Wave functions


by redtree
Tags: functions, wave
redtree
redtree is offline
#1
Oct10-08, 02:51 AM
P: 64
Why are wave functions, e.g., Schrodinger's, based on the complex exponential function (e[tex]^{}ix[/tex]) and not trigonometric functions (sine or cosine)?

See Euler's formula for their relationship: http://en.wikipedia.org/wiki/Euler%27s_formula

Furthermore, by using the complex exponential function, the probability amplitude becomes a complex-valued function (a + bi). Were sine or cosine used, the probability amplitude of the wave function would not be a complex-valued function. Is there a reason that the probability amplitude should be a complex-valued function?
Phys.Org News Partner Physics news on Phys.org
Better thermal-imaging lens from waste sulfur
Scientists observe quantum superconductor-metal transition and superconducting glass
New technique detects microscopic diabetes-related eye damage
Ben Niehoff
Ben Niehoff is offline
#2
Oct10-08, 04:29 AM
Sci Advisor
P: 1,563
You need complex-valued wavefunctions so that you can have standing waves of constant magnitude over time (such as in a potential well). Real-valued functions would have to oscillate in magnitude.
redtree
redtree is offline
#3
Oct10-08, 04:34 AM
P: 64
What do you mean by magnitude? Is it the same as amplitude?

koolmodee
koolmodee is offline
#4
Oct10-08, 08:41 AM
P: 55

Wave functions


Nothing keeps you from using sine and cos. That what Euler's identity says.

The time evolution of a state vector/ wave function must conserve the norm of the state vector, and it always has to be one.

Note e[tex]^{it}[/tex] e[tex]^{it}[/tex]=1.
akhmeteli
akhmeteli is offline
#5
Oct10-08, 10:11 AM
P: 582
Quote Quote by redtree View Post
Why are wave functions, e.g., Schrodinger's, based on the complex exponential function (e[tex]^{}ix[/tex]) and not trigonometric functions (sine or cosine)?

See Euler's formula for their relationship: http://en.wikipedia.org/wiki/Euler%27s_formula

Furthermore, by using the complex exponential function, the probability amplitude becomes a complex-valued function (a + bi). Were sine or cosine used, the probability amplitude of the wave function would not be a complex-valued function. Is there a reason that the probability amplitude should be a complex-valued function?
My post http://www.physicsforums.com/showpos...5&postcount=20 and some other posts in that thread may be relevant.
dx
dx is offline
#6
Oct10-08, 02:24 PM
HW Helper
PF Gold
dx's Avatar
P: 1,962
The simple answer is that standard quantum mechanics as we understand it requires a complex wavefunction [tex] \psi [/tex] defined on the configuration space. You can replace this with a function to [tex] \mathbb{R} \times \mathbb{R} [/tex] and change the equations accordingly, since [tex] \mathbb{C} [/tex] and [tex] \mathbb{R} \times \mathbb{R} [/tex] are isomorphic. But you can't replace [tex] \psi [/tex] with a function to just [tex] \mathbb{R} [/tex].

Also, Euler's formula doesn't turn a complex number into a real number. [tex] \cos \theta + i \sin \theta [/tex] is still a complex number, and the probability amplitude will still be a complex valued function.
redtree
redtree is offline
#7
Oct12-08, 03:47 PM
P: 64
Quote Quote by dx View Post
The simple answer is that standard quantum mechanics as we understand it requires a complex wavefunction [tex] \psi [/tex] defined on the configuration space. You can replace this with a function to [tex] \mathbb{R} \times \mathbb{R} [/tex] and change the equations accordingly, since [tex] \mathbb{C} [/tex] and [tex] \mathbb{R} \times \mathbb{R} [/tex] are isomorphic. But you can't replace [tex] \psi [/tex] with a function to just [tex] \mathbb{R} [/tex].
What properties of standard quantum mechanics require a complex wavefunction?
malawi_glenn
malawi_glenn is offline
#8
Oct12-08, 04:02 PM
Sci Advisor
HW Helper
malawi_glenn's Avatar
P: 4,739
[QUOTE=redtree;1912396]
Quote Quote by dx View Post
The simple answer is that standard quantum mechanics as we understand it requires a complex wavefunction [tex] \psi [/tex] defined on the configuration space. You can replace this with a function to [tex] \mathbb{R} \times \mathbb{R} [/tex] and change the equations accordingly, since [tex] \mathbb{C} [/tex] and [tex] \mathbb{R} \times \mathbb{R} [/tex] are isomorphic. But you can't replace [tex] \psi [/tex] with a function to just [tex] \mathbb{R} [/tex].QUOTE]


What properties of standard quantum mechanics require a complex wavefunction?
well the physics of spin 1/2 systems for example, see Sakurai - Modern Quantum Mechanics chapter 1
dx
dx is offline
#9
Oct12-08, 05:22 PM
HW Helper
PF Gold
dx's Avatar
P: 1,962
Quote Quote by redtree View Post
What properties of standard quantum mechanics require a complex wavefunction?
That's like asking what property of Newtonian physics requires the mechanical state of a particle to be position and momentum. It is possible to construct a theory where the time evolution of a particle depends only on initial position, but that's not the way nature is. It just happens to be so that the mechanical state of a quantum system needed to predict future probabilities is a complex function.


Register to reply

Related Discussions
wave functions Introductory Physics Homework 2
Wave functions Quantum Physics 17
wave functions Advanced Physics Homework 10
Wave functions Quantum Physics 14