# Physical interpretation of eliminating the real component when deriving the uncertainty principle

I've been working on some quantum information theory problems and I've revisited Griffith's Quantum Mechanics. On page 109, he derives the uncertainty principle. He goes through the steps:

1. For operators $$\hat{A}$$ and $$\hat{B}$$, defines : $$|f\rangle = ( \hat{A} - \langle \hat{A}\rangle)|\psi\rangle$$ and $$|g\rangle = (\hat{B} - \langle \hat{B}\rangle)|\psi\rangle$$
2. Defines variance : $$\sigma^{2}_{A} = \langle f | f\rangle$$ and $$\sigma^{2}_{B} = \langle g | g\rangle$$
3. Invokes Cauchy-Schwarz inequality: $$\sigma^{2}_{A} \sigma^{2}_{B} = \langle f | f \rangle \langle g | g \rangle \geq |\langle f | g \rangle|^{2}$$
4. Defines $$z$$ as a complex number : $$z = \langle f | g \rangle$$
5. Utilizes the magnitude of $$z$$ and discarding the real component (see equation 3.136) : $$|z|^{2} = (\text{Re}(z))^{2} + (\text{Im}(z))^{2} \geq (\text{Im}(z))^{2} = [\frac{1}{2i}(z-z^{*}]^{2}$$
6. He keeps only the imaginary component and plugs the result from equation 3.136 (step 5) on the right hand side of the Cauchy-Schwarz inequality (step 3). I realize that by discarding the real component he his not violating his inequality in equation 3.135 (step 3), so he can technically do this. Doing additional commutator math, he gets: $$\sigma_A^2\sigma_B^2 \geq \left|\frac{1}{2i} \langle[ \hat{A}, \hat{B}] \rangle\right|^2$$

Question:

1. In the Griffith's derivation, why did he discard the Real component? It seems by doing so you now no longer need to know anything about the wave function. I.e. you don't need to calculate $$\langle \psi | \hat{A} | \psi \rangle = \langle \hat{A} \rangle$$)? This would make the math easier from this aspect.

If you kept the real component, it seems like you'd have a stronger inequality. In fact Wikipedia gives such a derivation yielding : $$\sigma_A^2\sigma_B^2 \geq \left| \frac{1}{2}\langle\{\hat{A}, \hat{B}\}\rangle - \langle \hat{A} \rangle\langle \hat{B}\rangle \right|^2+ \left|\frac{1}{2i} \langle[ \hat{A}, \hat{B}] \rangle\right|^2$$ )

In the Griffith's derivation, why did he discard the Real component?

It's not really possible to know a textbook author's intent. However, there are several possible reasons why Griffiths chose to derive the Robertson uncertainty relation rather than the Schrödinger uncertainty relation:

• The derivation is simpler, as can be clearly seen by comparing the proofs. Given that Griffiths' textbook is introductory-level, simpler derivations are preferred.

• The derivation of the Robertson uncertainty relation does not utilize the anticommutator, which is not introduced at that point in Griffiths' textbook and is (as far as I recall) not particularly relevant for most of the material in the rest of the textbook.

• The Schrödinger uncertainty relation partially obscures a particular piece of insight which the Robertson uncertainty relation makes explicit: the role of the commutator in determining if operators are compatible with simultaneous measurement.

So the next question would be, "Why didn't Griffiths derive both, then?" The most likely answer is that the two inequalities contain basically the same conceptual content: the product of the uncertainties in the measurement of two operators can, under certain conditions, be constrained to be above some minimum value, and that minimum value involves the expectation value of the commutator. (I suspect that the cases in which two operators commute, but the other term is nonzero, don't come up in an introductory QM course).

It seems by doing so you now no longer need to know anything about the wave function. I.e. you don't need to calculate $$\langle\psi|\hat{A}|\psi\rangle=\langle A\rangle$$)?

But you do need to know about the wavefunction, though. After all, $$\langle [\hat{A},\hat{B}]\rangle=\langle\hat{A}\hat{B}-\hat{B}\hat{A}\rangle=\langle\psi|\hat{A}\hat{B}-\hat{B}\hat{A}|\psi\rangle$$.

• I'm not so sure that you need to explicitly know the form of the wave function. On the following page Griffiths uses a simple test function to derived the $[\hat{x},\hat{p}] = i\hbar$. He then happily puts this directly into 3.139 (step 6) to get the Heisenberg uncertainty principle (3.141). He does not make any conditions on the $|\psi\rangle$. Perhaps though this is a special case b/c $[\hat{x},\hat{p}]$ is a constant and thus $\langle AB - BA \rangle = \langle \psi | i\hbar | \psi \rangle = i\hbar \langle \psi | \psi \rangle = i\hbar$ due to the wavefunction being normalized? – irritable_phd_syndrom May 13 '19 at 10:43
• @irritable_phd_syndrom The commutator $[\hat{x},\hat{p}]$ is indeed a special case. In general the commutator of two operators will not be a constant, and therefore will depend on the wavefunction. For example, $[\hat{L}_x,\hat{L}_y]=i\hbar\hat{L}_z$, which is $\hbar^2\frac{\partial}{\partial\phi}$ in the position-space representation, so $\langle [\hat{L}_x,\hat{L}_y]\rangle=\int\hbar^2\psi^*(\vec{r})\frac{\partial\psi(\vec{r})}{\partial\phi}d^3\vec{r}$, which clearly depends on the wavefunction. – probably_someone May 13 '19 at 11:37