Quantum Queries on the Harmonic Oscillator

[T-7]

Hi folks,

I wonder if I could run a few things past the quantum gurus among you - I'm just not quite convinced of some of the results I've been deriving.

Homework Statement

Consider the ground state of the simple harmonic oscillator at t = 0 with the normalised wave function $$\sqrt[4]{\frac{m\omega}{\pi\hbar}}e^{-\frac{m\omega}{2\hbar} x^2}$$.

If the position uncertainty is defined as $$\Delta x = \sqrt{<\widehat{x}^{2}>-<\widehat{x}>^{2}}$$ and the momentum uncertainty as $$\Delta p = \sqrt{<\widehat{p}^{2}>-<\widehat{p}>^{2}}$$, determine the product $$\Delta x . \Delta p$$. Comment on the result.

The Attempt at a Solution

I begin by observing that $$<\widehat{x}>$$ and $$<\widehat{p}>$$ are both zero, by symmetry arguments. Thus $$\Delta x = \sqrt{<\widehat{x}^{2}>}$$ and $$\Delta p = \sqrt{<\widehat{p}^{2}>}$$.

I go on to calculate $$<\widehat{x}^{2}>$$ and $$<\widehat{p}^{2}>$$.

I find that $$<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}$$, a result I am pretty confident of (it was an easy integration).

My queries arise for $$<\widehat{p}^{2}>$$, which comes out as $$\hbar(m\omega - \frac{1}{2})$$, giving $$\Delta x . \Delta p = \sqrt{\frac{\hbar}{2m\omega}} . \sqrt{\hbar(m\omega - \frac{1}{2})} = \hbar\sqrt{1/2 - 1/4m\omega}$$ - not a very "nice" result. I'm suspicious.

I make $$<\widehat{p}^{2}>$$ to be

$$<\widehat{p}^{2}> = \sqrt{\frac{m\omega}{\pi\hbar}}\int^{+\infty}_{-\infty} e^{-\frac{m\omega}{2\hbar} x^2}.(-\hbar^{2}\frac{\partial^2}{\partial x^2}).(e^{-\frac{m\omega}{2\hbar} x^2}) dx = . . . = \sqrt{m^{3}\omega^{3}\hbar/\pi}.I_{2} - \sqrt{m^{3}\omega^{3}/\pi\hbar}.I_{1}$$

as

$$(\frac{\partial^2}{\partial x^2}).(e^{-\frac{m\omega}{2\hbar} x^2}) = ... = -\frac{m\omega}{\hbar}.e^{-m\omega x^2/2\hbar} + (\frac{m\omega}{\hbar})^2.x^2.e^{-m\omega x^2/2\hbar}$$,

where

$$I_{1} = \int^{+\infty}_{-\infty}x^2.e^{-m\omega x^2/\hbar} dx = ... = \sqrt{\frac{\pi\hbar^{3}}{4m^{3}\omega^{3}}}$$
$$I_{2} = \int^{+\infty}_{-\infty}e^{-m\omega x^2/\hbar} dx = ... = \sqrt{\frac{\pi\hbar}{m\omega}}$$

etc.

determining finally that $$\hbar(m\omega - \frac{1}{2})$$

Has something gone wrong here, perhaps, or should I just accept the result?

Cheers!

 

[nrqed]

Hi folks,

I wonder if I could run a few things past the quantum gurus among you - I'm just not quite convinced of some of the results I've been deriving.

Homework Statement

Consider the ground state of the simple harmonic oscillator at t = 0 with the normalised wave function $$\sqrt[4]{\frac{m\omega}{\pi\hbar}}e^{-\frac{m\omega}{2\hbar} x^2}$$.

If the position uncertainty is defined as $$\Delta x = \sqrt{<\widehat{x}^{2}>-<\widehat{x}>^{2}}$$ and the momentum uncertainty as $$\Delta p = \sqrt{<\widehat{p}^{2}>-<\widehat{p}>^{2}}$$, determine the product $$\Delta x . \Delta p$$. Comment on the result.

The Attempt at a Solution

I begin by observing that $$<\widehat{x}>$$ and $$<\widehat{p}>$$ are both zero, by symmetry arguments. Thus $$\Delta x = \sqrt{<\widehat{x}^{2}>}$$ and $$\Delta p = \sqrt{<\widehat{p}^{2}>}$$.

I go on to calculate $$<\widehat{x}^{2}>$$ and $$<\widehat{p}^{2}>$$.

I find that $$<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}$$, a result I am pretty confident of (it was an easy integration).

My queries arise for $$<\widehat{p}^{2}>$$, which comes out as $$\hbar(m\omega - \frac{1}{2})$$, giving $$\Delta x . \Delta p = \sqrt{\frac{\hbar}{2m\omega}} . \sqrt{\hbar(m\omega - \frac{1}{2})} = \hbar\sqrt{1/2 - 1/4m\omega}$$ - not a very "nice" result. I'm suspicious.

I make $$<\widehat{p}^{2}>$$ to be

$$<\widehat{p}^{2}> = \sqrt{\frac{m\omega}{\pi\hbar}}\int^{+\infty}_{-\infty} e^{-\frac{m\omega}{2\hbar} x^2}.(-\hbar^{2}\frac{\partial^2}{\partial x^2}).(e^{-\frac{m\omega}{2\hbar} x^2}) dx = . . . = \sqrt{m^{3}\omega^{3}\hbar/\pi}.I_{2} - \sqrt{m^{3}\omega^{3}/\pi\hbar}.I_{1}$$

there is something wrong here. Look below at your second derivative of x^2 times the exponential. The two terms you get from the second derivative have a relative factor of m omega/ hbar. and yet in the above expression th erelative factor is not m omega/hbar. Check that equation.

as

$$(\frac{\partial^2}{\partial x^2}).(e^{-\frac{m\omega}{2\hbar} x^2}) = ... = -\frac{m\omega}{\hbar}.e^{-m\omega x^2/2\hbar} + (\frac{m\omega}{\hbar})^2.x^2.e^{-m\omega x^2/2\hbar}$$,

where

$$I_{1} = \int^{+\infty}_{-\infty}x^2.e^{-m\omega x^2/\hbar} dx = ... = \sqrt{\frac{\pi\hbar^{3}}{4m^{3}\omega^{3}}}$$
$$I_{2} = \int^{+\infty}_{-\infty}e^{-m\omega x^2/\hbar} dx = ... = \sqrt{\frac{\pi\hbar}{m\omega}}$$

etc.

determining finally that $$\hbar(m\omega - \frac{1}{2})$$

Has something gone wrong here, perhaps, or should I just accept the result?

Cheers!

 

[T-7]

there is something wrong here. Look below at your second derivative of x^2 times the exponential. The two terms you get from the second derivative have a relative factor of m omega/ hbar. and yet in the above expression th erelative factor is not m omega/hbar. Check that equation.

You are right. The expression should have read:

$$<\widehat{p}^{2}> = \sqrt{\frac{m\omega}{\pi\hbar}}\int^{+\infty}_{-\infty} e^{-\frac{m\omega}{2\hbar} x^2}.(-\hbar^{2}\frac{\partial^2}{\partial x^2}).(e^{-\frac{m\omega}{2\hbar} x^2}) dx = . . . = \sqrt{m^{3}\omega^{3}\hbar/\pi}.I_{2} - \sqrt{m^{5}\omega^{5}/\pi\hbar}.I_{1}$$

This eventually leads to $$\hbar m\omega / 2$$, and then we do indeed derive Heisenberg's uncertainty relation in the form $$\hbar / 2$$

Many thanks.