- #1

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If so, why do the commutation relations hold in the continuum? I guess what I'm asking is, what is the analog of the trace in the continuous case and why does it not produce a contradiction in the continuous case?

Thanks!

- Thread starter "pi"mp
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- #1

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If so, why do the commutation relations hold in the continuum? I guess what I'm asking is, what is the analog of the trace in the continuous case and why does it not produce a contradiction in the continuous case?

Thanks!

- #2

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Do you mean that you are switching from matrices to the hole integral stuff?

- #3

aleazk

Gold Member

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Yes, if you try to realize the CCR on a finite dimensional Hilbert space, you get a contradiction, since you can realize the operators as nxn matrices and therefore take the trace of these matrices, but tr([x,p])=0 while tr(1)=n.

There's a notion of trace in infinite dimensions, but it only applies to certain operators: http://en.wikipedia.org/wiki/Trace_class

In particular, is not defined for the identity 1 in infinite dimensions, so you can't use the same 'trick' here.

Also, you can show very easily that at least one of the operators in the CCR has to be unbounded, so the thing becomes quite tricky because of domains considerations. This complication is avoided by postulating the Weyl relations as the fundamental relations, the CCR are then derived from them (by taking the generators in a suitable invariant dense domain).

BTW, if you are interested, that notion of trace in infinite dimensions is very important in QM, since according to Gleason's theorem, the trace-class operators characterize the possible quantum states.

There's a notion of trace in infinite dimensions, but it only applies to certain operators: http://en.wikipedia.org/wiki/Trace_class

In particular, is not defined for the identity 1 in infinite dimensions, so you can't use the same 'trick' here.

Also, you can show very easily that at least one of the operators in the CCR has to be unbounded, so the thing becomes quite tricky because of domains considerations. This complication is avoided by postulating the Weyl relations as the fundamental relations, the CCR are then derived from them (by taking the generators in a suitable invariant dense domain).

BTW, if you are interested, that notion of trace in infinite dimensions is very important in QM, since according to Gleason's theorem, the trace-class operators characterize the possible quantum states.

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- #4

tom.stoer

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As said the commutation relations cannot be realized in finite dimensional vector spaces.

I don't think that the trace is useful in the infinite-dimensional case since it's neither necessary not sufficient. It's not necessary b/c the commutation relations can be realized w/o a definition of the trace (simply b/c unity is not trace-class). And it's not sufficient b/c the commutation relations must hold as operator identifies on the whole Hilbert space.

I don't think that the trace is useful in the infinite-dimensional case since it's neither necessary not sufficient. It's not necessary b/c the commutation relations can be realized w/o a definition of the trace (simply b/c unity is not trace-class). And it's not sufficient b/c the commutation relations must hold as operator identifies on the whole Hilbert space.

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- #5

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Of course, we know that there is such a realization, e.g., in the Hilbert space of square integable functions ("wave mechanics"). So indeed the Heisenberg algebra is realizable on the infinitely dimensional separable Hilbert space.

Another thing, I'm not so sure about is Tom's statement, that the commutation relations must hold as operator identities on the whole Hilbert space. The position and momentum operators are not defined on the whole Hilbert space but only on a dense subspace. So how can the commutation relation hold on the whole Hilbert space? I guess, it's meant that the commutation relation holds on the dense subspace where the position and momentum operators are defined (e.g., in the Schwartz space of quickly falling [itex]C^{\infty}[/itex] functions that are square integrable), and then since the commutator is proportional to the identity operator on this sub space it can be continoued to the whole Hilbert space or something like this. My formal functional analysis is somewhat rusty, I must admit ;-)).

- #6

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[tex] [x,p] \subset i\hbar \hat{1}_{\mathcal{H}} [/tex]

because, due to the unboundedness of these 2 operators, x and p cannot be defined on all the Hilbert space. Putting now operator equality instead of operator inclusion, we can write:

[tex] [x,p] = i\hbar\hat{1}_{\mathcal{H}}|_{\mathcal{D}_{[x,p]}} [/tex]

- #7

tom.stoer

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Yes, this allowes us to write

[tex][x,p] \to [x, -i\partial_x]\,f(x) = i\,f(x)[/tex]

for suitable testfunctions f(x).

- #8

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I used to think that supposing that I had a finite dimensional representation of the canonical commutation relations and then taking the trace was enough to find a contradiction but actually, if we try to represent a Lie algebra with three elements that satisfy [P,X]=ih1, why would 1 have to be represented by the identity matrix ? It seems to me that we can only assume that it is a central element of the image of the representation. In this case, if the image of 1 by the representation is traceless, then there is no contradiction coming from taking the trace in this equation. Nevertheless, in a finite dimensional setup, its representation has to be a subalgebra of gl_r for a certain r and there it looks like asking [P,X]=ih1 AND [1,X]=[1,P]=0 is impossible. I can't find a contradiction though. Any help is welcome.

Goodnight

- #9

samalkhaiat

Science Advisor

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This is a mess, my friend. The canonical commutation relation of QM has

I used to think that supposing that I had a finite dimensional representation of the canonical commutation relations and then taking the trace was enough to find a contradiction but actually, if we try to represent a Lie algebra with three elements that satisfy [P,X]=ih1, why would 1 have to be represented by the identity matrix ? It seems to me that we can only assume that it is a central element of the image of the representation. In this case, if the image of 1 by the representation is traceless, then there is no contradiction coming from taking the trace in this equation. Nevertheless, in a finite dimensional setup, its representation has to be a subalgebra of gl_r for a certain r and there it looks like asking [P,X]=ih1 AND [1,X]=[1,P]=0 is impossible. I can't find a contradiction though. Any help is welcome.

Goodnight

The (3-dimensional) Heisenberg Lie algebra [itex]\mathfrak{h}_{3}[/itex] is the vector space [itex]\mathbb{R} \oplus \mathbb{R} \oplus \mathbb{R}[/itex] with the Lie bracket

[tex][ X , Y ] = Z , \ \ \ \ [ X , Z ] = [ Y , Z ] = 0 .[/tex]

Note that all Lie brackets of Z with anything else are zero. All Lie brackets of Lie brackets are also zero. So, the Heisenberg algebra is

In a local (real) coordinates [itex](x,y,z)[/itex], a general element [itex]A[/itex] of [itex]\mathfrak{h}_{3}[/itex] can be expanded in the basis [itex](X,Y,Z)[/itex] as

[itex]A = xX + yY + zZ[/itex] and the Lie bracket becomes

[tex]\left[ (x,y,z) , (\bar{x}, \bar{y},\bar{z}) \right] = ( 0 , 0 , x\bar{y} - y\bar{x}) .[/tex]

Like any other Lie algebra, [itex]\mathfrak{h}_{3}[/itex] has faithful matrix representation. In fact it is

[tex]

(x,y,z) \leftrightarrow \begin{pmatrix} 0 & x & z \\

0 & 0 & y \\

0 & 0 & 0

\end{pmatrix}

[/tex]

Under mild assumptions, [itex]\mathfrak{h}_{3}[/itex] can be exponentiated (using the BCH formula) to the Heisenberg Lie group [itex]H_{3}[/itex], the space [itex]\mathbb{R}^{2} \oplus \mathbb{R}[/itex] with multiplication

[tex]

\left( \begin{pmatrix}x \\ y \end{pmatrix}, z \right) \left( \begin{pmatrix}\bar{x} \\ \bar{y} \end{pmatrix} , \bar{z} \right) = \left( \begin{pmatrix} x + \bar{x} \\ y + \bar{y} \end{pmatrix}, z + \bar{z} + \frac{1}{2} ( x \bar{y} - y \bar{x}) \right) .

[/tex]

Note that [itex]H_{3}[/itex] has one-dimensional centre, spanned by [itex](0,0,1)[/itex]: Each one of the Lie algebra basis [itex](X,Y,Z)[/itex] generates a subgroup of [itex]H_{3}[/itex] isomorphic to [itex]\mathbb{R}[/itex]; the elements of the first two subgroups generate the full group, while the elements of the third one

Again, [itex]H_{3}[/itex] is isomorphic to the Lie group of upper triangular 3 by 3 real matrices with 1 on the diagonal:

[tex]

\left( \begin{pmatrix}x \\ y \end{pmatrix}, z \right) \leftrightarrow \begin{pmatrix} 1 & x & z + \frac{1}{2}xy \\

0 & 1 & y \\

0 & 0 & 1

\end{pmatrix}

[/tex]

This isomorphism means that [itex]H_{3}[/itex] has natural

[tex]d_{e}\Phi_{S}(X) = - i Q = -i q , \ \ d_{e}\Phi_{S}(Y) = - i P = - \frac{d}{dq},[/tex] [tex]d_{e}\Phi_{S}(Z) = - i \ \mbox{I}_{L^{2}(\mathbb{R})} = - i .[/tex] Namely

[tex]d_{e}\Phi_{S}\left( (x,y,z) \right) = x Q + y P - i z \ \mbox{I}_{L^{2}(\mathbb{R})}.[/tex]

For general group elements [itex](x,y,z) \in H_{3}[/itex] and all [itex]\Psi(q) \in L^{2}(\mathbb{R})[/itex], the Schrödinger representation [itex](\Phi_{S}, L^{2}(\mathbb{R}))[/itex] of [itex]H_{3}[/itex] is given by

[tex]\Phi_{S}\left( (x,y,z) \right) ( \Psi ) (q) = e^{- i (xq + z - \frac{1}{2}xy)} \Psi (q - y) .[/tex]

One can easily show that

1) [itex]\Phi_{S}[/itex] is a group homomorphism, i.e. representation.

2) the Schrödinger representation [itex](\Phi_{S}, L^{2}(\mathbb{R}))[/itex] of [itex]H_{3}[/itex] is irreducible.

And not so easy is the proof of the following famous

3) Every unitary representation [itex]\pi[/itex] of [itex]H_{3}[/itex] on a Hilbert space [itex]\mathcal{H}[/itex] on which the central element acts as [tex]e^{-i}\ \mbox{I}_{\mathcal{H}},[/tex] is isomorphic to a direct sum of copies of the Schrödinger representation [itex]\Phi_{S}[/itex] on [itex]L^{2}(\mathbb{R})[/itex]

[tex]( \pi , \mathcal{H}) = \bigoplus ( \Phi_{S}, L^{2}(\mathbb{R})) .[/tex]

Namely,

The so-called Weyl quantization procedure is obtained by extending [itex]\Phi_{S}[/itex] to the

[tex]\Phi_{S}(f) = \int d\alpha \ d\beta \ \hat{f}(\alpha , \beta) \ \Phi_{S}\left( e^{i (\alpha Q + \beta P)} \right) ,[/tex]

where

[tex]\hat{f}(\alpha , \beta) = (\frac{1}{2 \pi})^{2} \int dq \ dp \ f(q,p) \ e^{- i (\alpha q + \beta p)} ,[/tex] is the Fourier transform of [itex]f[/itex]. The operator [itex]\Phi_{S}(f)[/itex] is, at least formally, self-adjoint if [itex]f[/itex] is real-valued, and is well-defined and of trace class if [itex]f \in \mathcal{S}(\mathbb{R}^{2})[/itex], i.e., if [itex]f[/itex] is [itex]C^{\infty}[/itex] and decreases, together with all its derivatives, faster than the reciprocal of any polynomial at infinity. If [itex]f[/itex] is [itex]C^{\infty}[/itex] and grows at most polynomially at infinity, then [itex]\Phi_{S}(f)[/itex] will in general be a densely defined unbounded operator. And I should just stop here.

- #10

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I wasn't mixing between the two though.

In fact I did not know that the finite dimensional representation of h_3 you gave is faithful but non-hermitian, therefore useless from a physical point of view. I know understand that requiring the central element to act as a scalar fixes the unitary representation (up to isomorphism). But still one of the questions remains, is there any physical argument for the central element to ask as a scalar ?

Anyway, thanks for your time.

- #11

samalkhaiat

Science Advisor

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The point is this: once you say “canonical commutation relation”, then you are already dealing with the infinite-dimensional hermitian operator representation of the Heisenberg algebra.I wasn't mixing between the two though.

Up to a multiplicative scalar, the identity operator is the only available linear operator which is independent of P and Q and commutes with both of them.But still one of the questions remains, is there any physical argument for the central element to ask as a scalar ?

Anyway, thanks for your time.

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