Why Wave Functions Use Complex Exponentials vs Trigonometric Functions

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Why are wave functions, e.g., Schrodinger's, based on the complex exponential function (e^{}ix) and not trigonometric functions (sine or cosine)?

See Euler's formula for their relationship: http://en.wikipedia.org/wiki/Euler's_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?
 
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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.
 
What do you mean by magnitude? Is it the same as amplitude?
 
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^{it} e^{it}=1.
 
redtree said:
Why are wave functions, e.g., Schrodinger's, based on the complex exponential function (e^{}ix) and not trigonometric functions (sine or cosine)?

See Euler's formula for their relationship: http://en.wikipedia.org/wiki/Euler's_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 https://www.physicsforums.com/showpost.php?p=1539835&postcount=20 and some other posts in that thread may be relevant.
 
The simple answer is that standard quantum mechanics as we understand it requires a complex wavefunction \psi defined on the configuration space. You can replace this with a function to \mathbb{R} \times \mathbb{R} and change the equations accordingly, since \mathbb{C} and \mathbb{R} \times \mathbb{R} are isomorphic. But you can't replace \psi with a function to just \mathbb{R}.

Also, Euler's formula doesn't turn a complex number into a real number. \cos \theta + i \sin \theta is still a complex number, and the probability amplitude will still be a complex valued function.
 
dx said:
The simple answer is that standard quantum mechanics as we understand it requires a complex wavefunction \psi defined on the configuration space. You can replace this with a function to \mathbb{R} \times \mathbb{R} and change the equations accordingly, since \mathbb{C} and \mathbb{R} \times \mathbb{R} are isomorphic. But you can't replace \psi with a function to just \mathbb{R}.

What properties of standard quantum mechanics require a complex wavefunction?
 
redtree said:
dx said:
The simple answer is that standard quantum mechanics as we understand it requires a complex wavefunction \psi defined on the configuration space. You can replace this with a function to \mathbb{R} \times \mathbb{R} and change the equations accordingly, since \mathbb{C} and \mathbb{R} \times \mathbb{R} are isomorphic. But you can't replace \psi with a function to just \mathbb{R}.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
 
redtree said:
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.
 
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