- #1

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Can every such complex valued function be represented by:

f(z)=u(x,y)+iv(x,y)?

Also, is the limit of the conjugate such a function equal to the conjugate of the limit of the function?

Something like:

lim[conjugate[f]] (?)= conjugate[lim[f]]

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- Thread starter mathsciguy
- Start date

- #1

- 131

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Can every such complex valued function be represented by:

f(z)=u(x,y)+iv(x,y)?

Also, is the limit of the conjugate such a function equal to the conjugate of the limit of the function?

Something like:

lim[conjugate[f]] (?)= conjugate[lim[f]]

- #2

- 418

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Yes. You can always split a complex valued function into its real and imaginary parts. If you had f(z) as a function ℂ→ℂ, then you can define Re[f(z)]=u(z)=u(x,y) and Im[f(z)]=v(z)=v(x,y) for real-valued functions u and v.I have a question about complex valued functions, say f(z) where z=x+iy is a complex variable.

Can every such complex valued function be represented by:

f(z)=u(x,y)+iv(x,y)?

You can even write the real parts and imaginary parts of a complex expression w in a way that is easier to manipulate: (I like to use the notation w* to denote the conjugate of w.)

Re[w] = (w+w*)/2

Im[w] = (w-w*)/2

Let's apply the formula that you just suggested. We have:Also, is the limit of the conjugate such a function equal to the conjugate of the limit of the function?

Something like:

lim[conjugate[f]] (?)= conjugate[lim[f]]

[f(z)]*=[u(x,y) + i v(x,y)]* = u(x,y) - i v(x,y)

Therefore,

lim [(f(z))*]=lim[u(x,y) - i v(x,y)] = lim [u(x,y)] - lim [i v(x,y)] = lim[u(x,y))] - i lim[v(x,y)] = (lim[u(x,y)] + i lim[v(x,y)])*

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- #3

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Anyway, I wonder if you guys can answer a barely tangential question? Let's say I have a complex valued function f(x) of a real variable. If the limit f(x) as x-> infinity is zero, are the derivatives of f(x) as x-> infinity also zero?

- #4

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Well, when you ask about the derivative of a complex-valued function, you usually want the derivative to be defined in a special way that makes it independent of the direction along which you take the derivative. (Unlike say taking the gradient of a real-valued function.) Functions of this sort are called analytic, and I do think that if an analytic function approaches 0 as z→∞, then its derivative must as well. [PS: on second thought, I'm not totally sure of this and I can't find a quick reference. Maybe somebody could clarify on this particular point.]

However there are (non-analytic) real valued functions which disobey that. Like for example look at the function f(x) = sin(x^{2})/x. Clearly it converges to 0 but what about its derivative? You can just stick a z instead of x and you've constructed complex function which disobeys your suggestion.

However there are (non-analytic) real valued functions which disobey that. Like for example look at the function f(x) = sin(x

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- #5

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Hm, well, I also thought so. But see, I was trying to figure out how to show that an operator like L = i(d^3/dx^3) is hermitian for functions f(x) in the interval -(infinity)< x < (infinity) where as f approaches infinity, it also approaches zero. I tried using the straightforward method where I repeatedly do integration by parts but then I'd end up with boundary terms containing the derivatives of f at infinity (all the boundary terms should be zero, and L is hermitian).

Edit: Turns out I can just argue that since for the given boundary conditions, L = -i(d/dx) is hermitian, then L^3 = i(d^3/dx^3) is also hermitian.

Is it because:

if <f|L(g)> = <(L(f))|g> then <f|L^3(g)> = <L^3(f)|g> ?

Edit: Turns out I can just argue that since for the given boundary conditions, L = -i(d/dx) is hermitian, then L^3 = i(d^3/dx^3) is also hermitian.

Is it because:

if <f|L(g)> = <(L(f))|g> then <f|L^3(g)> = <L^3(f)|g> ?

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- #6

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This is one bit of mathematical trivia which often gets swept under the rug. There is a good discussion about this here:

http://mathoverflow.net/questions/1...aints-of-a-wave-function-in-quantum-mechanics

- #7

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So the way you would prove that L

(L

= L L L |ψ> = L

When you form matrix elements <φ|L|ψ>, you're usually not proving general things about the operators since you're projecting them onto a lower-dimensional subspace.

- #8

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Thanks, I think that kind of proof is something I would think of if I was working with matrices.

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