Uncertainty in position and kinetic energy

Question

How do you find the uncertainties for $x$ and $K$? Knowing that the general uncertainties =

$$ \sigma_A \sigma_B \geq 1/2\int \psi ^*[\hat A,\hat B] \psi dx\, $$

I figured out the commutator, for some function f(x) $$ [\hat x,\hat K][f(x)]=(\hbar/m)(df/dx) $$

Then plugging into the integral I got

$$ \sigma_x \sigma_K \geq \hbar ^2/(2mi)\int_{-\infty} ^\infty f^*( x) (df/dx) dx\, $$

Is there anyway to further reduce this? Am I correctly approaching this?

Answer 1

You are on the right track. Pushing one more step to the final answer may leave you disappointed: $\sigma_x \sigma_K$ can equal zero! To see this, I find it more helpful to think just in terms of $x$ and $K$ as linear operators satisfying certain commutation relations, rather than thinking explicitly in terms of integrals of wavefunctions.

Specifically, we have $K = \frac{p^2}{2m}$ and $[x,p] = i \hbar$. From these two equations it follows that \begin{eqnarray} [x,K] & = & \frac{1}{2m} [x,p^2] \\ & = & \frac{1}{2m} \left( [x,p]p + p[x,p] \right) \\ & = & \frac{1}{2m} \left( i \hbar p + p i \hbar \right) \\ & = & \frac{i \hbar}{m} p \end{eqnarray} So, for a state with zero momentum, the product $\sigma_x \sigma_K$ of uncertainties can be zero. This corresponds to the possibility that $\frac{df}{dx} = 0$ in your final equation. (The second equality in the above derivation uses the chain rule for commutators, which is a handy identity: For any operators $A$,$B$, and $C$ it holds that $[A,BC] = B[A,C] + [A,B]C$.)

It is only in the case that $[\hat{A},\hat{B}]$ is proportional to the identity that the state $\psi$ disappears from the right hand side of the general uncertainty relation $\sigma_A \sigma_B \geq 1/2 \int \psi^* [\hat{A}, \hat{B}] \psi dx$. The position and momentum operators have this relationship, as do any canonically conjugate pair. However, kinetic energy and position do not.

Answer 2

You do it as follows :

$$[X,K]=\frac{1}{2m}[X,P^2]=\frac{1}{2m}\left(P[X,P]+[X,P]P\right)=\frac{i\hbar P}{m}$$

The general Uncertainty principle for operators $\Omega$ and $\Lambda$ is given by : $$(\Delta \Omega)^2(\Delta \Lambda)^2\geq\frac{1}{4}\langle\psi|[\hat{\Omega},\hat{\Lambda}]|\psi\rangle^2+\frac{1}{4}\langle\psi|\Gamma|\psi\rangle^2$$ where $[\hat{\Omega},\hat{\Lambda}]=i\Gamma$. The first term is positive and thus for any $|\psi\rangle$ $$(\Delta \Omega)^2(\Delta \Lambda)^2\geq\frac{1}{4}\langle\psi|\Gamma|\psi\rangle^2=\frac{\hbar^2 }{4m^2}\langle\psi|P|\psi\rangle^2=\frac{\hbar^2 }{4m^2}\langle P\rangle^2$$

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Note that this might not be useful for case where $\langle P \rangle = 0$.