I'm currently following Quantum Computation and Quantum Information by Nielsen & Chuang. I'm struggling to understand the derivation of The Heisenberg Uncertainty Principle in Box 2.4 page 89. I understand how they arrive to the following equation:

$$|\langle v|[A,B]|v\rangle|^2 \leq 4\langle v|A^2|v\rangle\langle v|B^2|v\rangle.$$

But the problem I have is with the following step. After this, they set $A = C - \langle C\rangle$ and $B = D - \langle D\rangle$ to arrive at

$$\Delta(C)\Delta(D) \geq \frac{|\langle v|[C,D]|v\rangle|}{2}.$$

What I don't understand is how you can subtract C, a matrix, with $\langle C\rangle$, a value, and set it equal to A, a Hermitian. Earlier in the book they define $\langle M \rangle = \langle v|M|v\rangle$ which is an inner product and thus a value and not a matrix. How can you subtract a value from an operator/matrix?

Edit: I now see that $\langle C \rangle$ is $\langle C \rangle 1$, but am still confused about how this substitution gets us from the first inequality to the 2nd inequality.

  • 3
    $\begingroup$ Note that $$A=C-\langle C\rangle = A-\langle C\rangle \mathbb{1}$$ $\endgroup$
    – Quiver
    Mar 24, 2020 at 17:45
  • $\begingroup$ I'm still confused, even with this substitution, on how to get from the first inequality to the second one. Like if I just substitute those two expressions for A and B, it gets messy and doesn't seem to simplify to the 2nd inequality. $\endgroup$ Mar 24, 2020 at 17:56
  • $\begingroup$ Try to use Schwartz inequality $$\langle\alpha|\alpha\rangle\langle\beta|\beta\rangle \geq |\langle\alpha|\beta\rangle|^2 $$ and see if you can get there. In case you don't, I'll write the full calculation. $\endgroup$
    – Quiver
    Mar 24, 2020 at 18:20
  • $\begingroup$ I used the Schwartz inequality to get to the first inequality, but I'm not sure how it can be used to simplify the substitution and get to the 2nd inequality. $\endgroup$ Mar 24, 2020 at 18:41

1 Answer 1


I'll write down the whole calculation for completion. Firstly we define $$A = C-\langle C\rangle \qquad B = D-\langle D \rangle$$

We firstly evaluate the following $$\frac{\langle v|A^2|v\rangle}{\langle v | v \rangle} = \frac{\langle v|C^2-2C\langle C\rangle+\langle C\rangle^2|v\rangle}{\langle v | v \rangle} = \frac{\langle v|C^2|v\rangle}{\langle v | v \rangle}-2\langle C\rangle\frac{\langle v|C|v\rangle}{\langle v | v \rangle}+\left(\frac{\langle v|C|v\rangle}{\langle v | v \rangle} \right)^2\\ = \frac{\langle v|C^2|v\rangle}{\langle v | v \rangle} -2\left(\frac{\langle v|C|v\rangle}{\langle v | v \rangle} \right)^2+\left(\frac{\langle v|C|v\rangle}{\langle v | v \rangle} \right)^2\\ = \frac{\langle v|C^2|v\rangle}{\langle v | v \rangle} -\left(\frac{\langle v|C|v\rangle}{\langle v | v \rangle} \right)^2 = \Delta A^2 $$ which is precisely the definition of standard deviation. The same also works for $B$ clearly.

We now define two states $$|\alpha\rangle = A|v\rangle \qquad |\beta\rangle = B|v\rangle $$ notice that $$\langle\alpha|\alpha\rangle = \langle v|A^2|v\rangle = \Delta A^2\langle v|v\rangle$$ and the same for $|\beta\rangle$. What we want to find is an inequality on the product of the standard deviation $$\Delta A^2\Delta B^2 = \frac{\langle v|A^2|v\rangle}{\langle v|v\rangle}\frac{\langle v|B^2|v\rangle}{\langle v|v\rangle}\geq\frac{|\langle v |B^\dagger A|v \rangle|^2}{|\langle v|v\rangle|^2} = \frac{|\langle v |B A|v \rangle|^2}{|\langle v|v\rangle|^2}$$ where we used Schwartz's inequality and the hermiticiy of the operators (remember that we're talking about measurable quantities here).

We invoke now the following equality $$BA = \frac{\{A,B\}}{2}+\frac{[A,B]}{2}$$ Plugging this back in we get $$\Delta A^2\Delta B^2 \geq \frac{\left|\left\langle v \left|\frac{\{A,B\}}{2}+\frac{[A,B]}{2}\right|v \right\rangle\right|^2}{|\langle v|v\rangle|^2}\\ =\frac{1}{4}\frac{|\langle v |[A,B]|v \rangle|^2}{|\langle v|v\rangle|^2}+\frac{1}{4}\frac{|\langle v |\{A,B\}|v \rangle|^2}{|\langle v|v\rangle|^2} $$ Since now both terms are positive and real, clearly omitting one of them does not affect the inequality, so we discard the anticommutator one and get the result $$\Delta A^2 \Delta B^2\geq\frac{1}{4}\frac{|\langle v |[A,B]|v \rangle|^2}{|\langle v|v\rangle|^2}$$ or, by taking the square root of both members $$\Delta A\Delta B\geq\frac{1}{2}\frac{|\langle v |[A,B]|v \rangle|}{|\langle v|v\rangle|}$$ In your result is implied that $\langle v|v\rangle = 1$, so that the states are normalized.

  • 1
    $\begingroup$ I have down voted because, while correct, this answer doesn't address what the OP is asking about. $\endgroup$ Mar 24, 2020 at 19:48
  • $\begingroup$ Check the comments under my question -- I have now edited my question to reflect the change in what I was confused about. $\endgroup$ Mar 24, 2020 at 19:57
  • $\begingroup$ Another question here. In the first step, why is $-2\frac{\langle v|C\langle C\rangle|v\rangle }{\langle v|v\rangle}$ equal to $-2\langle C\rangle \frac{\langle v|C|v\rangle }{\langle v|v\rangle}$? Because <C> is not a value but rather a matrix, are we allowed to just take it out of the inner product like that? $\endgroup$ Mar 24, 2020 at 22:15
  • 1
    $\begingroup$ Yeah, that's exactly it! Just remember that there's always an identity matrix, but this does not change the result, and that's why we do not bother to write it down most of the times. $\endgroup$
    – Quiver
    Mar 24, 2020 at 22:58

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