Is This Wave Function an Eigenfunction of Energy and Momentum Operators?

  • Thread starter Thread starter Craptola
  • Start date Start date
  • Tags Tags
    Eigenvalues Qm
AI Thread Summary
The discussion centers on determining if the wave function Ψ(x,t) = exp(-i(kx + ωt)) is an eigenfunction of the energy and momentum operators. The energy operator is defined as \(\hat{E} = i\hbar \frac{\partial}{\partial t}\), and the calculation shows that \(\hat{E}\Psi = \hbar\omega \Psi\), confirming that Ψ is indeed an eigenfunction with eigenvalue E = ħω. However, confusion arises regarding the calculation of a specific numerical value for E, as it seems to depend on constants k and ω rather than being a fixed number. The conversation emphasizes that k and ω are constants provided in the problem, which clarifies the misunderstanding about the variability of E. Overall, the thread highlights the process of verifying eigenfunctions in quantum mechanics and the importance of recognizing constants in calculations.
Craptola
Messages
14
Reaction score
0
I've been wrestling with this question for a while and can't seem to find anything in my notes that will help me.

Homework Statement


Determine whether the wave function \Psi (x,t)= \textrm{exp}(-i(kx+\omega t)) is an eigenfunction of the operators for total energy and x component of momentum, and if it is, calculate the eigenvalues.

Homework Equations


Condition for an eigenfunction:
\hat{E}\Psi =k\Psi
Where K is the eigenvalue
Energy operator:
\hat{E}=i\hbar\frac{\partial }{\partial t}

The Attempt at a Solution


Determining that psi is an eigenfunction is easy enough.
\hat{E}\Psi =i\hbar\frac{\partial }{\partial t}[\textrm{exp}(-i(kx+\omega t))]
=-i\hbar i\omega \Psi =\hbar\omega \Psi =\frac{h}{2\pi }2\pi f\Psi =hf\Psi =E\Psi

I can't figure out how to calculate the value of E from this information alone. I imagine the same method works for momentum when I figure out what it is.
 
Physics news on Phys.org
hmm? You have just calculated the value of E. Now, yeah it is pretty much a similar method to find out if it is also an eigenstate of momentum, once you remember what the operator looks like :)
 
  • Like
Likes 1 person
Hi Craptola! :smile:

hmmm …
Craptola said:
\hat{E}\Psi =i\hbar\frac{\partial }{\partial t}[\textrm{exp}(-i(kx+\omega t))]
=-i\hbar i\omega \Psi =\hbar\omega \Psi

stop there? :wink:
 
I assumed that the question wanted me to calculate an actual number for the eigenvalue, which is what confused me as it seems that E could be anything depending on other variables. This is all stuff we covered fairly recently so I wasn't sure if there was some kinda law that wasn't in the lecture notes which limited the possible values of E.
 
Craptola said:
… it seems that E could be anything depending on other variables.

Ah, but k and ω aren't variables, they're your given constants! :wink:
 
  • Like
Likes 1 person
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top