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 '15 at 21:34

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 '15 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. $\endgroup$ – Muhsin Ibn Al Azeez Dec 3 '15 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 '15 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.