Is the normalisation constant of a wavefunction real?

Click For Summary
SUMMARY

The discussion centers on the normalization constant of a wavefunction, specifically the factor ##c## in the equation ##\Psi (x, t) = c\ \psi (x) e^{-iEt/ \hbar}##. It is established that the normalization condition ##\int | \Psi (x, t) |^{2} dx = 1## leads to the conclusion that ##c## can be chosen as a real number. This is due to the fact that quantum mechanics allows for the redefinition of phase factors without affecting the physical outcomes, thus enabling the normalization constant to be real.

PREREQUISITES
  • Understanding of wavefunctions in quantum mechanics
  • Familiarity with normalization conditions in quantum states
  • Knowledge of complex numbers and their properties
  • Basic principles of quantum phase factors
NEXT STEPS
  • Study the implications of normalization in quantum mechanics
  • Explore the role of phase factors in quantum state interference
  • Learn about the mathematical properties of complex numbers in quantum theory
  • Investigate different forms of wavefunctions and their normalization
USEFUL FOR

Students of quantum mechanics, physicists exploring wavefunction properties, and anyone interested in the mathematical foundations of quantum theory.

spaghetti3451
Messages
1,311
Reaction score
31

Homework Statement



Consider the wavefunction ##\Psi (x, t) = c\ \psi (x) e^{-iEt/ \hbar}## such that ##\int | \Psi (x, t) |^{2} dx = 1##.

I would like to prove to myself that the normalisation factor ##c## is a real number.

Homework Equations



The Attempt at a Solution



##\int | \Psi (x, t) |^{2} dx = 1##

##\int c\ \psi (x)\ e^{-iEt/ \hbar}\ c^{*}\ \psi^{*}(x)\ e^{iEt/ \hbar}\ dx = 1##

##\int c\ \psi (x)\ c^{*}\ \psi^{*}(x)\ dx = 1##

##\int |c|^{2}\ |\psi (x)|^{2}\ dx = 1##

This isn't getting me anywhere, though! :frown:
 
Physics news on Phys.org
You cannot prove this, quantum mechanics is fine with redefining all phases by the same phase factor and the physics does not depend on the phase of a state. It only depends on the relative phase of interfering states. However, you may choose the normalisation constant to be real for this very reason.
 
  • Like
Likes   Reactions: Noctisdark
Thanks!

I knew that the phase factors are tunable, but I was not able to see that this could account for the plausibility of a real normalisation constant.

Now, it's all clear!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
4
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K