Uncertainty Principle & Ground State of Harmonic Oscillator

[alejandrito29]

For the ground state of harmonic oscillator i have

\Delta p \Delta x =\frac{\hbar}{2}

why if i do

\frac{1}{F} \Delta p \Delta x \cdot F =\frac{\hbar}{2}

\Delta t \Delta E = \frac{\hbar}{2}

\Delta E = \frac{\hbar}{2} \frac{1}{\Delta t}

\Delta E = \frac{\hbar}{2} f

but my answer should be \Delta E = \frac{\hbar}{2} \omega ?

 

[Simon Bridge]

The delta-terms are measurements of uncertainty.
i.e. $\Delta E$ represents the precision to which energy is measured.

 

[alejandrito29]

The delta-terms are measurements of uncertainty.
i.e. $\Delta E$ represents the precision to which energy is measured.

ok, but $<\Delta E^2>= <E^2>-<E>^2= (<p^2>/(2m) + 0.5 m \omega^2 <x^2> )^2- ( <p>^2/(2m)+ 0.5 m \omega^2 <x>^2 ) ^2$

$=E_0^2-0^2=E_0 ^2 \to \Delta E = E_0$

then, my aswer should be $\Delta E= \hbar \omega /2$, and not $\Delta E = \hbar f/2$

 

[dlgoff]

This is a very significant physical result because it tells us that the energy of a system described by a harmonic oscillator potential cannot have zero energy. Physical systems such as atoms in a solid lattice or in polyatomic molecules in a gas cannot have zero energy even at absolute zero temperature. The energy of the ground vibrational state is often referred to as "zero point vibration". The zero point energy is sufficient to prevent liquid helium-4 from freezing at atmospheric pressure, no matter how low the temperature.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc4.html

 

[alejandrito29]

ok, but my question is about the value of $\Delta E$, i don't ask about $E_0$

$\Delta E$ is $\hbar \omega /2$ or $\hbar f /2$ ?

 

[HomogenousCow]

The mean deviation of energy in an energy eigenstate is obviously zero.
The manipulations you made in the original post are not correct, the deltas are not differences or infinitesimals

 

[alejandrito29]

The mean deviation of energy in an energy eigenstate is obviously zero.
The manipulations you made in the original post are not correct, the deltas are not differences or infinitesimals

look this

http://books.google.cl/books?id=Pk9...&q=δe harmonic oscillator uncertainty&f=false

or this

http://books.google.cl/books?id=3rx...q=uncertainty principle energy time v&f=false

 

[Simon Bridge]

ok, but $<\Delta E^2>= <E^2>-<E>^2= (<p^2>/(2m) + 0.5 m \omega^2 <x^2> )^2- ( <p>^2/(2m)+ 0.5 m \omega^2 <x>^2 ) ^2$

I think you are still very confused.

$\langle (\Delta E)^2\rangle$ reads: "expectation of the variance of E", and you put it equal to the variance, which is an odd thing to say ... but I think you meant to write:

$(\Delta E)^2 = \langle E^2\rangle-\langle E\rangle^2$ would be the definition of variance.

For an energy eigenstate $\langle H\rangle = E_n$
You seem to want to work on the ground state:

$\cdots =E_0^2-0^2=E_0 ^2 \to \Delta E = E_0$

then, my aswer should be $\Delta E= \hbar \omega /2$, and not $\Delta E = \hbar f/2$

Who says that $\Delta E = \hbar f/2$ is correct?
Where are you getting these ideas from?

Checking:
For the ground state:

$(\Delta E)^2 = \langle E^2\rangle-\langle E\rangle^2 = \langle E^2\rangle-\langle E\rangle^2 = E_0^2-E_0^2=0\implies \Delta E = 0$

... which means that the energy eigenstates of the harmonic oscillator are infinitely thin
... the energy of the lowest eigenstate is exactly $E_0$ and so on.

(I think you missed out a term in your subtraction.)

Since $\Delta t \propto 1/\Delta E$, this means that the lifetime of the state is infinitely uncertain. i.e. the state lasts forever.

It may be easier of you write the relations out as: $$\sigma_x\sigma_p\geq \frac{\hbar}{2},\; \sigma_E\sigma_t\geq \frac{\hbar}{2}$$

Or it may be that you are referring to something else - I can't really tell.
Please tell us where these ideas are coming from.

Meantime, the statistics of the harmonic oscillator are detailed in:
http://academic.reed.edu/physics/courses/P342.S10/Physics342/page1/files/Lecture.9.pdf