Calculating Energy Uncertainty for a Particle Described by a Wave Function

In summary, the energy uncertainty for a particle described by a wave function is \sqrt{<E^2> - <E>^2}.
  • #1
Uniquebum
55
1

Homework Statement


Determine the energy uncertainty [itex]\Delta E = \sqrt{<E^2> - <E>^2}[/itex] for a particle described by a wave function
[itex]\Psi (x) = c_1 \psi (x)_1 + c_2 \psi (x)_2[/itex]
where [itex]\psi_1[/itex] and [itex]\psi_2[/itex] are different (orthonormal) energy eigenstates with eigenvalues [itex]E_1[/itex] and [itex]E_2[/itex].


Homework Equations


I'd presume you need to know
[itex]\hat E = i \hbar \frac{\partial}{\partial t}[/itex]
[itex]\hat E_{kin} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}[/itex]

The Attempt at a Solution


First, I'm not sure whether i should throw in the kinetic or total energy operator. If i put in the total energy operator, i'll have to derivate the function in respect to time which in this case would result in 0. If i put in the kinetic energy operator it just might work but I'm not sure how i work with those expectation values when they're of the form <E^2> or <E>^2.

Assuming i'd use the total energy operator, should it look like
[itex]<E^2> = \int_{-\infty}^{\infty} \psi^* (x) i^2 \hbar^2 \frac{\partial^2}{\partial t^2} \psi (x)[/itex]
[itex]<E>^2 = \int_{-\infty}^{\infty} \psi^* (x)^2 i^2 \hbar^4 \frac{\partial^2}{\partial t^2} \psi (x)^2[/itex]
?

Any help would be appreciated.
 
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  • #2
Note that
[tex]\langle E \rangle_\Psi \equiv \langle \Psi \mid E \mid \Psi \rangle[/tex]

Since you are given that the wave function is a linear combination of orthonormal energy
eigenstates you can work out what <E> is in terms of c1,2 and E1,2. Squaring that will give you <E>2.

You can do a similar trick for

[tex]\langle E^2 \rangle_\Psi \equiv \langle \Psi \mid E^2 \mid \Psi \rangle
= \left( \langle \Psi \mid E \right) \left( E \mid \Psi \rangle \right)
[/tex]
 
  • #3
Hmm... so in this case [itex]\hat E_{12} = E_{12}[/itex]? Meaning i don't have to mess around with the operators but i can just put in the constants E_1 and E_2?
 
  • #4
Sorry, what do you mean by [itex]\hat E_{12}[/itex]?

Maybe I wasn't entirely clear with:
you can work out what <E> is in terms of c1,2 and E1,2.

What I meant is:
you can work out what <E> is in terms of c1 and of c2 and of E1 and of E2.

But yes, you can work it out in terms of eigenvalues rather than using the explicit form of the operator - but of course you will still need to work it out to some extent (you can't just call the energy E, you do need to get it in terms of the given constants)

[edit]
Just to be completely clear on what I mean: can you tell me (from the given information) what [itex]\hat E { |\Psi\rangle}[/itex] is?
 
  • #5
[itex]\int_{-\infty}^{\infty} \Psi^* \hat E \Psi dx[/itex], where [itex]\hat E = i \hbar \frac{\partial}{\partial t}[/itex]

But i figured i can just ignore the operator and use E instead. Which would lead into

[itex]<E> = \int_{-\infty}^{\infty} \Psi^* \hat E \Psi dx = \int_{-\infty}^{\infty} (c_1^* \psi (x)_1^* + c_2^* \psi (x)_2^* )\cdot(c_1 E_1 \psi (x)_1 + c_2 E_2 \psi (x)_2)dx[/itex]
 
  • #6
You can't just ignore stuff. That's a sign that you're making a mistake. In this case, you're using the wrong operator to find the energy. You should be using the Hamiltonian [itex]\hat{H}[/itex] because your wave function is a solution to the time-independent Schrodinger equation [tex]\hat{H}\psi(x) = E\psi(x)[/tex] where E is a constant, not an operator. The operator [tex]\hat{E} = i\hbar\frac{\partial}{\partial t}[/tex] works on solutions [itex]\Psi(x,t)[/itex] to the time-dependent Schrodinger equation [tex]\hat{H}\Psi(x,t) = i\hbar\frac{\partial}{\partial t}\Psi(x,t).[/tex]
The expected value of the energy E is then
[tex]\langle E \rangle = \int_{-\infty}^\infty \psi^* \hat H \psi \, dx = \int_{-\infty}^\infty [c_1^* \psi_1^*(x) + c_2^* \psi_2^*(x)][c_1 E_1 \psi_1(x) + c_2 E_2 \psi_2(x)]\,dx[/tex]
 
  • #7
Ok, so the way i thought you'd do it led to the right equation but it was pure luck since i just guessed the way it'd go. This definitely helped even more than just considering the solution to the main question.
Anyway, the rest should be fairly simple since i just have to take into account that the wave functions are orthonormal. Which mainly means that the wave functions will disappear since they'll be multiplying each others and thus be 1 or 0. I'll give it a shot. Thanks!
 

1. What is uncertainty for delta E?

Uncertainty for delta E is a measure of the range of possible values for the change in energy (delta E) of a system. It takes into account the potential errors and variability in measurements and calculations.

2. Why is uncertainty for delta E important in scientific research?

Uncertainty for delta E is important because it allows scientists to assess the reliability and precision of their results. It also helps to determine the level of confidence that can be placed on the conclusions drawn from the data.

3. How is uncertainty for delta E calculated?

Uncertainty for delta E is typically calculated using statistical methods, such as error propagation or standard deviation. It involves taking into account the uncertainties associated with each measurement or data point, and combining them to determine the overall uncertainty.

4. What factors can contribute to uncertainty for delta E?

Several factors can contribute to uncertainty for delta E, including limitations in measurement equipment, human error in taking measurements, and assumptions made in calculations. External factors, such as environmental conditions, can also affect the uncertainty.

5. Can uncertainty for delta E be reduced?

Yes, uncertainty for delta E can be reduced by improving the accuracy and precision of measurements, using more advanced equipment, and minimizing sources of error. However, it is impossible to completely eliminate uncertainty, so it is important for scientists to report and interpret their results with the associated uncertainties in mind.

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