Can someone give me a list of the commutation relations between the
L_i and p_i and x_i in ordinary QM
[L_i, p_j] ij=123
[L_i, x_j]
If not tell me where such a list may be found? I'm deriving
them myself, but would like a reference for the future and in
case I make some dyslexic error.
Thanks
a student
Jul6-08, 06:00 AM
On Jul 3, 6:18 am, pirillo <ultraman2...@hotmail.com> wrote:
> Can someone give me a list of the commutation relations between the
> L_i and p_i and x_i in ordinary QM
>
> [L_i, p_j] ij=123
> [L_i, x_j]
>
> If not tell me where such a list may be found? I'm deriving
> them myself, but would like a reference for the future and in
> case I make some dyslexic error.
>
> Thanks
For any vector operator (A_1, A_2, A_3), one has
[ L_r, A_s ] = -i hbar eps_{rst} A_t,
where eps_{rst} is the Levi-Civita symbol. See, eg, Merzbacher,
"Quantum Mechanics".
This relation arises because L is the generator of rotations, i.e.,
the unitary operator
U(n) = exp[ i L.n/hbar ]
has to rotate any given observable vector A, i.e.
U(n)* A U(n) = R(n) A,
where R(n) denotes rotation in the direction n/|n| by an angle |n|.
In particular, for infinitesimal n, one can expand both sides of this
relation, to first order in n, to give
(-i/hbar) [ L.n, A] = n x A,
and thence obtain the above commutation relation.
Igor
Jul6-08, 06:00 AM
On Jul 2, 4:18 pm, pirillo <ultraman2...@hotmail.com> wrote:
> Can someone give me a list of the commutation relations between the
> L_i and p_i and x_i in ordinary QM
>
> [L_i, p_j] ij=123
> [L_i, x_j]
>
> If not tell me where such a list may be found? I'm deriving
> them myself, but would like a reference for the future and in
> case I make some dyslexic error.
>
> Thanks
The rules are as follows:
[L_i, x_j] = i hbar eps(ijk) x_k
[L_i, p_j] = i hbar eps(ijk) p_k
where eps(ijk) is the Levi Civita symbol.
pirillo
Jul13-08, 06:00 AM
I asked this question, because I thought it would help me to prove
something.
Unfortunately I am not any closer...I think. I want to show that the
average values
of position and of momentum get rotated after we apply e^{iL_z
theta} .
In other words I want to show that :
< e^{iL_z theta} v, x_k e^{iL_z theta} v > = taking the vector
made by the <v, x_j v > where j=123 give the components of
the vector, and rotating this vector arount the z axis by theta
and then taking its kth-component.
However, I want the proof of this.
Thanks
pirillo
Jul14-08, 06:00 AM
I am amazed by this statement:
> For any vector operator (A_1, A_2, A_3), one has
> [ L_r, A_s ] = -i hbar eps_{rst} A_t,
> where eps_{rst} is the Levi-Civita symbol. See, eg, Merzbacher,
> "Quantum Mechanics".
Just seems weird even though It seems to work in all the
cases I can think of right now. It would seem to me that
the properties of the L_i rotating things, or even the commutation
relations would depend heavily on the type of observable one chose.
Unless I'm not understanding the definition of a vector operator. Is it
just 3 commuting observables? Or is it 3 observables whose average
values rotate correctly under a rotation? Either way I think the answer
to my reply to Igor, will clear up this point.
Hendrik van Hees
Jul16-08, 06:00 AM
pirillo wrote:
> I am amazed by this statement:
>
>> For any vector operator (A_1, A_2, A_3), one has
>> [ L_r, A_s ] = -i hbar eps_{rst} A_t,
>> where eps_{rst} is the Levi-Civita symbol. See, eg, Merzbacher,
>> "Quantum Mechanics".
>
> Just seems weird even though It seems to work in all the
> cases I can think of right now. It would seem to me that
> the properties of the L_i rotating things, or even the commutation
> relations would depend heavily on the type of observable one chose.
>
> Unless I'm not understanding the definition of a vector operator. Is
> it just 3 commuting observables? Or is it 3 observables whose average
> values rotate correctly under a rotation? Either way I think the
> answer to my reply to Igor, will clear up this point.
The statement is just a definition of the term "vector operator". A
vector is an object whose components wrt. a Euclidean basis transform
under rotations by application of the pertinent SO(3) matrix (the
fundamental representation of the rotation group). The "infinitesimal
version" of such a rotation are given by the above written commutator
relations between the (total) angular momentum and the quantity in
question. Note that the angular-momentum operators are the generators
of rotations in the sense of Lie groups, i.e., the rotation is given by
R(phi,n)=exp(i n.J phi),
where n is the unit vector giving the direction of the rotation axis in
the sense of the right-hand rule, and phi is the rotation angle.
Of course, you can generalize these ideas to more complicated objects,
e.g., tensor operators or spinor operators. A good treatment of these
topics can be found, e.g., in
J.J. Sakurai, Modern Quantum Mechanics, Addison-Wesley.
On Jul 12, 10:35 pm, pirillo <ultraman2...@hotmail.com> wrote:
> I asked this question, because I thought it would help me to prove
> something.
> Unfortunately I am not any closer...I think. I want to show that the
> average values
> of position and of momentum get rotated after we apply e^{iL_z
> theta} .
> In other words I want to show that :
>
> < e^{iL_z theta} v, x_k e^{iL_z theta} v > = taking the vector
> made by the <v, x_j v > where j=123 give the components of
> the vector, and rotating this vector arount the z axis by theta
> and then taking its kth-component.
>
> However, I want the proof of this.
The desired transformation properties of x_k already hold as operator
identities. It necessarily follows that the same identities are
satisfied when expectation values are taken.
I'll illustrate with the single component x_1, as an infinitesimal
rotation about the z-axis (using L_3) is effected. You should be able
to complete the proof.
Take the relevant commutator from (the other) Igor's post:
[L_3, x_1] = i hbar x_2 .
Expand the rotation operators to first order in theta:
Doing the same thing for x_2 and x_3, and then exponentiating this to
a finite rotation, you'll get
(rotated x_1) = cos(theta) x_1 + sin(theta) x_2 .
Which is precisely what you'd expect for the x-axis component of a 3-
vector, when it is rotated through angle theta about the z-axis. Once
this identity is proved, it's easy to take expectation values of both
sides and notice that the identity still holds.
Hope this helps.
Igor
P.S.: pirillo, if you post through Google Groups, you should either
manually terminate your lines at 71 characters, or make manual breaks
only between paragraphs. This will remove spurious line breaks from
your posts and make them easier to read.
Rock Brentwood
Jul19-08, 06:00 AM
On Jul 2, 3:18 pm, pirillo <ultraman2...@hotmail.com> wrote:
> Can someone give me a list of the commutation relations between the
> L_i and p_i and x_i in ordinary QM
>
> [L_i, p_j] ij=123
> [L_i, x_j]
>
> If not tell me where such a list may be found? I'm deriving
> them myself, but would like a reference for the future and in
> case I make some dyslexic error.
>
> Thanks
[L_i, V_j] = V_k for (i,j,k) = (1,2,3), (2,3,1), (3,1,2) for any 3-
vector V formed out of P, L, K, and the energy H by use of cross
products, dot products, multiplication by scalars and addition. [L_i,
S] = 0 for any scalar S formed out of P, L, K and H by the same set of
operations.
If you use vectors-valued dummy arguments to index the generators
with; i.e. L_u = L_1 u^1 + L_2 u^2 + L_3 u^3; similarly for K_u, P_u,
and V_u above, then [L_u, V_v] = V_{uxv}; [L, S] = (vector 0).
For ordinary QM, one has [K_u, K_v] = 0 for the boost generators; and
[K_u, P_v] = m (u.v). H is interpreted a kinetic energy, and one has
[K, H] = P.
For relativistic QM, one has [K_u, K_v] = -(1/c)^2 L_{uxv}, [K_u, P_v]
= M (u.v), where M = E/c^2 = m + H/c^2. H is the kinetic energy and E
is the total energy, here. The distinction between H and E, in this
context, is superfluous, but helps show what's going on in the non-
relativistic limit (1/c^2) -> 0.
The invariants in both cases are: m = M - H/c^2 (or just m = M for the
non-relativistic case); P^2 - 2MH + (1/c)^2 H^2 = P^2 - E^2/c^2 +
(mc)^2 (or just P^2 - 2MH for the non-relativistic case). Plus you
have the invariant W^2 - (1/c)^2 W_0^2 (or just W^2 if (1/c)^2 -> 0),
where W = ML + PxK and W_0 = P.L gives you mass x spin. This is
present both relativistically AND non-relativistically.
A consequence of the points raised above:
[L_i, W_j] = W_k where (i,j,k) = (1,2,3), (2,3,1) and (3,1,2); or
[L_u, W_v] = W_{uxv};
and
[L_i, W_0] = 0.
It's a remarkable, but little-known, fact that the same process used
to convert the invariant P^2 - E^2/c^2 + (mc)^2 into the Dirac
equation can be used for P^2 - 2MH + k H^2 for any k. The resulting
algebra is independent of what k is! (In fact, it's just the
complexified Dirac algebra gamma_0, gamma_1, gamma_2, gamma_3,
gamma_5). The "Dirac" equation for k = 0 is equivalent to the
Schroedinger equation. Though quadratic, it "reduces" to linear in H
because M reduces to a constant, m. So you more commonly see it as H =
P^2/2m, when really you ought to be writing P^2 - 2mH = 0 to get a
better picture of what's going on here.