11
$\begingroup$

In the first answer of this question it is said that every quantum observable, let's say $\hat{A}$, can be represented as a function of position and momentum observables. In other words, as I understand it, if $H$ is the Hilbert space spanned by position eigenkets, then every observable could be represented as: \begin{equation} \hat{A}=\hat{A}(\hat{x},\hat{p}) \end{equation} But, what are the mathematical (or physical?) reasons for this to be true?

Assuming this is true, then what happens if we consider the spin part of Hilbert space? Will there appear another pair of operators $(\hat{s_1},\hat{s_2})$ such that every observable in spin space can be represented as functions of $(\hat{s_1},\hat{s_2})$?

I know that this are a many questions, but I am mainly interested on the first one. Thanks!

$\endgroup$
  • 4
    $\begingroup$ The spin isn't a function of position and momenta $\endgroup$ – Annibale Oct 2 '17 at 8:34
  • $\begingroup$ Yes, but spin is a function of position, momentum and spin $\endgroup$ – adiselann Oct 2 '17 at 8:46
  • $\begingroup$ Yes, I misunderstood the division between the two spaces. Now I suggest not to divide in two spaces, but two options: with or without spin. If there is spin in our space also exist observables that are functions of position, momentum and spin $\endgroup$ – Annibale Oct 2 '17 at 9:17
  • 1
    $\begingroup$ I think OP is asking a valid question: Are all observables on the Hilbert space spanned by $|x\rangle$ a function of $\hat x$ and $\hat p$. I would think the answer is "yes for all practical purposes" but I don't know a proof. $\endgroup$ – Noiralef Oct 2 '17 at 9:19
  • $\begingroup$ In practice, what $A(x,p)$ are there beyond the trivial $x$ and $p$, and then the Hamiltonian? For the latter, it comes from the quantisation recipe and the trivial fact that for a classical system, the Hamiltonian system is by definition function of $x$ and $p$ (not the operators this time). $\endgroup$ – user154997 Oct 2 '17 at 9:29
13
$\begingroup$
  1. In classical physics, the definition of an observable is that it is a (reasonably smooth) function of position and momentum. Therefore, quantum systems obtained purely by quantization of a classical system also have that all their observables are functions of position and momentum.

  2. In quantum physics, an observable is just an operator designate to belong to the "algebra of observables". There is no a priori reason to even have position and momentum operators, and indeed, systems with finite-dimensional state spaces (such as qubits or other system without a positional degree of freedom) don't have a position or momentum operator to begin with because the CCR $[x,p] = \mathrm{i}\hbar$ cannot hold on a finite-dimensional space.

  3. In relativistic quantum field theory (which is a type of quantum theory, in the end), there are no position operators, the closest are the so-called Newton-Wigner operators, so the question dissolves because it doesn't make sense anymore. Additionally, the momentum operator is a function of the field operators, but not vice versa - you cannot restore the field from the total momentum operator alone, like you cannot restore a function from the value of a definite integral.

$\endgroup$
  • $\begingroup$ I don't think this answers the question if you read beyond the title. OP does not ask if every observable is a function of $x$ and $p$ in every Hilbert space (as you say, there is no reason to have $x$ and $p$ in an arbitrary Hilbert space). The way I understand the question is: Take $H = L^2(\mathbb R)$ with the usual definitions of $x$ and $p$ and let $A$ be a self-adjoint operator on $H$. Under which conditions / In what sense can $A$ be expressed as a function of $x$ and $p$? (I'm being insistent because I'd actually like to see a good answer, and you can probably give it :)) $\endgroup$ – Noiralef Oct 2 '17 at 13:52
  • 4
    $\begingroup$ @Noiralef The answer to that is no, not all self-adjoint operators on $L^2(\mathbb{R})$ are functions of multiplication and differentiation. Take any Hilbert basis $v_i$ of that space and define $v_i \mapsto \begin{cases} v_i & ,i \text{ even} \\ 2v_i & ,i\text{ odd}\end{cases}$. This operator is diagonal in this basis and hence bounded self-adjoint, but it's neither a multiple of the identity nor can you build it from the unbounded $x$ and $p$, at least not in any way I can see. $\endgroup$ – ACuriousMind Oct 2 '17 at 14:50
-4
$\begingroup$

Ok, maybe this is a proof; it is not complete, but almost.

In quantum mechanics each self adjoint operator is an observable, and so also the projector on position eigenstates $| x \rangle \langle x'|$. Moreover, if you think to operators in $\hat{x}$ basis you can represent each one with it's matrix elements in this basis (is a bijective mapping), that is: all operators are decomposable in a sum (continuous sum $\rightarrow$ an integral) of projectors:

$$\hat{A} = \int \text{d} x~\text{d} x' \alpha(x,x') |x'\rangle \langle x|$$

So we now try to construct all the projectors in terms of $\hat{x}$ and $\hat{p}$ operators.

Restrict the case to one particle in one dimension so you have one position operator and one momentum operator (for easyness, but is general).

Diagonal elements

For the diagonal projector we can consider the operator $C ~\delta(\hat{x} - x)$, with $C$ a normalization constant. I claim that:

$$ C ~\delta(\hat{x} - x) = |x\rangle \langle x|$$

if we can choose that constant $C$ in the right way to "normalize" the delta function (you can prove that, aside from the normalization, the previous equations is true apllying it to the position eigenstates). In this way all the diagonal element are found.

Off-diagonal elements (momentum operator)

For the off-diagonal elements $|x'\rangle \langle x|~, x \neq x'$ we can't use the only $\hat{x}$, because it has only diagonal elements but searching the matrix elements of $\hat{p}$ we find:

$$ \langle x' | \hat{p} | \alpha \rangle = - i \hbar \frac{\partial}{\partial x'} \langle x' | \alpha \rangle $$

and imposing $| \alpha \rangle = | x \rangle$ we obtain:

$$ \langle x' | \hat{p} | x \rangle = - i \hbar \frac{\partial}{\partial x'} \langle x' | x \rangle = - i \hbar \frac{\partial}{\partial x'} \delta(x-x') $$

that, if we interpret the delta as a function (that is null on all the point but $0$), means there aren't diagonal elements for $\hat{p}$, but this not sounds to me (because if true it commutes with $\hat{x}$, and this is not).

On this point I suggest:

Generic operator

Because I can't find a univocal answer at the previous point I proceed with a two branch proof:

  • $\hat{p}$ has only diagonal elements:

This case is clearly false, as I say previously, I continue this way to find another absurdum.

So if you have an operator $\hat{A}$ and we assume that is a function only of $\hat{x}$ and $\hat{p}$ we can expand in Taylor series:

$$ \hat{A} = A(\hat{x},\hat{p}) = \sum A_{\alpha, \beta} ~\hat{x}^\alpha ~\hat{p}^\beta $$

we can do this, because even if $\hat{x}$ and $\hat{p}$ don't commute their commutator is a scalar, so if we have a mixed term $\hat{x} \hat{p} \hat{x} \dots$ we can reorder it finding a term with the same number of operators and other terms with less operators (of lower "order", that is with lower number of operators), and we can repeat this algorithm to obtain a series with only "ordered" terms, as the one that we have write above.

Now if we calculate the matrix elements of $\hat{A}$ we find:

$$ \langle x' | A | x \rangle = \sum A_{\alpha, \beta} ~\langle x' | \hat{x}^\alpha ~\hat{p}^\beta | x \rangle = \int \text{d} x'' \sum A_{\alpha, \beta} ~\langle x' | \hat{x}^\alpha |x''\rangle \langle x'' | \hat{p}^\beta | x \rangle = \\ = \sum A_{\alpha, \beta} \int \text{d} x'' ~x'^\alpha ~\delta(x'' - x') \langle x'' | \hat{p}^\beta | x \rangle = \sum A_{\alpha, \beta}x'^\alpha ~ \langle x' | \hat{p}^\beta | x \rangle $$

But if $\hat{p}$ has not off-diagonal elements, this demonstrates that also $\hat{A}$ can't have diagonal elements, so the number of operators writable as a function of $\hat{x}$ and $\hat{p}$ are limited to the operators diagonal in position basis, that is that one that commutes with $\hat{x}$.

This is absurdum, and we proof one more time that $\hat{p}$ has to have diagonal elements in position representation.

  • $\hat{p}$ has off-diagonal elements:

that is the only realistic case.

So I can write the expression of $\hat{p}$ in terms of position projectors, and I can select only one of them in a way similar as before:

$$ \delta(\hat{x} - x') \hat{p} \delta(\hat{x} - x) = \delta(\hat{x} - x') \left(\int \text{d} y ~\text{d} y' ~a(y,y') ~|y'\rangle \langle y| \right) \delta(\hat{x} - x) = \\ = \int \text{d} y ~\text{d} y' ~a(y,y') ~\big[\delta(\hat{x} - x') |y'\rangle \big]\big[\langle y| \delta(\hat{x} - x)\big] = \\ = \int \text{d} y ~\text{d} y' ~a(y,y') ~\delta(y'- x') |y'\rangle \langle y| \delta(y - x) = a(x,x') |x'\rangle \langle x|$$

so with the operators $\delta(\hat{x} - x') \hat{p} \delta(\hat{x} - x)$ we can isolate all the off-diagonal elements presents in $\hat{p}$.

Conclusion

The last part of the proof is show that $\hat{p}$ operators contains all the off-diagonal elements; I give you one attempt, but is a little bit sloppy (and maybe the statement is false, but I don't find literature about, so I'll ask another question on this point):

  • $\hat{p}$ has at least one off-diagonal element, else it commutes with $\hat{x}$ and it isn't;
  • $\hat{p}$ is invariant under reference translations (think on it's eigenstates $\langle x | p \rangle = \varphi_p (x) = exp(\frac{ixp}{\hbar})$), so it matrix elements depends only on the "distance" $x - x'$ (and it's sign, because it has to be hermitian)
  • under scale transformations of positions $\hat{p} \rightarrow \lambda \hat{p}$ (think again on the action on its eigenstates)

so if one elements is not null also the others has to be not null, because their distance can be reduce to the one of the not-null elements with a scale transformation. So all the off-diagonal elements of $\hat{p}$ are not null, and I can extract them to construct the off-diagonals projectors.

Since I have construct all the projectors with only $\hat{x}$ and $\hat{p}$ now I can construct all the operators as sum, dependent only on that two operators.

$\endgroup$
  • 3
    $\begingroup$ 1. "In quantum mechanics each self adjoint operator is an observable" - This is not true - all observables are self-adjoint, but not every self-adjoint operator must correspond to an observable. 2. The "states" $\lvert x\rangle$ do not exist within the Hilbert space since they are not normalizable, hence the operators $\lvert x\rangle\langle x\rvert$ also don't really exist on the Hilbert space. Proofs involving these dubious entities must be carried out very carefully. $\endgroup$ – ACuriousMind Oct 2 '17 at 14:55
  • 2
    $\begingroup$ 3. $\delta(\hat{x} - x)$ is meaningless - you can use functional calculus to apply any measureable function to an operator, but $\delta$ is not a function. You have to carefully define what you mean by this if you want to use this. 4. In the infinite-dimensional case, there is no direct correspondence between "matrix representations" and operators. The operators which have reasonable matrix representations are Hilbert-Schmidt operators and position and momentum are not among these. $\endgroup$ – ACuriousMind Oct 2 '17 at 14:55
  • $\begingroup$ @ACuriousMind For the point 3. the answer is easy: instead of $\delta$ consider the function $f(x) = \begin{cases}1 \qquad x=0\\0 \qquad x \neq 0\end{cases}$ $\endgroup$ – Annibale Oct 2 '17 at 16:38
  • $\begingroup$ @ACuriousMind For the point 4. I answer with a question: two equal operators has the same matrix elements (is trivial), what you say is that even if they have the same matrix elements they could be different, but can you show an example of this? Maybe you're right, but for me is not trivial $\endgroup$ – Annibale Oct 2 '17 at 16:44
  • $\begingroup$ @ACuriousMind I have a better argument for point 3.: if I approximate delta with analytical mollifiers I obtain a family of regular operators (it's granted by analicity cause I can define them with a series expansion) and then I take the limit of operators. Maybe it's something wrong (completeness of operators space, not convergent series expansion,...), can you help me to understand? $\endgroup$ – Annibale Oct 3 '17 at 8:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.