What does it mean for a function to be spherically symmetric?
I have searched but I couldn't find much. I was asked to show if a function $f=f(x,y,z)$, say, is spherically symmetric. Does this mean if we rotate $x,y,z$ axises then the function will stay the same?
Like, if $f=x^2+y^2+z^2$ then this is symmetric spherically? but if $f=x+y+z$ then this is not spherically symmetric? (Just two easy examples I thought of)
Any help is appreciated!
Bit more context:
This was a wave equation I got when doing quantum mechanics, $\psi(x,y,z)=Ae^{-\frac{m\omega (x^2+y^2+z^2)}{2\hbar}}$, which I can see that this is just depending on $|\vec r|$ really but I am struggling to explain in greater detail as I don't seem to have an exact definition. (A is just a constant)
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$\begingroup$With regard to a function in the context given, the phrase spherically symmetric means that the function, which is a function of a vector, depends only on the magnitude of that vector. That is,$$ f(x) = f(y) \qquad\text{whenever}\qquad \|x\| = \|y\|. $$There are other equivalent ways of describing this notion, which can be stated at various levels of rigor (in each statement, it might help to imagine that $n=3$, as that fits the context of the question):
- A function $f : \mathbb{R}^n \to \mathbb{R}$ is spherically symmetric if it is constant on any sphere centered at the origin, i.e.$$\renewcommand{\vec}[1]{\mathbf{#1}} f(\{ \vec{x} : \|\vec{x}\| = r \} ) = c(r), $$where $c$ is some constant which depends only on $r$. Note that this is really just a rephrasing of the definition I gave above.
- A function $f : \mathbb{R}^n \to \mathbb{R}$ is spherically symmetric if it is invariant under the action of an orthogonal (unitary) transformation. That is,$$ f(O\vec{x}) = f(\vec{x}), $$where $O$ is any orthogonal matrix. In particular, this means that $f$ is invariant under "rotation", though I am somewhat hesitant to use that language, as the precise definition of "rotation" is a little unclear in anything other than two dimensions[1], whereas "orthogonal transformation" is clear in any dimension.
- Via a slight abuse of notation, authors will often say that a function $f : \mathbb{R}^n \to \mathbb{R}$ is spherically symmetric if$$ f(\vec{x}) = f(\|\vec{x}\|). $$I say that this is an abuse of notation since the function $f$ on the left-hand side is a function of a vector, while the function $f$ on the right-hand side is a functon of a single real variable. Since the domains don't match, these cannot be the same function, hence it is a slight abuse of notation to name the function $f$ on both sides of the equation. However, such an abuse of notation is common. One might also see$$ f(\vec{x}) = f(r), $$where $r = \|\vec{x}\|$. In three dimensions, this means that$$ f(x,y,z) = f(r), $$where $r^2 = x^2 + y^2 + z^2$.
Given the context provided in the question, it may be worth mentioning that the wave equation in three dimensions is solved by a function which is spherically symmetric in the sense given above.
[1] I think that, in general, "rotation" is equivalent to "$SO(n)$ action", i.e. a rotation is the action of an element of the special orthogonal group, which is basically the set of $n\times n$ dimensional matrix with determinant $1$. The orthogonal group consists of matrices with determinant $\pm 1$. These matrices give rotations (if the determinant is $1$), plus rotations with reflection (if the determinant is $-1$). A spherically symmetric function is invariant under reflection so, hence the description above.
$\endgroup$ $\begingroup$As Xander says in the comments, my best guess at what's meant is that $f$ is invariant under the symmetries of the sphere, i.e. $f(Rx) = f(x)$ for any rotation $R$.
I've sometimes seen similar terminology for other kinds of symmetry, i.e. "$f(x,y,z)=x^4+y^4+z^4$ has octahedral symmetry".
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