Chain rule for functions of operators?

[pellman]

This is strictly a math question but I figured that since it is something which would show up in QM, the quantum folks might be already familiar with it.

Suppose we have an operator valued function A(x) of a real parameter x and another function f, both of which have well defined derivatives.

consider $$\frac{d}{dx}f(A(x))$$

Does this equal

$$\frac{df}{dA}\frac{dA}{dx}$$

or

$$\frac{dA}{dx}\frac{df}{dA}$$

or something else? Of course, if A and dA/dx commute, then either expression is good. But it is not clear to me that A and dA/dx would necessarily commute.

 

[Fredrik]

How do you define df/dA? (I don't think it's something we even want to define).

 

[pellman]

How do you define df/dA? (I don't think it's something we even want to define).

If f(u) is R --> R and has a Taylor series representation

$$f(u)=\Sigma \frac{1}{n!}f_n u^n$$

where the f_n are just coefficients. Then

$$f'(u)=\Sigma \frac{1}{n!}f_{n+1} u^n$$


We can similarly put

$$f(A)=\Sigma \frac{1}{n!}f_n A^n$$

$$f'(A)=\Sigma \frac{1}{n!}f_{n+1} A^n$$

For some f this may not work, it may not converge, blah, blah, blah. Let's just assume f is a function for which this works. The actual function I am interested in is $$f(A)=e^A$$, so f(A) = df/dA anyway.

 

[pellman]

Maybe I shouldn't be so general. My problem is this:

Suppose we have time-dependent Hamiltonian H(t). Then we can no longer write

$$|\Psi(t)\rangle = e^{-iHt}|\Psi(0)\rangle$$

because dH/dt != 0 . What we need is an operator Q(t) such that dQ/dt=H .

Then we would have

$$i\frac{d}{dt}|\Psi(t)\rangle =i\frac{d}{dt}e^{-iQ(t)}|\Psi(0)\rangle$$

$$=He^{-iQ(t)}|\Psi(0)\rangle$$

or would it be

$$=e^{-iQ(t)}H|\Psi(0)\rangle$$?

If H does not commute with Q, then the latter means we are not dealing with a solution to the Schrödinger equation. So what is

$$\frac{d}{dt}e^{-iQ(t)}=?$$

 

[Fredrik]

(I wrote this before I saw your last post).

I think that's a directional derivative in the direction of A

$$\lim_{t\rightarrow 0}\frac{f(A+t\frac{A}{\|A\|})-f(A)}{t}$$

Both the df/dA notation and the f'(A) notation seem very inadequate for directional derivatives. You could use something like $D_X f(A)$ for the directional derivative in direction X, at A. Your df/dA would then be $D_A f(A)$. However, when we take the derivate of exponentials, don't we always do it with respect to a parameter? For example, when we prove that A is self-adjoint if U=exp(itA) is unitary:

$$U^\dagger U=I$$

$$U^\dagger=U^{-1}$$

$$e^{-itA^\dagger}=e^{-itA}$$

Now apply $$\frac{d}{dt}\bigg|_0$$ to both sides, and we're done.

Added after I read your post #4: If Q(t) commutes with Q(s) for all t and s, then Q'(t) commutes with Q(t) and therefore with exp(iQ(t)), so the two options are equivalent. I need to think about the possibility that Q(t) doesn't commute with Q(s).

 

[dextercioby]

If your A is either self-adjoint or unitary in a (rigged) Hilbert space, then you can easily define a function f(A) by the means of the spectral decomposition of A. Then you can compute a derivative, but, of course, under tight conditions of convergence.

 

[Fredrik]

I need to think about the possibility that Q(t) doesn't commute with Q(s).

I don't think there are any simple formulas in this case. Note e.g. that d/dt Q(t)²=Q'(t)Q(t)+Q(t)Q'(t). So if we try to apply d/dt to each term of the exponential, things are already weird in the second order term.

 

[pellman]

I see what you mean. Darn. I was hoping this would have a simple answer.

 

[strangerep]

If your A is either self-adjoint or unitary in a (rigged) Hilbert space, then you can easily define a function f(A) by the means of the spectral decomposition of A. Then you can compute a derivative, but, of course, under tight conditions of convergence.

Umm,... how does this work when one is dealing is a continuous family of operators
such as A(t) ?

E.g., for a given time, we have an operator $A_0 = A(t=0)$, (assumed to self-adjoint, say),
then we can spectral-decompose in terms of its eigenvalues and eigenstates:

$$f(A_0) ~=~ \int da_0 f(a_0) |a_0\rangle \langle a_0| ~~.$$

But each A(t) will have a different set of eigenvalues and eigenstates in general,

$$f(A_t) ~=~ \int da_t f(a_t) |a_t\rangle \langle a_t| ~~.$$

so how does one take the t derivative of the LHS without first computing
the time-dependent eigenvalues and eigenstates explicitly?

(Or did I misunderstand you?)