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Consider the case of a free particle with wavefunction $$\Psi(x, t=0) = \left(\frac{a}{\pi}\right)^{1/4} e^{-ax^2/2} \quad a>0$$ Prove that the uncertainty product $(\Delta x)(\Delta p)$ is parabolic close to $t=0$.

Attempt: I basically want to show that for small $\delta t$ we have $(\Delta x)_{\delta t}(\Delta p)_{\delta t} = \frac{\hbar}{2} + b\cdot \delta t^2$, for $b>0$. My only thought at this point is to consider the Taylor expansion of the uncertainty product as a function of time and show that $\frac{d}{dt}(\Delta x)(\Delta p)\; \big|_{t=0} = 0$ and that $\frac{d^2}{dt^2}(\Delta x)(\Delta p)\;\big|_{t=0} > 0$ (or just $\neq 0$ since the uncertainty principle will guarantee that it's positive). Is this the right approach? If so, how do I proceed?

I also tried to explicitly calculate $\Psi(x,t)$ and calculate the product, but the calculations involved are very tedious.

I'm sorry if my question is not appropriate, but I'm quite new to quantum mechanics (and physics in general).

Níckolas Alves
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Giorgos Giapitzakis
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1 Answers1

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I am not sure if you have been taught the simpler, Heisenberg picture, so I'll stick to the Schroedinger one, born messier, as you observe.

  1. As WP suggests, show $$\Psi(x, t) = \left(\frac{a}{\pi}\right)^{1/4}\frac{ e^{-ax^2/2(1+i\hbar at/m)} }{\sqrt{1+i\hbar at/m}}. $$ There are 39.7 methods to do this, but my favorite is the free propagator.

  2. Show $\langle x\rangle = \langle p\rangle =0$.

  3. Show $$\langle x^2 \rangle_0 \langle p^2\rangle_0 = \hbar^2/4 .$$

  4. "Compute" $$ \frac{\langle x^2 \rangle \langle p^2\rangle }{\langle x^2 \rangle_0 \langle p^2\rangle_0 }= 1+(\hbar at/m)^2. $$ Just scale the variables. You need not compute $\int\!\!dx ~e^{-x^2} x^2$ anymore.

  5. Take the square root and expand in t to lowest order.

None of my business, but the 500kg Gorilla in the room is the behavior at large t, where the variance product, uncertainty, increases linearly with t.

Cosmas Zachos
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