Partial derivatives treating a complex number and its complex ...

Partial derivatives treating a complex number and its complex
conjugate as independent variables: Why the trick works
David Foster
October 27, 2012
Think of a complex number z. Now add a small change dz to z, such that the value of z changes
(obviously) but the value of its complex conjugate z ∗ does not change. You quickly realize this is impossible.
There is no way that z can change while z ∗ is held constant. And yet, there it is, in your textbook or your
lecture notes, something like:
. . . since 〈ψ|H − λ|ψ〉 is variational in |ψ〉, we can take the functional partial derivative with
respect to 〈ψ| and set it equal to zero to get the eigenvalue equation
H|ψ0〉 = λ|ψ0〉.
We essentially just took the (functional) 1 partial derivative with respect to 〈ψ| while treating |ψ〉 as an
unchanging, independent variable. It appears that would entail changing (say we go to the real space basis)
ψ ∗ (x) without changing ψ(x), which you know is impossible.
While such a variation is not OK, we can proceed as if it were because the resulting equations are true.
The reason has something to do with the fact that there are two independent degrees of freedom of ψ at any
point, the real and imaginary parts, and that variation with real and imaginary parts leads to two separate
equations.
To prove that the result is correct, it’s easiest to leave the bra-ket notation and just deal with ψ and ψ ∗ ,
which may be either functions of a continous variable or finite dimensional vectors. We then suppose that
we have a functional quantity, F that is variational about some special function ψ0. But F is expressed as a
functional of two arguments: F [ψ, ψ ∗ ]. We further suppose that F is “analytic” in both of its arguments, that
is, that the functional derivatives δF/δψ and δF/δψ ∗ are well defined. This is the case of interest, because
the goal of the trick is to set δF/δψ (and/or δF/δψ ∗ ) to zero and this would be useless if those expressions
were not well defined or unknown. An example of a “non-analytic” F would be an explicit functional of its
arguments that took the complex conjugate of one of it’s arguments somewhere in it’s evaluation. The most
common physical example of an “analytic” functional would be ψ ∗ (x)f(x)ψ(x)dx.
We now proceed to properly write δF = 0 as
δψ · δF
δψ + δψ∗ · δF
= 0 (1)
δψ∗ 1 You may also encounter a direct example with standard partial derivatives. My first encounter with this trick was the
sentence, “Note that it is convenient now to read P as a function of the two independent variables α and α ∗ ”, in the book “An
Open Systems Approach to Quantum Optics” by Howard Carmichael (Springer, 1991).
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where the dot indicates the appropriate inner product (integral or vector contraction). At this point it is not
clear that δF = 0 implies the two functional derivatives in the equation are separately equal to zero. First
we write the real and imaginary parts of this equation separately.
Re δψ · Re δF
δF
− Im δψ · Im
δψ δψ + Re δψ∗ · Re δF
δψ∗ − Im δψ∗ · Im δF
= 0
δψ∗ (2)
Re δψ · Im δF
δF
+ Im δψ · Re
δψ δψ + Re δψ∗ · Im δF
δψ∗ + Im δψ∗ · Re δF
= 0
δψ∗ (3)
Now we use the facts that Re δψ∗ = Re δψ and Im δψ∗ = −Im δψ. (Of course we only do this for numerator
quantities). The above equations become
Re δψ · Re δF δF
+ Re
δψ δψ∗
+ Im δψ · Im δF
δF
− Im = 0 (4)
δψ∗ δψ
Re δψ · Im δF δF
+ Im
δψ δψ∗
+ Im δψ · Re δF δF
− Re
δψ δψ∗
= 0 (5)
Multiplying Eq. (5) by i and adding to Eq. (4) gives an equation with the real and imaginary parts of the
partial derivatives recombined
δF δF
Re δψ · +
δψ δψ∗
δF δF
= −i Im δψ · −
δψ δψ∗
(6)
Since the real and imaginary parts of δψ can be varied independently, both quantities in square brackets
must be zero. This quickly leads to the desired equations
δF
= 0
δψ
(7)
δF
= 0
δψ∗ (8)
By starting with the proper differential of F in Eq. (1), we have derived something that looks improper.
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