I'm confused in finding the condition for minimum uncertainty, The author in the book I refer goes on saying that


is the condition for minimum uncertainity

for some constant $c$

Where $|f\rangle=(\hat{A}-\langle A\rangle)\psi$

and $|g\rangle=(\hat{B}-\langle B\rangle)\psi$

where $A$ and $B$ are any observables

This is acceptable due to the Schwarz inequaltity (since it becomes an equality when the angle between 2 vectors are zero), the author further says that this equality results if

${\operatorname{Re}\langle f|g\rangle}={0}$

I dont understand why this should be the condition , so if according to the author the angle should be zero, shouldn't

${\operatorname{Im}\langle f|g\rangle}$ be equal to ${0}$ ?

since for any varialbles $a$ and $b$ the argument of $z=a+ib$ can be zero only if the

$\tan^{-1}\frac{b}{a}=0$ which implies $b=0$ or $\operatorname{Im}{z}=0$

Kindly help me understand, any help is appreciated

  • $\begingroup$ Seems totally unacceptable, let c=1, $\hat A=\hat B=\hat\sigma_z$ then you claim any state is minimum uncertainty, but only he eigenstates of $\hat\sigma_z$ have minimum (zero) uncertainty. And there is nothing orthogonal in this example. Further if c=0 is allowed then you could be minimum uncertainty in B but as huge an uncertainty in A as you want. $\endgroup$
    – Timaeus
    Nov 12, 2015 at 21:34

2 Answers 2


The generalized uncertainty principle relating two operators $A$ and $B$ is $$\sigma_A^2 \sigma_B^2 \geq {\left( \frac{1}{2i} \langle\left[ A,B\right] \rangle\right)} ^2$$ where $[A,B]=AB-BA$ is the commutator of the operators. This relation was derived using two inequalities. The first is the Schwarz inequality which, in bra- ket notation is $$\langle{f|f}\rangle\langle{g|g}\rangle \geq |\langle {f|g}\rangle |^2$$ The second inequality is condition on a complex number $z=x+iy$ : $$|z|^2 \geq y^2$$
It’s interesting to see what happens if we require these two relations to be equalities rather than inequalities. This will give us a condition on the functions f and g that gives the minimum uncertainty between them. In the case of the Schwarz inequality, we want $\langle{f|f}\rangle\langle{g|g}\rangle =|\langle{f|g}\rangle |^2$ .

To see what this implies, we can examine the proof of the Schwarz inequality. To this end, we’ll introduce the function, $$|h\rangle =|g\rangle -\frac{\langle{f|g}\rangle}{\langle{f|f}\rangle} |f\rangle$$ since, $\langle{h|h}\rangle \geq 0$ ,
by taking the product with itself, $$\langle{h|h}\rangle =\langle{g|g}\rangle -\frac{\langle{f|g}\rangle}{\langle{f|f}\rangle} \langle{g|f}\rangle -\frac{\langle{g|f}\rangle}{\langle{f|f}\rangle} \langle{f|g}\rangle +\left( \frac{|\langle{f|g}\rangle |}{\langle{f|f}\rangle}\right)^2\langle{f|f}\rangle\\ =\langle{g|g}\rangle -\frac{|\langle{f|g}\rangle |^2}{\langle{f|f}\rangle}\\ \geq 0\\ \implies \langle{f|f}\rangle \langle{g|g}\rangle \geq |\langle{f|g}\rangle |^2$$
In order for this inequality to be replaced with an equality, we would need $|h\rangle =0$ . Thus, from the definition, we have $|g\rangle=\frac{\langle{f|g}\rangle}{\langle{f|f}\rangle} |f\rangle$ That is, the Schwarz inequality becomes an equality if one function is a scalar multiple of the other: $|g\rangle=c|f\rangle$ where c is, in general, a complex scalar. The second inequality is an equality if x=0 so that z is purely imaginary. In our original derivation of the uncertainty principle, we used $z=\langle{f|g}\rangle$ , so if we’re requiring equality, we get $Re(z)=0$


Here $z=\langle{f|g}\rangle$ is imaginary, because for minimum uncertainity we need the complex number to be puerly imaginary, as I stated early. And also $|g\rangle =c|f\rangle$ . Thus, $z=c\langle{f|f}\rangle$ . Since $\langle{f|f}\rangle$ is always real, c must be complex number.

  • $\begingroup$ Maybe I have to reword my question, It's pretty straightforward that if $c$ is complex then $Re(z)=0$ since $z=\langle{f|g}\rangle$ is complex, So my question basically boils down to why $c$ needs to be complex. $\endgroup$
    – Courage
    Dec 1, 2015 at 16:11
  • $\begingroup$ @Vishwaas I can't understand you. $z$ is a complex number because of the equality $|z|^2=y^2$ and this yield,that the number c must be complex. Take a look to my edit. $\endgroup$ Dec 3, 2015 at 6:14
  • 1
    $\begingroup$ I think you are going in circles in proving that $c$ is complex, nevertheless I came to a conclusion that $c$ has to be a complex because that is the most general form a number can take, Thank you. $\endgroup$
    – Courage
    Dec 3, 2015 at 6:29

I think I found out the answer to my own question

$|f\rangle$ and $|g\rangle$ are defined in Hilbert space which is an inner product space, and one of the properties of this space is $$\langle f|f \rangle \geq 0$$

So $\langle f|f \rangle$ should be real,

In the derivation of the uncertainty principle we take

$$(\text {Re}(z))^2+(\text {Im}(z))^2 \geq (\text {Im}(z))^2$$

where $z=\langle f|g\rangle$

The above inequality can be an equality when $\text {Re}(z)=0$, i.e $\text {Re}(\langle f|g\rangle)=0$

So we take $|f\rangle=c|g\rangle$ (due to Schwarz inequality), which gives $\text {Re}(c\langle f|f\rangle)=0$, since $\langle f|f \rangle$ is real (as stated above), $c$ must be complex


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