrical object. A graph can be undirected or directed (vectored). At least I think you can do this (I can't be sure there isn't some class of geometrical objects that don't fit).

[Frud11]

Z, the set of all integers... is such that for any a and b in the set a+b is [also in the set]. Are a, b and a+b called automorphs of Z?

P.S. With the operator/operation concept. You can't have an operator without the idea of an operation, i.e. what it does. Even if it does nothing, like the identity operator. Distinctions between what it is and what it does aren't necessarily meaningful.

P.P.S I like to push the lingo envelope, and see if someone comes along to rescue the situation, as it were.