My question is about the derivation for the uncertainty of a quantum harmonic oscillator in the non-zero ground state energy. In my textbook A modern Approach to Quantum Mechanics by John S. Townsend there is a discussion about the position uncertainty $(\Delta x)^2= \big<x^2 \big>+\big<x\big>^2$ for a harmonic oscillator in the ground state energy.

We have established for a harmonic oscillator $\hat x={\sqrt {\hbar\over 2m\omega}}(\hat a+{\hat a^\dagger})$ so $\big<x\big>=\big<n \big|{\sqrt {\hbar\over 2m\omega}}(\hat a+{\hat a^\dagger}) \big|n\big>$ which gives us $\big<x^2\big>=\big<n\big|{\hbar\over 2m\omega}(\hat a+{\hat a^\dagger})^2\big|n\big>$. And in a similar fashion $\big<x\big>^2=\big<n \big|{\sqrt {\hbar\over 2m\omega}}(\hat a+{\hat a^\dagger}) \big|n\big>^2$.

To me this means that $(\Delta x)^2=\big<n\big|{\hbar\over 2m\omega}(\hat a+{\hat a^\dagger})^2\big|n\big> + \big<n \big|{\sqrt {\hbar\over 2m\omega}}(\hat a+{\hat a^\dagger}) \big|n\big>^2$. However, the book seems to drop (with no explanation) $\big<x\big>^2$ and comes up with $(\Delta x)^2=\big<0\big|{\hbar\over 2m\omega}(\hat a+{\hat a^\dagger})^2\big|0\big>$. I know we are talking about ground state so I'm assuming that's why the $n$ was replaced with $0$, however as far as I can tell we just have $\big<x^2\big>$. Why is the other term dropped? Is this something to do with the oscillator being in the ground state?



I remember being baffled by how this works out mathematically myself, and it's one of those "I can't believe it's that simple!" ones.

Three facts:

  • $\hat{a}$ and $\hat{a}^\dagger$ are the lowering and raising ladder operators: they take a numbered state to one numbered either one less or one more that the starting state.
  • The numbered states are a set of eigenstates, so they are orthogonal to one another.
  • Expand $\left( \hat{a} + \hat{a}^\dagger \right)^2$, and see why it has a very different character than $\left( \hat{a} + \hat{a}^\dagger \right)$.
  • $\begingroup$ I just expanded them and I see exactly what you mean, I need to be more careful about how I use operators. Looking at this I can't see any way $<x>^2$ would ever contribute to the uncertainty, no matter what energy state it is in because of the lowering and raising ladder operations. Is that the case? $\endgroup$ – matryoshka Oct 13 '18 at 22:11
  • 2
    $\begingroup$ Yup. Or at least, "Yup for the harmonic oscillator". The situation where $\hat{x}$ is composed of a sum of simple raising and lowering operators is special to the SHO. $\endgroup$ – dmckee Oct 13 '18 at 22:14

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