# How do we get the x-representation of the momentum operator?

This answer to this question will go a long way towards improving my understanding of representations of Hilbert space in general.

How do we show that the position representation of p is -i &part;/&part;x?

Here is what I have (using Dirac's notation of priming eigenvalues, unprimed are operators):

Using &int; |p'><p'| dp' = 1 we have

<x'|p|&psi;> = &int; <x'|p'> <p'|p|&psi;> dp'

= &int; <x'|p'> p' <p'|&psi;> dp'

Now if I write <x'|p'> = exp(ix'p'), we have

= &int; p' exp(ix'p') <p'|&psi;> dp'

= -i
&part;/&part;x' &int; exp(ix'p') <p'|&psi;> dp'

= -i
&part;/&part;x' &int; <x'|p'> <p'|&psi;> dp'

= -i
&part;/&part;x' <x'|&psi;>

= -i
&part;/&part;x' &psi;(x')

But how do we show that <x'|p'> = exp(ix'p')? It should follow directly from [x,p] = i. Anybody see how? Or is there a way to get this result more directly from [x,p] = i?

i m going to use capitals for operators, and lowercases for eigenkets and eigenvalues, OK?

here we go:

[X,P]=i

<x'|XP-PX|x>=i<x'|x>=i&delta;(x'-x)

x'<x'|P|x>-x<x'|P|x>=(x'-x)<x'|P|x>=i&delta;(x'-x)

<x'|P|x>=i&delta;(x'-x)/(x'-x)

next, we have to use some identities about the dirac delta function...

d&delta;(x)/dx=-&delta;(x)/x

want to verify that identity for me? it s your homework.

OK, so then

<x'|P|x>=-i&delta;'(x'-x)

now, let s see how <x'|P (which is just P in the x' representation), acts on a general ket

|&psi;>=&int;dx|x><x|&psi;>=&int;dx &psi(x)|x>

<x'|P|&psi;>=-i&int;dx &psi(x)<x'|P|x>=-i&int;dx &psi(x)&delta;'(x'-x)

next integrate by parts:

<x'|P|&psi;>=i[&psi;(x)&delta;(x'-x)-&int;dx &psi;'(x)&delta;(x'-x)]=-i&psi;'(x')

here, the boundary term disappears, because at any x different from x' the delta function is zero.

OK, so is that the result you wanted? are we done here? you can use this method to also calculate that <x|p>=exp(ixp). it all follows directly from the commutator.

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Originally posted by lethe
OK, so is that the result you wanted? are we done here?
Yes! Thanks, lethe!

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lemme show you that calculation too.

P|p>=p|p>
<x|P|p>=p<x|p>

-ip'(x)=p*p(x)

simple differential equation. most general solution is:

<x|p>=p(x)=Aexp(ipx)

you can fix A, up to a phase, by normalization requirements.

ganz einfach

Originally posted by pellman
Probably. I'll play with it some more tonight. Thanks, lethe!

sure. just come on back, if you have any questions with this calculation.

I got all that. Thanks again.

pellman

since you seem to be interested in thinking about the relationship between commutators and differential operators, i have something else i want you to think about.

in the proof above, i started with the commutator, and used it to prove the form of the differential operator for P.

let me call to your attention one fact about commutators: [X,P+X]=[X,P]+[X,X]=[X,P]. so if [X,P]=i, then [X,P+X] is also =i. which means that if i had two operators X and P that satisfy the required commutation relation, i can always construct a new X' and P' that also satisfy this commutation relation.

in fact, define X'=X, P'=P+f(X), where f is any function, these new operators will still satisfy [X',P']=i

on the one hand, i know that if my two operators satisfy this commutation relation, then P in the x basis is -id/dx, and has eipx as an eigenfunction.

on the other hand, i know that X, P'=P+X satisfy the same commutation relation, so the proof above should still apply. yet i know that eipx is not an eigenfunction of -id/dx+x, as you can easily check. i know that -id/dx [x=]-id/dx+x, even though my proof tells me that these facts follow from the commutation relation.

so what went wrong? is there an error in my proof? can you find it? what is the physical meaning of changing the P operator like that?

it s a subtle point, and if you can really understand what s going wrong here, it will go a long way towards helping you understand quantum mechanics.

arivero
Gold Member
refs

As for the notion of "operator", it was Eckart the one who suggested to incorporate it into quantum mechanics. Here is the paper (univ. access only)
Or course the substitution trick was already know by the fathers.

As for the notion of "quantum derivation" related to the commutator, it was due to Dirac, see the article in
http://gallica.bnf.fr/scripts/ConsultationTout.exe?E=0&O=n056202.htm [Broken]
esp. page 605. By the way, the other papers form dirac in the proceedings are also in gallica, just move up and down the number in the url.

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Originally posted by lethe
it s a subtle point, and if you can really understand what s going wrong here, it will go a long way towards helping you understand quantum mechanics.
Very interesting, lethe. Of course, rewriting the Hamilitonian in terms of P' leaves the position representation of the Schrod equation unchanged.

P2/2m -> (P' - F(X))2/2m

and using X = x, P' = id/dx + F(x), gives the same differential eq. for &psi;(x). Does this obervation mean simply that it is the differential equations that have physical significance, while the abstract representation in terms of rays |&psi;> and operators X and P has an ambiguity with no physical significance, akin to the gauge freedom in EM?

If you have a reference to a text that covers this, I'd be interested.

arivero
Gold Member
uh? I supposse that you change P->P' directly, without compensating in schroedinger equation, do you? Of course, you will change it too in all the observables.

In any case, I can not see the point. Commutation rules are not about P and X but about any pair of canonical coordinates. Any canonical transformation will preserve the commutation rules.

I guess you are really asking if:
a) P-> p+x, X->x is a canonical transformation.
b) if any pair of canonical coordinates are related by fourier transform.

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Originally posted by arivero
uh? I supposse that you change P->P' directly, without compensating in schroedinger equation, do you?
Changing P to P' in the Hamiltonian changes the physical problem since it is no more than changing Hamiltonians to one with with a different vector potential.

The assumption that I had prior to this thread was that (for spin-0 particles) the three pieces

[X,P] = i
the form of H(X,P)
H|&psi;> = id|&psi;>/dt

were sufficient to describe any problem in the Schrodinger picture. However, this seems not to be the case because it leaves which actual differential equation to use ambiguous.

Oh, leeeethhhe!

Originally posted by pellman
Very interesting, lethe. Of course, rewriting the Hamilitonian in terms of P' leaves the position representation of the Schrod equation unchanged.

P2/2m -> (P' - F(X))2/2m

and using X = x, P' = id/dx + F(x), gives the same differential eq. for &psi;(x). Does this obervation mean simply that it is the differential equations that have physical significance, while the abstract representation in terms of rays |&psi;> and operators X and P has an ambiguity with no physical significance, akin to the gauge freedom in EM?

this isn t quite right.

i mean, you re basically changing to X' and P', and then changing back to X and P, so of course you get the same equations.

what i meant was, let s decide to use X' and P' as our canonical variables. in the hamiltonian formalism, the only assumption is that X and P are operators that satisfy my commutation relation, so there is no a priori way to know which operator is the "real" physical momentum operator. in principle, either set is valid. if i use the X eigenbasis to write the P eigenvalue equation as above:

-idp(x)/dt=p*p(x)

whose solutions are as above, plane wave solutions, like eipx

however, X' and P' are equally as valid as X and P since they satisfy the same commutation relation. one would expect that they yield the same solutions and the same physics. let s see what the momentum eigenvalue equation looks like using the X'=X basis and P'=P+f(X)

[-id/dx+f(x)]p(x)=p*p(x)

obviously, this new operator will not have the same eigenfunctions as before. the solutions to schrödinger s equation will also look different:

-1/2m[d/dx+f(x)]2&psi;n(x)=En&psi;n(x)

obviously, this differential equation will not have the same solutions. i m using P' here as the canonical momentum here, so H=(P')2/2m.

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Originally posted by pellman

If you have a reference to a text that covers this, I'd be interested.

this whole discussion is from shankar. which is where i first saw the resolution to this apparent paradox. i first thought of this problem when i learned a little field theory.

in field theory, one starts from only a commutator, and never sees a differential operator form for the operators. it was therefore rather disconcerting to me that there is this seeming complete arbitrariness in the form of the operators.

but i looked for a while, and shankar discusses the matter, although he left some of it as homework. i was relieved when i found it there, because this question had been bugging me for quite some time.

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i guess i will tell you what s going on here.

this is what s known as a canonical transformation in classical mechanics. a canonical transformation is any transformation that leaves the poisson brackets unchanged.

hamilton s equations describe the same physical system under a canonical transformation.

in quantum mechanics, one way of describing a system is hamilton s equation of motion, which is

dO/dt=i[O,H], where O is any operator.

it is obvious from this starting point, that any system where [X',P']=i will also give the same commutator with O' and H', and hence the same equations of motion.

however, when we do the calculations to go from the commutator quantum mechanics to schrödinger wave mechanics, the two representations seem to be very different, and the differential equations (e.g. schrödinger s equation) change considerably. we get different wavefunctions as solutions.

where did we go wrong?

well there is one part of the calculation that i did above that i sort of glossed over. let s look at it a bit more closely.

to prove that P=-id/dx, we assumed that <x'|X|x>=x&delta;(x'-x). this is simply the orthonormality relation for the eigenbasis of the hermitian X operator.

however, if that orthonormality relation is satisfied for some set of kets |x>, then it is also satisfied for the set ei&Phi;|x>, for some hermitian operator &Phi;. in other words, there is a freedom of choice of phase in our calculation, and we shall see that that corresponds exactly to our freedom of choice of f(X) in P'=P+f(X). P+f(X) is a canonical transformation, and it produces a unitary change of basis transformation in the |x> kets, ei&Phi;, where &Phi;=g(X), g(x)=&int;dx f(x).

label our new kets
|y>=ei&int;dx f(X)|x> = ei&int;dx f(x)|x> = eig(x)|x>

then
<y|&psi;> = &psi;(y) = e-ig(x)<x|&psi;> = e-ig(x)&psi;(x)

first, notice that these |y> are still orthonormal eigenkets of the X'=X operator.

next, let s see how P' operator acts on a |&psi;> ket in the y representation:

[-id/dx+f(x)]e-ig(x)&psi;(x) = -f(x)e-ig(x)&psi;(x) - ie-ig(x)d&psi;(x)/dx + f(x)e-ig(x) =-ie-ig(x)d&psi;(x)/dx

and there you see, that P' is simply -id/dx, multiplied by a phase.

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arivero
Gold Member
Hmm, to round the thing, it should be nice to look which the analogue of Fourier Trasform is, now.

IE, in the usual representation we go from f(x) to f(p) by fourier transforming, f(p)= \int exp(ipx) f(x) dx

This is because we see f(x) as the coeffients of the expansion in the |x> orthogonal basis, and same with f(p), so that we really do an unitary chage of basis |p>= \int |x><x|p> dx.
The usual exp(ipx) comes from < &delta;(u-x)| exp(ipu) >. If the exp are not anymore the eigenvectors, then the kernel of the transformation also changes. Is it easy to calculate the new one?

Originally posted by lethe
this isn t quite right.

i mean, you re basically changing to X' and P', and then changing back to X and P, so of course you get the same equations.

...

this is what s known as a canonical transformation in classical mechanics.
I see where you're going. But in a canonical transformation, say from (X,P) to (X,P') as we are discussing here, there is a corresponding change of Hamiltonian from H(X,P) to K(X,P') such that it leaves the physics unchanged. Canonical transformations are precisely those transformations that are allowed such that we can ultimately get equivalent solutions to Hamilton's equations of motion. In the quantum case, this means identical differential equations.

Transforming P without changing the Hamiltonian is equivalent to changing the Hamiltonian while leaving P = id/dx, i.e., considering a different physical problem.

For example, it is true that changing P results in a change of phase with x-dependence. But in the time-dependent solution, the energy will also have an x-dependence. So the predicted energies no longer match experiment.

Note that your observation that it rests in the choice of phase of the |x> is equivalent, I believe, to my original observation that deriving P = id/dx from the commutator depends on the choice that <x|p> = exp(ipx). And I still don't see any reason to conclude that choice is anything but arbitrary.

It seems that the only basis for the choice P = id/dx is that is the choice that matches experiment, e.g., gives correct energies, and that is choice for which the expectation values approach the classical equations of motion for h->0.

Is that Shankar reference Principles of Quantum Mechanics or Quantum Field Theory and Condensed Matter : An Introduction?

Originally posted by pellman
I see where you're going. But in a canonical transformation, say from (X,P) to (X,P') as we are discussing here, there is a corresponding change of Hamiltonian from H(X,P) to K(X,P') such that it leaves the physics unchanged. Canonical transformations are precisely those transformations that are allowed such that we can ultimately get equivalent solutions to Hamilton's equations of motion. In the quantum case, this means identical differential equations.

hmmm... yes, i suppose you re right. i think i did have my canonical transformation not right.

well, just consider the momentum eigenvalue equation then, that s what s in shankar, anyway.

Is that Shankar reference Principles of Quantum Mechanics or Quantum Field Theory and Condensed Matter : An Introduction?

Principles of QM. chapter 7.

Thanks, lethe and arivero. This thread has given me a much stronger understanding of this.

arivero
Gold Member
units

Me too, pellman.

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Just another detail... It is even funnier if we put units into the play. Then we can not move, say, P---> P+X; we need both terms having the same units, so really P---> P + &lambda; X. The units of lambda depend on the pair of canonical variables we choose.

If we choose the popular pair (p, x), then the units of x/p are the ones of G/c^3, G being Newton constant.

If we choose the action-angle pair, {J, &theta;}, then the units of the quotient are units of action, ie, the ones of Plank constant.

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A lot of wild spec has grown from this, in all the years of stagnation of the theory.

Plank length and Plank momentum are defined from the first pair by asking both px=h and x/p=G/c^3. Also, Plank mass can be defined if we impose c=1, which automatically happens in the natural units system.

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Originally posted by arivero
Just another detail... It is even funnier if we put units into the play. Then we can not move, say, P---> P+X; we need both terms having the same units, so really P---> P + &lambda; X. The units of lambda depend on the pair of canonical variables we choose.

If we choose the popular pair (p, x), then the units of x/p are the ones of G/c^3, G being Newton constant.

If we choose the action-angle pair, {J, &theta;}, then the units of the quotient are units of action, ie, the ones of Plank constant.
The transformation P---> P + &lambda; X is a special case of P---> P + F(X), which amounts to adding a vector potential to the Hamiltonian, e.g., P---> P - eA(X) as in EM.

If you do it right, then changing to new dynamical variables will give J---> J + G(&theta;), I would think, with the units working out all right. But I guess I'm stating the obvious.

I don't really know what would be significant in the different units of &lambda; in the two cases.

- Todd