$f$ and $g$ are two quadratially integrable functions. It is shown from Schwarz inequality, def of variance and an identity of complex numbers that

$$(\Delta A)^2(\Delta B)^2=\langle f|f\rangle \langle g|g\rangle~\geq~|\langle f|g\rangle|^2~\geq~[\frac{(\langle f|g\rangle-\langle g|f \rangle)}{2i}]^2$$

Then the proof says: Use the definition of $f$ and $g$ and the normalization of $\Psi$ to verify that

$$\left\langle f|g> = <g|f \right\rangle^*=\langle\Psi|\hat A \hat B|\Psi\rangle-\langle A \rangle \langle B \rangle \tag{I}$$

Which leads to the formula:

$$(\Delta A)^2(\Delta B)^2~\geq~\frac{1}{4}|\langle \Psi| [\hat A, \hat B] | \Psi \rangle|^2 $$

I don't get the equality in (I). I also don't get the notation on the left side. Can anyone help out?

The proof is taken from problem 7.60 in Ira Levine quantum mechanics.

  • $\begingroup$ I mistakely used () for <>. I have updated that. In (I) the first < and the last > should be larger then the one in the middle on the left side I don't know how to type it more clear unfortunately $\endgroup$
    – torgny
    Jan 4, 2016 at 20:42
  • $\begingroup$ by principle, a principle doesn't need to be proved $\endgroup$
    – user46925
    Jan 4, 2016 at 23:18

1 Answer 1


If $\langle A\rangle=\langle \Psi| \hat A| \Psi \rangle$ exists and is finite. And $\langle B\rangle=\langle \Psi| \hat B| \Psi \rangle$ exists and is finite. Then we can define $|f\rangle=\hat A |\Psi\rangle-\langle A\rangle|\Psi\rangle$ and define $|g\rangle=\hat B |\Psi\rangle-\langle B\rangle|\Psi\rangle.$

If $f$ and $g$ are square integrable then you can compute the integrals of their squares (which are $\langle f|f\rangle$ and $\langle g|g\rangle$ respectively). By simplifying the results you'll see they equal the variance of $A$ and $B$ respectively. Then you can combine with Cauchy-Schwarz to get:

$$(\Delta A)^2(\Delta B)^2=\langle f|f\rangle\langle g|g\rangle~\geq~|\langle f|g\rangle |^2.$$

Now $\langle f|g\rangle = \langle g|f\rangle^*$ is just the generalization of $\vec a \cdot \vec b=\vec b\cdot \vec a$ to complex vectors where instead of being symmetric it is conjugate symmetric. For example, if $\langle f|g\rangle=\int f^*g$ and $\langle g|f\rangle=\int g^*f$ then $\langle f|g\rangle = \langle g|f\rangle^*$ since:

$$\begin{array} & \int f^*g &=\left(\int f^*g\right)^{**}\\&=\left(\int (f^{*}g)^*\right)^*\\&=\left(\int f^{**}g^*\right)^*\\&=\left(\int g^*f\right)^*.\end{array}$$

And to get the right hand side you do the same tricks as showing they equaled the variance, bring constants outside of the integrals.

  • $\begingroup$ You would not derive the last equality of (I) also? thanks! $\endgroup$
    – torgny
    Jan 4, 2016 at 21:19
  • $\begingroup$ @torgny It is against policy to do your homework for you. I should have provided enough details for you to do the rest by yourself. Treat the operators like matrices, the vectors like vectors and pull constants through and compute the parts. If you could show the variance part you should be able to do this part. You might need to use that the operators are observables. $\endgroup$
    – Timaeus
    Jan 4, 2016 at 21:27

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