# I “Vector” operators

1. Jul 23, 2017

### David Olivier

I have always seen “vector” operators, such as the position operator $\vec R$, defined as a triplet of three “coordinate” operators; e.g. $\vec R = (X, Y, Z)$. Each of the latter being a bona fide operator, i.e. a self-adjoint linear mapping on the Hilbert space of states $\mathcal H$.

(I am putting aside here completely all the subtleties discussed in a recent thread.)

I'd like to be able to express the object $\vec R$ without a reference to a particular coordinate system.

What could such an operator be? A linear mapping $\mathcal H \to \mathcal E \otimes \mathcal H$, with $\mathcal E$ the vector space of the three-dimensional “physical” space? This would allow us to write $\vec R | \vec r > = \vec r \otimes | \vec r >$

However, $\mathcal E$ is normally a real vector space, while $\mathcal H$ is a complex one. Perhaps it will do if we first convert $\mathcal E$ into a complex vector space?

Operators are to be self-adjoint. How do we express this condition for such a vector operator?

What about “pseudovectors” such as the angular momentum operator? Would we view them as antisymmetric elements of $\mathcal E \otimes \mathcal E$?

2. Jul 23, 2017

### andrewkirk

It looks like in your post $\mathcal E$ denotes $\mathbb R^3$. If so then, unless I have misunderstood your question, $\mathcal E$ plays no role in the definition. A three-dimensional position operator is a linear map from $\mathcal H$ to $\mathcal H$ where $\mathcal H=\mathcal H_0\otimes\mathcal H_0\otimes\mathcal H_0$ and $\mathcal H_0$ is an infinite-dimensional Hilbert space that is isomorphic to a suitably-restricted subspace of $W$, the space of complex distributions on $\mathbb R$.

$W$ is a bit like the space of all functions from $\mathbb R$ to $\mathbb C$ but, in addition to an isomorphic copy of that space, it contains additional members that are not functions, such as the Dirac Delta.

The analog in $\mathcal H$ to the point $(x,y,z)\in \mathbb R^3$ is the tensor product
$$\delta_x\otimes\delta_y\otimes\delta_z$$
where $\delta_a$ is the Dirac Delta distribution that is 'zero every where except at $a$'.

3. Jul 23, 2017

### David Olivier

Well, I'm trying to have a formulation that doesn't refer to a coordinate system. So $\mathcal E$ is a three-dimensional vector space, but I precisely don't want to call it $\mathbb R^3$, which would assume that I have determined the x, y and z axes.

If I understand you, your $\mathcal H_0$ is the one-dimensional spinless state space; which is consistent with a three-dimensional state space $\mathcal H = \mathcal H_0 \otimes \mathcal H_0 \otimes \mathcal H_0$. However, the position operator cannot be a linear map from $\mathcal H \to \mathcal H$. In the usual, coordinate system dependent, description, we have three position operators, $X$, $Y$ and $Z$, each of which is such a linear map; the combination of the three doesn't make a linear map $\mathcal H \to \mathcal H$, but might be seen as a linear map $\mathcal H \to \mathcal E \otimes \mathcal H$. This is what I was getting at in my OP.

4. Jul 23, 2017

### andrewkirk

Why do you think it isn't a linear map?

Say $B$ is a basis for $\mathcal H_0$. Then the Cartesian product $B'=B\times B\times B$ is a basis for $\mathcal H$. The action of $X\otimes Y\otimes Z$ on an element $|x\rangle\otimes|y\rangle\otimes|z\rangle$ of the ($B$-induced) basis of $\mathcal H$, for $|x\rangle,|y\rangle,|z\rangle\in\mathcal B$, is

$$(X\otimes Y\otimes Z)(|x\rangle\otimes|y\rangle\otimes|z\rangle) \triangleq X|x\rangle\otimes Y|y\rangle\otimes Z|z\rangle$$

and this is then extended to linear combinations of basis elements to make a linear map from $\mathcal H$ to itself.

IIRC it is reasonably straightforward to show that this map is consistent with each of the component maps.

5. Jul 24, 2017

### David Olivier

Thanks for the answer, but I don't see that it works.

When I say that the position operator is not a linear map $\mathcal H \to \mathcal H$, the issue is not that it is not linear, but that the target set cannot be $\mathcal H$.

Let's consider the action of the solution you give on the ket $| \vec r_0 \rangle$representing a particle localized in $\vec r_0$ with coordinates $(x_0, y_0, z_0)$; in terms of wavefunction, this ket represents $\delta(\vec r - \vec r_0)$. Following your construction, we have $| \vec r_0 \rangle = | x_0 \rangle \otimes | y_0 \rangle \otimes | z_0 \rangle$, or, in terms of wavefunctions, $\delta(\vec r - \vec r_0) = \delta(x - x_0) \delta(y - y_0) \delta(z - z_0)$. But then the action of the operator you describe on this ket gives:

$(X \otimes Y \otimes Z) | \vec r_0 \rangle = x_0 y_0 z_0 | \vec r_0 \rangle$

Thus $X \otimes Y \otimes Z$ would be an operator with eigenvalue $x_0 y_0 z_0$ on $| \vec r_0 \rangle$ ‑ which is nothing like what we want, and doesn't even transform properly if we change our coordinate system.

What we need is for our eigenvalue to be a three-dimensonal vector, representing, for a localized particle, its position; which means that our three-dimensional position operator cannot be $X \otimes Y \otimes Z$ mapping $\mathcal H$ to itself, but must map $\mathcal H$ to some larger space, which I suggested might be $\mathcal E \otimes \mathcal H$, with $\mathcal E$ the three-dimensional ordinary vector space, perhaps made into a complex vector space.

6. Jul 24, 2017

### andrewkirk

It looks like I've misunderstood what you are after. I'll have another try.

It sounds like you are trying to define a linear map between Hilbert spaces that corresponds to an instantaneous measurement of the 3D position of a particle, in the same way that a measurement operator (a linear map from $\mathcal H$ to $\mathcal H$) corresponds to the measurement of a scalar amount. Is that right?

If so, the problem that looms is that multiple measurements - at least in the QM I've seen - are never simultaneous. One measurement gives one result - a single real number. If we want to measure 3D position we need to make three separate measurements, and they will not be simultaneous, so they cannot be encapsulated in a single operator.

Consider, for example, the measurement of spin. A central theme in the Bell theorem is that we can only measure spin in relation to a given axis. We cannot make a single measurement that tells us a particle's pre-measurement 3D spin vector.

If it were possible to make three scalar measurements simultaneously - or to measure a 3D vector simultaneously, which is the same thing - I suspect the formulation of Bell's Theorem would be very diffferent from what it is.

7. Jul 25, 2017

### David Olivier

Thanks.

Quite right.

However, your objection ‑ that it is impossible to measure simultaneously the values of $x$, $y$ and $z$ ‑ does not hold. QM tells us that we cannot measure simultaneously the values of operators that do not commute. Because, for instance, $[X, P_x] = i$, one cannot measure both $x$ and $p_x$ together; there is no state that is an eigenstate for both operators $X$ and $P_x$. But the position operators do commute; for instance, $[X, Y] = 0$. One can measure simultaneously all three spatial coordinates, in other words the exact position of a particle (if indeed it is in such an exact position, which in principle is possible; that is what the eigenstate $| \vec r \rangle$ is about).

Your objection does apply, for instance, if I was attempting to form an operator for the angular momentum vector (or pseudovector), since, as you say for spin, the three coordinate operators $J_x$, $J_y$ and $J_z$ do not commute. It would not however keep us from wanting to form the corresponding vector “operator”, which would be a linear mapping, and have some property replacing the auto-adjoint requirement, but would have no eigenstates. This is indeed a difficulty, that I hadn't thought of. But in the case of position, it simply isn't there.

8. Jul 25, 2017

### andrewkirk

The obstacle that I was imagining was practical rather than theoretical. I expect the two are related. It's not clear to me how, but non-commutativity was not what I had in mind.

The only two methods that immediately come to my mind to measure the 3D location of an object are triangulation and radar/sonar. Triangulation consists of two separate 1D measurements and radar/sonar is a rapid sequence of measurements as the beam sweeps out a sector. So in each case only 1D measurements are ever made.

Do you have an experimental procedure in mind whereby one would measure the 3D location of an object in an instant, where that measurement cannot be interpreted as a 1D measurement?

9. Jul 26, 2017

### David Olivier

Photographic plates record photons one by one. When they record a photon, you have measured the three coordinates simultaneously. I don't see how that could be interpreted as a 1D measurement.

But this is irrelevant to my problem, which was: we have a certain notion, called “position”, that, in a given coordinate system, is expressed as three operators, $X$, $Y$ and $Z$. (Or alternatively, a notion called “angular momentum” that is expressed as three operators $J_x$, $J_y$ and $J_z$.) Why is it that one cannot express this notion independently from the coordinate system, using only the basic tools of the 3D vector space and other objects such as the Hilbert space that derive from it?

(By the way, to say that a vector space is 3D does not mean that we have chosen some coordinate system. It has the property of being 3D independently from any coordinate system. For instance, it is a vector space in which one can embed a Möbius strip but not a Klein bottle.)

10. Jul 26, 2017

### Staff: Mentor

That isn't a vector space, it's a metric space. Vector spaces have an inner product, but not a metric.

11. Jul 26, 2017

### David Olivier

Vector spaces with an inner product are metric spaces.

12. Jul 26, 2017

### Staff: Mentor

They can be given a metric, induced by the norm that is derived from the inner product. But that metric is not logically required by the vector space or inner product.

A better way of defining the dimension of a vector space, which does not require using any metric space properties, is to count the number of vectors required to form a basis. This is also coordinate-independent.

13. Jul 27, 2017

### David Olivier

Specifying the inner product is tantamount to specifying the metric.

Anyway, the issue of Möbius strips and Klein bottles isn't even a metric one. It is a topological issue, and on a finite-dimensional vector space there is but one topology that is compatible with the vector space structure. It is thus a property of the vector space itself whether you can or cannot embed in it a given structure such as a Möbius strip or a Klein bottle.

Specifying whether or not you can insert a Klein bottle or a Möbius strip does not require any metric properties.

Counting the number of vectors to form a basis is the obvious method. I only mentioned the question of what you can or cannot embed in the space to stress that the dimension of the space is part of the very fabric of the space independently from any choice of coordinates.

Anyway, this is a secondary issue. My initial question still stands: how can we define a “vector operator” without referring to a given coordinate system?

14. Jul 27, 2017

### vanhees71

If the inner product is positive definite, it implies a metric, but there are metric vector space where the metric is not induced by an inner product.

15. Jul 27, 2017

### yossell

It's unclear to me whether your worry is one about coordinate systems or vectors.

If you want to avoid using coordinate systems in expressing the laws, why does moving to a vector space help? What is `the' vector space of 3 dimensional space (that you mention in your first post)?

After all, which position a given vector picks out is still relative to a choice of origin. We would still need to make sure that the laws were invariant under the relevant transformations.

16. Jul 27, 2017

### andrewkirk

What a spot on the plate tells us is that, at some time during the exposure period, a photon struck that spot. I don't see that as a 3D measurement, since it is implicit when we talk about measurements that we know the time at which the measurement is taken. With the plate we do not know the time.

17. Jul 28, 2017

### stevendaryl

Staff Emeritus
Getting back to the original question, it seems to me that in most treatments of quantum mechanics, only scalar-valued operators are considered. So we talk about $L_z$ and $L^2$ but not $\vec{L}$.

Just thinking out loud (or at the keyboard)...

If you have a vector-valued operator $\vec{V}$ and you apply it to a state $|\psi\rangle$, you don't get an element of the Hilbert space back, you get something like a triple of states: $V_x |\psi\rangle$, $V_y |\psi\rangle$, $V_z |\psi\rangle$. So is there a way to work with vectors that doesn't involve breaking it up into components?

Well, if you sandwich $\vec{V}$ between two states, you get $\langle \phi|\vec{V}|\psi \rangle$. If $\mathcal{E}$ is your vector space, then this will be an element of the complexification of $\mathcal{E}$. I don't know of there is a standard term for that, but let me call it $\bar{\mathcal{E}}$. So $\vec{V}$ has type: $\mathcal{H} \times \mathcal{H} \rightarrow \bar{\mathcal{E}}$, and is anti-linear in the first argument and linear in the second argument. In tensor terms, I think that's isomorphic to $\mathcal{H}^* \otimes \mathcal{H} \otimes \bar{E}$. In terms of Dirac notation, we could write:

$\vec{V} = \sum_{\alpha \beta j} C_{\alpha \beta j} \mathbf{e_j} |\psi_\alpha\rangle \langle \psi_\beta |$

where $C_{\alpha \beta j} = \langle \psi_\alpha|V_j|\psi_\beta \rangle$, and where $e_j$ is a basis vector for $\mathcal{E}$, and $|\psi_\alpha\rangle$ is a complete orthonormal basis for the Hilbert space $\mathcal{H}$.

18. Jul 28, 2017

### vanhees71

No, $\vec{L}$ is a vector operator, $L_z$ is a vector component and not a scalar. By definition a vector operator obeys the commutation relations
$$[L_i,V_j]=\mathrm{i} \epsilon_{ijk} V_k,$$
where $L_i$ are the components of $\vec{L}$ and $V_k$ the components of $\vec{V}$ with respect to an arbitrary Cartesian basis.

This implies that the expectation values of vector operators transform as vectors, as it should be.

All this of course generalizes to tensor operators of higher rank.

19. Jul 28, 2017

### David Olivier

I am looking for an expression of the physical laws of quantum mechanics that doesn't depend on the choice of the three ordinary spatial directions. My vector space $\mathcal E$ is the ordinary three dimensional real (i.e. over field $\mathbb R$) vector space. I want only expressions with vectors from that vector space, time, and objects built on it (such as $\mathcal H = L^2[\mathcal E]$, the Hilbert space of states).

In classical mechanics, for instance, you have expressions such as $\vec f = m \vec a$, which involve vectors, not scalars. Of course, one can project this equation onto any coordinate system (i.e. choose a basis $(\vec i, \vec j, \vec k)$, and obtain three equations about scalars, but that is artificial and useless.

I would like an expression in quantum mechanics that, like those of classical mechanics, doesn't care about the basis you have chosen.

Checking that laws are invariant under various transformations is all very well, but it is better still if the relevant invariances were incorporated in the formalism itself. If you have expressions involving $x$, $y$, $z$, $f_x$ and so on, you may want to check that any physical results vary properly under rotation; but if you have expressions only with vectors, there is nothing to check.

Your point about the origin is a good one, because one would also want to express everything independently from a specific origin. For instance, expressing the position (in classical mechanics) by a position vector $\vec r$ is better than with a triplet $(x, y, z)$, but still implies an artificial choice of origin. It would be better still to have expressions directly in an affine space (the space of points). My question is about how to do the first step in quantum mechanics; leaving the second step for after.

Another issue is that of movement. It would be good to have expressions that don't depend on the choice between different Galilean frames of reference. But that too is an additional step, that can be left for later. For the moment, I would be happy if we could get rid of the pesky $(X, Y, Z)$.

20. Jul 28, 2017

### David Olivier

If we have other kinds of detectors, we may know the time too, and have all three coordinates at a precise time. But as I said, this is not very relevant to my issue.