# Gateaux derivative

1. Mar 15, 2008

### braindead101

suppose A->H is a symmetric operator for some Hilbert Space H, define $$\varphi$$: H->R by $$\varphi$$(u) = <Au,u>. Compute the Gateaux derivative of $$\varphi$$

I know the definition of the Gateaux derivative, I just don't know how to use the information given to compute it.
The definition of the Gateaux derivative of J of y in direction v is, ep is epsilon: gateaux J(y,v) lim ep->0 (J(y+epv) - J(y)) / ep

2. Mar 15, 2008

### Hurkyl

Staff Emeritus
Where exactly are you stuck in the computation? As far as I can tell, it should be a straightforward plug-and-chug.

3. Mar 15, 2008

### braindead101

i've never done a question where you plug in an inner product. my attempt would be just this:
lim ep->0 <Au+ep,u+ep> - <Au,u> / ep
how do i go from there?

and i know that the symmetric operator tells u that <Au,u> = <u,Au>

4. Mar 15, 2008

### Hurkyl

Staff Emeritus
What happened to v? And to the parentheses around the numerator?

Don't you know any rules for simplifying / expanding expressions involving inner products? (They are essentially the same rules you use when the product is ordinary multiplication)

5. Mar 15, 2008

### braindead101

i'm sorry i don't know anything about inner product, i will be learning it next week for my linear algebra class, but it will be too late for this class as this assignment is due tuesday.
if you can inform me about these rules, i will try to apply them and do this problem.

Sorry, i missed the v, but is this correct?
lim ep->0 <Au+epv,u+epv> - <Au,u> / ep
Can i subtract this so it becomes this:
lim ep->0 <epv, epv> / ep
lim ep->0 ep<v,v> / ep
<v,v>?
i think this is incorrect as i did not get to use the fact that <Au,u>=<u,Au> anywhere.

6. Mar 15, 2008

### Hurkyl

Staff Emeritus
You haven't learned any of the properties of < , > in your quantum course?

As its name suggests, it's a kind of product -- one of the more relevant points being that it has the distributive property, and (to an extent) the associative property.

I noticed another error in your plugging in: when you replaced u with $u + \epsilon v$ in the definition of $\varphi$, you again forgot parentheses where necessary.

7. Mar 15, 2008

### braindead101

ok let me try again!
no i have not learn this because the teacher expected it to be known as this is a 4th year course and you learn it in the 1st year course (which i am taking at the same time, and this section has not came up yet)

my attempt once again, please correct me as I really do not know what I am doing:
lim ep->0 <A(u+epv),u+epv> - <Au,u> / ep
lim ep->0 <Au+Aepv,u+epv> - <Au,u> / ep
From online site: <ax+by,z> = a<xz> + b<yz> so can i do this:
lim ep->0 A<u(u+epv)> + A<epv(u+epv)> - <Au,u> / ep

ok ... can u please give me a starting point or the next step as i am really stuck.

8. Mar 16, 2008

### Hurkyl

Staff Emeritus
Well, it might be worth looking ahead in your algebra book... but I'll spell out some of its relevant properties:

u,v,w are vectors
a is a scalar
T is an operator

distributivity:
<u+v, w> = <u, w> + <v, w>
<u, v + w> = <u, v> + <u, w>

anti-symmetry:
<u, v> = <v, u>*

associativity:
<au, v> = a* <u, v>
<u, av> = a <u, v>
<T*u, v> = <u, Tv>

Pay attention to the similarities and differences between these and ordinary multiplication -- they share the most important properties, and so much of your intuition about ordinary multiplication still applies here. But also don't forget to pay heed of what does not apply. (e.g. Operators associate in a limited way)

Incidentally, you can always rewrite an inner product in terms of vector algebra: each inner product defines a transpose operation, such that

$\langle u, v \rangle = u^* v$

In particular, if H is finite-dimensional, then relative to a certain basis, the coordinates of v would be a column vector, those of u would be a row vector, and this would be the usual conjugate-transpose operation. The coordinates of T would be a square matrix. Try rewriting each of the above identities in this fashion, to see what you think of it.

Last edited: Mar 16, 2008
9. Mar 16, 2008

### braindead101

associativity:
<u, av> = <u, av>

is there a typo here or are you trying to say that there's nothing you can do to that?

ok, i see what i did wrong, here is my revised attempt at the problem.

lim ep->0 <A(u+epv),u+epv> - <Au,u> / ep
lim ep->0 <A(u+epv),u>+<A(u+epv),epv>-<Au,u> / ep
lim ep->0 <Au,u> + <Aepv,u> + <Au,epv> + <Aepv,epv> - <Au,u> / ep
lim ep->0 <Aepv,u> + <Au,epv> + <Aepv,epv> / ep

ok, so now i know i have to pull out the ep to cancel with the bottom.
since ep is a scalar. but i am not sure how you pull it out when it is in this location, < , ep> as in your associativity property, it doesn't get pulled out. so i left it inside, and pulled it out for the ones where i could use the first associativity property.

lim ep->0 ep<Av,u> + <Au,epv> + ep<Av,epv> / ep

Using symmetric operator:
you gave me anti-symmetric. and i stated previously that i thought symmetric was <Au,u>=<u,Au> but you did not say whether this is correct or not, so i will assume i can use this.

lim ep->0 ep<Av,u> + <epv,Au> + ep<Av,epv> / ep
lim ep->0 ep<Av,u> + ep<v,Au> + ep<Av,epv> / ep
cancelling out the ep from top and bottom.

lim ep->0 <Av,u> + <v,Au> + <Av,epv>
so when ep->0 i am left with just
<Av,u> + <v,Au>

is this correct?

10. Mar 16, 2008

### Hurkyl

Staff Emeritus
It was a typo. I also found another typo: <au,v> = a* <u,v>. I've corrected these in my post.

11. Mar 16, 2008

### Hurkyl

Staff Emeritus
<Au,u>=<u,Au> does not necessarily mean <Au,v>=<u,Av> nor <Au,v>=<v,Au>! Symmetric means, IIRC, that <Au,v>=<u,Av> -- it doesn't let you swap the vectors around with it.

But aside from that error, you have the right idea here.

And if you were so inclined, you could apply symmetry to simplify this.

12. Mar 16, 2008

### braindead101

okay, so using your updated associativity rules, my answer is:

lim ep->0 <A(u+epv),u+epv> - <Au,u> / ep
lim ep->0 <A(u+epv),u>+<A(u+epv),epv>-<Au,u> / ep
lim ep->0 <Au,u> + <Aepv,u> + <Au,epv> + <Aepv,epv> - <Au,u> / ep
lim ep->0 <Aepv,u> + <Au,epv> + <Aepv,epv> / ep
associativity:
lim ep->0 ep*<Av,u> + ep<Au,v> + ep*<Av,epv> / ep

now i am confused as to whether i can cancel ep* out?
so if i can cancel out the ep* and the ep
it would be just
<Av,u> + <Au,v>
and since you said i cannto swap the vectors around, i guess that is just the final answer.. i don't see how i can simplify it.

how about now?

13. Mar 16, 2008

### Hurkyl

Staff Emeritus
If ep is a real variable, then ep* = ep. I suppose it's probably meant to be complex though -- you can always normalize it. Remember that (ep*)ep = |ep|^2

Antisymmetry of the inner product lets you swap the factors around; symmetry of the operator lets you shift the operator itself around. Equivalently, A* = A for your A, and you can apply that rule....

14. Mar 16, 2008

### braindead101

lim ep->0 <A(u+epv),u+epv> - <Au,u> / ep
lim ep->0 <A(u+epv),u>+<A(u+epv),epv>-<Au,u> / ep
lim ep->0 <Au,u> + <Aepv,u> + <Au,epv> + <Aepv,epv> - <Au,u> / ep
lim ep->0 <Aepv,u> + <Au,epv> + <Aepv,epv> / ep
associativity:
lim ep->0 ep*<Av,u> + ep<Au,v> + ep*<Av,epv> / ep
so ep*(ep) = |ep|^2

lim ep->0 |ep|^2/ep <Av,u> + ep<Au,v> + |ep|^2/ep <Av,epv> / ep
can i then simplify |ep|^2/ep to ep.

lim ep->0 ep<Av,u> + ep<Au,v> + ep<Av,epv> / ep
so,
lim ep->0 <Av,u> + <Au,v> + <Av,epv>
lim ep->0 <Av,u> + <Au,v> + ep<Av,v>
<Av,u> + <Au,v>
anti-symmetry:
<u, v> = <v, u>*
<Av,u> + <Av,u>*
can i then bring this * into the inner product:
<Av,u> + <A*v*,u*>
since: A=A*
<Av,u> + <Av*,u*>

is there anything else i can do? and have i done any steps that are incorrect?

15. Mar 16, 2008

### Hurkyl

Staff Emeritus
This you cannot do. The expression doesn't simplify in the way I thought, but it does simplify.

As you did:

<Av,u> + <Au,v>
=
<Av, u> + <v,Au>* (don't forget this step!)
=
<Av, u> + <Av, u>*

then this is just a complex number added to its complex conjugate.

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