Solving Dirac's Notation Homework: Energy-Time Uncertainty Principle

  • Thread starter Thread starter jg370
  • Start date Start date
Click For Summary
SUMMARY

The forum discussion centers on the derivation of the Energy-Time Uncertainty Principle using Dirac's notation. The user struggles with the differentiation of the expectation value of an operator, specifically \langle{Q}\rangle, and its relation to the Schrödinger equation \imath\hbar\frac{\partial\psi}{\partial t} = \hat{H}\psi. The response confirms that the user's assumption regarding the sign change when factoring out \frac{1}{\imath\hbar} from the bra and ket is correct, leading to the commutator between the Hamiltonian \hat{H} and the operator \hat{Q}.

PREREQUISITES
  • Understanding of Dirac notation and quantum mechanics
  • Familiarity with the Schrödinger equation
  • Knowledge of operators in quantum mechanics
  • Basic grasp of commutators and their significance in quantum theory
NEXT STEPS
  • Study the derivation of the Energy-Time Uncertainty Principle in quantum mechanics
  • Learn about the properties and applications of commutators in quantum mechanics
  • Explore the implications of the Schrödinger equation on quantum states
  • Investigate the role of expectation values in quantum mechanics
USEFUL FOR

Students of quantum mechanics, physicists working with quantum theory, and anyone interested in the mathematical foundations of quantum mechanics.

jg370
Messages
16
Reaction score
0

Homework Statement


jg370 said:
Hi,

My textbook provides me with the steps to derive the Energy-Time Uncertainty Principle; while I can follow most of it, I have problem with one particular step.

[tex]\frac{d}{dt}\langle{Q}\rangle =\frac{d}{dt}\langle\psi\lvert\hat{Q}\psi\rangle[/tex]

[tex]\frac{d}{dt}\langle{Q}\rangle =\langle\frac{\partial \psi}{\partial t}\lvert\hat{Q}\psi\rangle +\langle\psi\lvert\frac{\partial\hat{Q}}{\partial t}\psi\rangle +\langle\psi\lvert\hat{Q}\frac{\partial\psi}}{\partial t}\rangle[/tex]

Now, the Schrödinger equation says:

[tex]\imath\hbar\frac{\partial\psi}{\partial t} = \hat{H}\psi[/tex]

So, I deduced:

[tex]\frac{\partial\psi}{\partial t} = \frac{1}{\imath\hbar}*\hat{H}\psi[/tex]

Substituting this in the main equation, we have:

[tex]\frac{d}{dt}\langle{Q}\rangle = \langle(\frac{1}{\imath\hbar})\hat{H}\psi\lvert\hat{Q}\psi +\langle\psi\lvert\frac{\partial\hat{Q}}{\partial t}\psi\rangle+\langle\psi\lvert\hat{Q}(\frac{1}{\imath\hbar})\hat{H}\psi\rangle[/tex]

[tex]\frac{d}{dt}\langle{Q}\rangle =- \frac{1}{\imath\hbar}\langle\hat{H}\psi\lvert\hat{Q}\psi +\langle\psi\lvert\frac{\partial\hat{Q}}{\partial t}\psi\rangle+\frac{1}{\imath\hbar}\langle\psi\lvert\hat{Q}\hat{H}\psi\rangle[/tex]


Homework Equations


In the last equation, I have factored [tex]\frac{1}{\imath\hbar}[/tex] from the "bra" (first term} and from the "ket" (last term) of above equation and assumed that the sign would be respectively negative and positive? I assumed so because the "bra" is the conjugate of the "ket".

The Attempt at a Solution



Have I assumed corretly? I really so not see any oher possibility. I thank you for your kind assistance

jg370
 
Physics news on Phys.org
What you did is ok. Indeed the scalar picks up a minus and a plus and that's how you end up with the commutator between H and Q.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
Replies
3
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
770
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
5K
Replies
2
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K