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