Proving Complex Wave Function: |ψ1 + ψ2|^2 in [0, 4α]

Click For Summary
SUMMARY

The discussion focuses on proving that the expression |ψ1 + ψ2|^2 can take any value in the range [0, 4α], where α = |ψ1|^2 = |ψ2|^2. The wave functions are defined as ψ1(x) = √a exp(ik1x + iφ1) and ψ2(x) = √a exp(ik2x + iφ2), with k1, k2, φ1, and φ2 being real numbers. The solution involves understanding the properties of complex wave functions and their moduli, particularly through the use of triangular identities.

PREREQUISITES
  • Understanding of complex wave functions
  • Familiarity with the properties of modulus in complex numbers
  • Knowledge of exponential functions in quantum mechanics
  • Basic grasp of trigonometric identities and their applications
NEXT STEPS
  • Study the properties of complex wave functions in quantum mechanics
  • Learn about the application of triangular identities in complex analysis
  • Explore the implications of constant modulus in wave functions
  • Investigate the role of phase differences in interference patterns
USEFUL FOR

Quantum physicists, students of quantum mechanics, and anyone interested in the mathematical foundations of wave functions and their interactions.

jc09
Messages
42
Reaction score
0
Show that for ψ1 , ψ2 ∈ C,
|ψ1 + ψ2 |^2
can take any value in [0, 4α], where
α = |ψ1 |^2 = |ψ2 |^ 2

I think the solution has something to do with triangular identies but I am not sure how to start this problem at all.
 
Physics news on Phys.org
I think you can get away without using the triangle inequality. If [itex]\psi_1[/itex] and [itex]\psi_2[/itex] both have constant modulus [itex]\sqrt{a}[/itex] then the functions must be of the following form (assuming 1 spatial dimension and time independence).

[tex]\psi_1(x)=\sqrt{a}\exp\left(ik_1x+i\phi_1\right)[/tex]

[tex]\psi_1(x)=\sqrt{a}\exp\left(ik_2x+i\phi_2\right)[/tex]

Here [itex]k_1,k_2,\phi_1,\phi_2\in\mathbb{R}[/itex].

Can you do something with that?
 

Similar threads

Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
8K
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
8
Views
2K