Calculate the commutator of the vector fields $A=x\partial_y-y\partial_x$ and $B=x\partial_x+y\partial_y$

Question

Q: Given the vector fields

$A=x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}$,

$B=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}$

Calculate the commutator $\left[A,B\right]$.

My thoughts: I immediately assumed we had the following

$x\frac{\partial}{\partial y}\left(x\frac{\partial}{\partial x}\right)-x\frac{\partial}{\partial x}\left(x\frac{\partial}{\partial y}\right)+x\frac{\partial}{\partial y}\left(y\frac{\partial}{\partial y}\right)-y\frac{\partial}{\partial y}\left(x\frac{\partial}{\partial y}\right)+…$

where I've used the standard commutation relations. However I'm now unsure as to whether it should actually be

$x\frac{\partial}{\partial y}\left(x\right)\frac{\partial}{\partial x}-x\frac{\partial}{\partial x}\left(x\right)\frac{\partial}{\partial y}+x\frac{\partial}{\partial y}\left(y\right)\frac{\partial}{\partial y}-y\frac{\partial}{\partial y}\left(x\right)\frac{\partial}{\partial y}+…$

This has confused me slightly because in Quantum Mechanics I would usually have done the former calculation and perhaps applied this to some function $f$. This is a differential geometry question where we're considering vector fields and (I'm assuming) these objects commute. I'm now tending towards the latter though because it makes the calculation a lot easier!

So help with this particular problem would be great. But also an explanation of whether this would differ if we had operators (in QM for instance) and whether the correct method of the two above is $\textit{always}$ the one to use (if we're considering the commutations of differentials $\frac{\partial}{\partial x^i}$) would be perfect.

Answer 1

My somewhat rusty knowledge says that: $$ \left[A,B\right] = A B - B A = \left( x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right) \left( x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}\right)- \left( x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}\right) \left( x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right) $$ The following basic formula from Operator Calculus will be employed repeatedly: $$ \frac{d}{dx} f(x) = f(x) \frac{d}{dx} + \frac{df}{dx} $$ Evaluate $(A B)$: $$ x\frac{\partial}{\partial y} x\frac{\partial}{\partial x} + x\frac{\partial}{\partial y} y\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} x\frac{\partial}{\partial x} - y\frac{\partial}{\partial x} y\frac{\partial}{\partial y} =\\ x\frac{\partial}{\partial y} x\frac{\partial}{\partial x} + \left[ y\frac{\partial}{\partial y} x\frac{\partial}{\partial y} + x\left( \frac{\partial y}{\partial y} \right) \frac{\partial}{\partial y} \right] - \left[ x\frac{\partial}{\partial x} y\frac{\partial}{\partial x} + y\left( \frac{\partial x}{\partial x} \right) \frac{\partial}{\partial x} \right] - y\frac{\partial}{\partial x} y\frac{\partial}{\partial y} =\\ x\frac{\partial}{\partial y} x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y} x\frac{\partial}{\partial y} + x\frac{\partial}{\partial y} - x\frac{\partial}{\partial x} y\frac{\partial}{\partial x} - y\frac{\partial}{\partial x} - y\frac{\partial}{\partial x} y\frac{\partial}{\partial y} $$ Evaluate $(B A)$: $$ x\frac{\partial}{\partial x} x\frac{\partial}{\partial y} - x\frac{\partial}{\partial x} y\frac{\partial}{\partial x} + y\frac{\partial}{\partial y} x\frac{\partial}{\partial y} - y\frac{\partial}{\partial y} y\frac{\partial}{\partial x} =\\ \left[x\frac{\partial}{\partial y} x\frac{\partial}{\partial x} + x\left(\frac{\partial x}{\partial x}\right)\frac{\partial}{\partial y}\right] - x\frac{\partial}{\partial x} y\frac{\partial}{\partial x} + y\frac{\partial}{\partial y} x\frac{\partial}{\partial y} - \left[y\frac{\partial}{\partial x} y\frac{\partial}{\partial y} + y\left(\frac{\partial y}{\partial y}\right)\frac{\partial}{\partial x}\right] =\\ x\frac{\partial}{\partial y} x\frac{\partial}{\partial x} + x\frac{\partial}{\partial y} - x\frac{\partial}{\partial x} y\frac{\partial}{\partial x} + y\frac{\partial}{\partial y} x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} y\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} $$ We see that $A B = B A$ . Thus the commutator is zero.
This must be so because rotations and scaling around the origin in the plane commute.
More of the mathematics can be found in: Lie Groups and Variances .

Answer 2

The easiest way I know to evaluate the commutator or Lie bracket $[A, B]$ of two vector fields such as

$A=x\dfrac{\partial}{\partial y}-y\dfrac{\partial}{\partial x} \tag{1}$

and

$B=x\dfrac{\partial}{\partial x}+y\dfrac{\partial}{\partial y} \tag{2}$

is to apply it to some differentiable funcction $f$, and then simply work out the result by successive differentiation and application of the Leibniz rule for products, etc. This method works because $[A, B]$ is determined by its action on functions, it is simple to apply and to remember. So if $f$ is any sufficiently (probably twice) differentiable function, we have

$A[f] = (x\dfrac{\partial}{\partial y}-y\dfrac{\partial}{\partial x})[f] = xf_y - yf_x, \tag{3}$

$B[f] = (x\dfrac{\partial}{\partial x}+y\dfrac{\partial}{\partial y})[f] = xf_x + yf_y, \tag{4}$

where we have used the subscript notation for partial derivatives, $f_x = \partial f / \partial x$ etc. Then (3) and (4) show that $A[f]$ and $B[f]$ are themselves functions, we can apply $B$ to (3) and $A$ to (4), obtaining

$BA[f] = B[A[f]] = (x\dfrac{\partial}{\partial x} + y\dfrac{\partial}{\partial y})[xf_y - yf_x] = x\dfrac{\partial}{\partial x}[xf_y - yf_x] + y\dfrac{\partial}{\partial y}[xf_y - yf_x]$ $=x(f_y + xf_{xy} - yf_{xx}) + y(xf_{yy} - f_x - yf_{yx}) \tag{5}$

and

$AB[f] = A[B[f]] = (x\dfrac{\partial}{\partial y}-y\dfrac{\partial}{\partial x})[xf_x + yf_y] = x\dfrac{\partial}{\partial y}[xf_x + yf_y] - y\dfrac{\partial}{\partial x}[xf_x + yf_y]$ $= x(xf_{yx} + f_y + yf_{yy}) - y(f_x + xf_{xx} + yf_{yx}); \tag{6}$

it is now a relatively simple algebraic maneuver to subtract the right-hand sides of (5) and (6), obtaining

$[A, B][f] = (AB - BA)[f] = 0, \tag{7}$

or

$[A, B] = 0, \tag{8}$

since as we have said vector fields are determined by their application to functions; thus (7) implies (8), and the commutator $[A, B]$ of $A$ and $B$ vanishes. Of course, the preceding calculation is a little bit of a grind and I for one have to credit Han de Bruijn for having the insight to realize that we must have $[A, B] = 0$ for purely geometrical reasons.

As for the "difference" between the first and second methods of calculation mentioned in the question, a careful scrutiny of the two equations shows that these methods are in fact the same, once on realizes that the second-order derivative operators entirely wash out of the first expression, leaving only first order operators behind to form $[A, B]$. Indeed, expanding out the first two terms of the first equation yields

$x\dfrac{\partial}{\partial y}\left(x\dfrac{\partial}{\partial x}\right)-x\dfrac{\partial}{\partial x}\left(x\dfrac{\partial}{\partial y}\right) = x\dfrac{\partial}{\partial y}(x)\dfrac{\partial}{\partial x} + x^2\dfrac{\partial^2}{\partial y \partial x} - x\dfrac{\partial}{\partial x}(x)\dfrac{\partial}{\partial y} - x^2\dfrac{\partial^2}{\partial x \partial y}$ $= x\dfrac{\partial}{\partial y}(x)\dfrac{\partial}{\partial x} - x\dfrac{\partial}{\partial x}(x)\dfrac{\partial}{\partial y}, \tag{9}$

in agreement with the second. The reason is, of course, that the second order operators $\partial^2 / \partial y \partial x$ and $\partial^2 / \partial x \partial y$ agree on any sufficiently differentiable function $f$; since vector fields are defined by their action on such functions, it is clear the second method of calculation is essentially the same as the first as far as vector fields are concerned.

Vector fields do not in general commute. To see an easy example, just consider $X =\partial / \partial x$ and $Y = x \partial / \partial y$; then for any suitable function $g$ we have

$[X, Y]g = X[Y[g]] - Y[X[g]] = \dfrac{\partial}{\partial x}(xg_y) - x\dfrac{\partial}{\partial y}g_x = g_y + xg_{xy} - xg_{yx} = g_y; \tag{10}$

since (10) holds for all sufficiently smooth $g$, we see that in fact

$[X, Y] = \dfrac{\partial}{\partial y}. \tag{11}$

Coordinate vector fields, however, do commute, for example

$[\dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}] = 0 \tag{12}$

by virtue of the fact that $g_{xy} = g_{yx}$. Furthermore, it is in fact possible to calculate the commutator of any two vector fields $V = V^x\frac{\partial}{\partial x} + V^y\frac{\partial}{\partial y}$ and $W = W^x\frac{\partial}{\partial x} + W^y\frac{\partial}{\partial y}$ where $V^x, V^y, W^x, W^y$ are functions of $x$ and $y$:

$[V, W]g = V[W[g]] - W[V[g]]$ $= (V^x\dfrac{\partial}{\partial x} + V^y\dfrac{\partial}{\partial y}) (W^x\dfrac{\partial}{\partial x} + W^y\dfrac{\partial}{\partial y})[g] - (W^x\dfrac{\partial}{\partial x} + W^y\dfrac{\partial}{\partial y})(V^x\dfrac{\partial}{\partial x} + V^y\dfrac{\partial}{\partial y})[g]$ $= (V^x\dfrac{\partial}{\partial x} + V^y\dfrac{\partial}{\partial y})(W^xg_x + W^yg_y) - (W^x\dfrac{\partial}{\partial x} + W^y\dfrac{\partial}{\partial y})(V^xg_x + V^yg_y)$ $= V^x\dfrac{\partial}{\partial x}(W^xg_x + W^yg_y) + V^y\dfrac{\partial}{\partial y}(W^xg_x + W^yg_y)$ $- W^x\dfrac{\partial}{\partial x}(V^xg_x + V^yg_y) - W^y\dfrac{\partial}{\partial y}(V^xg_x + V^yg_y)$ $=V^x(W^x_xg_x + W^xg_{xx} + W^y_xg_y + W^yg_{yx}) + V^y(W^x_yg_x + W^xg_{xy} + W^y_yg_y + W^yg_{yy})$ $-W^x(V^x_xg_x + V^xg_{xx} + V^y_xg_y + V^yg_{yx}) - W^y(V^x_yg_x + V^xg_{xy} + V^y_yg_y + V^yg_{yy}). \tag{13}$

A careful inspection of the terms occurring on the extreme right (last two lines) of (13) reveals that every one containing a second derivative of $g$, $g_{xx}$, $g_{yx}$, etc., cancels out and we are left with terms containing only the first derivatives of $g$; such cancellation, of course, depends on the fact that $g_{xy} = g_{yx}$, which is one reason we stipulate that $f, g$ be at least $C^2$ functions. If we gather the first derivative terms of (13) together and group them by independent variable, then we will obtain an expression for $[V, W]$ in terms of the basis vector fields $\partial / \partial x$ and $\partial / \partial y$; indeed we have

$[V, W]g = V[W[g]] - W[V[g]]$ $= (V^xW_x^x + V^yW_y^x - W^xV^x_x - W^yV^x_y)g_x + (V^xW^y_x + V^yW_y^y - W^xV^y_x - W^yV_y^y)g_y$ $= ((V[W^x] - W[V^x])\dfrac{\partial}{\partial x} + (V[W^y] - W[V^y])\dfrac{\partial}{\partial y})[g], \tag{14}$

in which the expressions $V[W^x]$ etc. are simply derivatives of the $x$ and $y$ component functions of $V$ and $W$ in the $V$ and $W$ directions, as $V[g]$ is that of $g$ in the direction $V$. (14) shows that

$[V, W] = (V[W^x] - W[V^x])\dfrac{\partial}{\partial x} + (V[W^y] - W[V^y])\dfrac{\partial}{\partial y} \tag{15}$

itself is in fact a first order differential operator or vector field. The calculations (13)-(14), a rather long-winded, grungy-but-you-might-as-well-see-the-whole-mess-at-least-once lot, may in fact be considerably streamlined if one adopts certain standard identities which apply to the Lie bracket or commutator operation:

$[X + Y, Z] = [X, Z] + [y, Z] \tag{16}$

and

$[fX, Y] = f[X, Y] - Y[f]X; \tag{17}$

of these, the first is virtually self-evident and so I leave its demonstration to the reader. The second is almost as easily seen if we apply $[X, Y]$ to some function $g$:

$[fX, Y][g] = fX[Y[g]] - Y[fX[g]]$ $= fX[Y[g]] - Y[f]X[g] - fY[X[g]] = f[X, Y][g] - Y[f]X[g]; \tag{18}$

it should further be noted that if we negate each side of (17) and then use the identity $[V, W] = -[W, V]$ we obtain

$[Y, fX] = f[Y, X] + Y[f]X; \tag{19}$

we apply (16), (17), (19) to $[V, W]$ with $V = V^x\frac{\partial}{\partial x} + V^y\frac{\partial}{\partial y}$:

$[V, W] = [ V^x\dfrac{\partial}{\partial x} + V^y\dfrac{\partial}{\partial y}, W] = [V^x\dfrac{\partial}{\partial x}, W] + [V^y\dfrac{\partial}{\partial y}, W]$ $= V^x[\dfrac{\partial}{\partial x}, W] - W[V^x]\dfrac{\partial}{\partial x} + V^y[\dfrac{\partial}{\partial y}, W] - W[V^y]\dfrac{\partial}{\partial y}, \tag{20}$

and now we repeat the process with $W = W^x\frac{\partial}{\partial x} + W^y\frac{\partial}{\partial y}$:

$[\dfrac{\partial}{\partial x}, W] = [\dfrac{\partial}{\partial x}, W^x\dfrac{\partial}{\partial x} + W^y\dfrac{\partial}{\partial y}]$ $= [\dfrac{\partial}{\partial x}, W^x\dfrac{\partial}{\partial x}] + [\dfrac{\partial}{\partial x},W^y\dfrac{\partial}{\partial y}] = W_x^x\dfrac{\partial}{\partial x} + W_x^y\dfrac{\partial}{\partial y}, \tag{21}$

and likewise

$[\dfrac{\partial}{\partial y}, W] = W_y^x\dfrac{\partial}{\partial x} + W_y^y\dfrac{\partial}{\partial y}, \tag{22}$

(21) and (22) holding since $[\frac{\partial}{\partial x}, \frac{\partial}{\partial y}] = 0$, and $[Z, Z] = 0$ for any vector field $Z$, always. Bringing together (20), (21), and (22) we see that

$[V, W]$ $= V^x(W_x^x\dfrac{\partial}{\partial x} + W_x^y\dfrac{\partial}{\partial y}) - W[V^x]\dfrac{\partial}{\partial x} + V^y(W_y^x\dfrac{\partial}{\partial x} + W_y^y\dfrac{\partial}{\partial y}) - W[V^y]\dfrac{\partial}{\partial y}; \tag{23}$

if the terms of (23) are gathered together and regrouped then it is easy to arrive at

$[V, W] = (V[W^x] - W[V^x])\dfrac{\partial}{\partial x} + (V[W^y] - W[V^y])\dfrac{\partial}{\partial y}, \tag{24}$

i.e., (15). Systematic deployment of the identities (16), (17), (19) allows us to find the expression (15), (24) for $[V, W]$ in terms of the coordiante basis $\partial / \partial x$, $\partial / \partial y$ in a somewhat more streamlined manner than the derivation (13), (14) which takes everything back to basic definitions in terms of the differential operators $\partial / \partial x$, $\partial / \partial y$.

The above remarks pertain, of course, to vector fields on manifolds in the context of differential topology/geometry. When one turns to quantum mechanics, however, the situation is somewhat different. Though both the Lie theory we have discussed above and the theory of operators on Hilbert spaces, which is the framework for much of quantum mechanics, have much in common, there are significant differences. Consder things from the point of view of the nature of the operators and the spaces on which they are defined. In the Lie approach to vector fields (at least in the stream-lined version presented here), they are construed to be first-order differential operators on an appropriate function space, which here is taken to be $C^\infty(\Bbb R^2, \Bbb R)$; in this way we assure the necessary property $f_{xy} = f_{yx}$ which allows the theory to fly in the sense that then $[X, Y]$ will be a first-order operator if $X$ and $Y$ are, as has hopefully been made (perhaps painfully) evident in the above discussion. In the quantum mechanical case, however, the underlying space is a Hilbert space, which may be given concrete form by taking it to be, for example, $L^2(\Bbb R^2, \Bbb C)$. The operators are then linear maps defined either on all of $L^2(\Bbb R^2, \Bbb C)$ or on some dense subspace, as is the case with $p_x = i \hbar (\partial / \partial x)$ or $H = -(\hbar / 2m)\nabla^2$ etc. And though in either case we may define bracket operations $[A, B]$, the precise definitions differ, though there are evident similarities. So from a purely computational point of view, it is likely best to stick with the first method, which keeps track of all derivatives until the very end, rather than the second, which uses a short-cut which as far as I can tell depends on the domain being $C^\infty$. Indeed, since $L^2$ contains non-$C^2$ functions, it is not clear exactly how the commutator of two first derivative maps will be one itself, e.g. what is $[p_x, xp_y]$ going to do, exactly, to a non-$C^2$ element of $L^2$? Though I think the quantum mechanics (and here I refer to the practitioners of quantum mechanics, the subject) have developed answers which depend on the theory of unbounded operators on Hilbert spaces. And that's as far as I can take these things in this post.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Answer 3

Here is the general formula for calculating commutator of vector fields, taken from do Carmo's "Riemannian Geometry", page 27: If $$ X=\sum_{i} a_i \frac{\partial}{\partial x_i}, $$ $$ Y=\sum_{i} b_i \frac{\partial}{\partial x_i}, $$ then $$ [X,Y]= \sum_{i,j} (a_i \frac{\partial b_j}{\partial x_i} - b_i \frac{\partial a_j}{\partial x_i}) \frac{\partial}{\partial x_j}. $$ Now, you can do your computation.