Form of Wavefunctions: Exploring Outside of e^ikx

In summary: BACKGROUND FYIWave equations were important in physics long before the invention of quantum mechanics. A wave equation is a function which satisfies the second order differential equation ∂²F/∂t² = c²∇²F. Probably, you haven't seen the ∇² ( ) operator before. An operator acts on something (a number or a vector or a function or...) You're familiar with the addition "operator" +, and the cosine operator cos( ) and probably the derivative operator d( ). The Laplacian operator ∇² (del squared) is also symbolized as ∇⋅∇ or sometimes Δ but since it must act on something, it isn't something that should
  • #1
Hi all!
I'm currently watching MIT 8.04 (Quantum Physics I) on MIT open courseware. I have just finished lecture 5. In the past 2 lectures, they introduced opperators, specifically momentum, energy and position. To prove/derive (I'm not sure what the correct term is) the momentum and energy opperators, they operated on ##e^{ikx}## and got back to the De Broglie equations. So, my question is, do wavefunctions have to be in the form ##Ae^{ikx}##? I would think no, but if I am correct, would the momentum and energy opperators work for other types of wavefunctions? Why is it enough to prove it just with ##e^{ikx}##? This especially confuses me because ##e^{ikx}## isn't even normalizeable, and thus can't even be a wavefunction.
Thanks!
 
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  • #2
Isaac0427 said:
do wavefunctions have to be in the form AeikxAeikxAe^{ikx}?
A wavefunction is any functions satisfying the Schroedinger equation. The specific form you have there is an eigenfunction of a free particle (##V(x)=C<E##).
Isaac0427 said:
would the momentum and energy opperators work for other types of wavefunctions?
By "work", do you mean operate or act on? Yes they can, except for a special case of infinite well where the momentum operator is not defined.
Isaac0427 said:
This especially confuses me because eikxeikxe^{ikx} isn't even normalizeable, and thus can't even be a wavefunction.
The free particle's eigenfunction does not correspond to a realizable wavefunction, but it can serve as the eigenfunction of a realizable (normalizable) wavefunction.
 
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  • #3
blue_leaf77 said:
A wavefunction is any functions satisfying the Schroedinger equation. The specific form you have there is an eigenfunction of a free particle (##V(x)=C<E##).

By "work", do you mean operate or act on? Yes they can, except for a special case of infinite well where the momentum operator is not defined.

The free particle's eigenfunction does not correspond to a realizable wavefunction, but it can serve as the eigenfunction of a realizable (normalizable) wavefunction.
Ok, so my big question is "why can you say that if it works on ##e^{ikx}## it works on any wavefunction (with a few exceptions but I don't want to focus on those)?"
 
  • #4
Isaac0427 said:
Ok, so my big question is "why can you say that if it works on ##e^{ikx}## it works on any wavefunction (with a few exceptions but I don't want to focus on those)?"

I'm not exactly sure what you mean by "work", but every wave function [itex]\psi(x)[/itex] can be written as a superposition of exponentials:

[itex]\psi(x) = \sqrt{\frac{1}{2\pi}} \int dk \tilde{\psi}(k) e^{ikx}[/itex]

Since the operators of quantum mechanics are linear (typically), you can (usually) take them inside the integral:

[itex]\hat{p} \psi(x) = \sqrt{\frac{1}{2\pi}} \int dk \tilde{\psi}(k) \hat{p} e^{ikx} = \sqrt{\frac{1}{2\pi}} \int dk \tilde{\psi}(k) (\hbar k) e^{ikx} [/itex]
 
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  • #5
stevendaryl said:
I'm not exactly sure what you mean by "work", but every wave function [itex]\psi(x)[/itex] can be written as a superposition of exponentials:

[itex]\psi(x) = \sqrt{\frac{1}{2\pi}} \int dk \tilde{\psi}(k) e^{ikx}[/itex]

Since the operators of quantum mechanics are linear (typically), you can (usually) take them inside the integral:

[itex]\hat{p} \psi(x) = \sqrt{\frac{1}{2\pi}} \int dk \tilde{\psi}(k) \hat{p} e^{ikx} = \sqrt{\frac{1}{2\pi}} \int dk \tilde{\psi}(k) (\hbar k) e^{ikx} [/itex]
Oh, so if I'm getting you correctly, if something applies to ##e^{ikx}## then it applies to all wavefunctions because a wavefunction can be expressed in terms of ##e^{ikx}##.
 
  • #6
Isaac0427 said:
Oh, so if I'm getting you correctly, if something applies to ##e^{ikx}## then it applies to all wavefunctions because a wavefunction can be expressed in terms of ##e^{ikx}##.

Yes, in principle, if you know how an operator acts on [itex]e^{ikx}[/itex], you can figure out how it works on any wavefunction.
 
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  • #7
BACKGROUND FYI
Wave equations were important in physics long before the invention of quantum mechanics. A wave equation is a function which satisfies the second order differential equation ∂²F/∂t² = c²∇²F. Probably, you haven't seen the ∇² ( ) operator before. An operator acts on something (a number or a vector or a function or...) You're familiar with the addition "operator" +, and the cosine operator cos( ) and probably the derivative operator d( ). The Laplacian operator ∇² (del squared) is also symbolized as ∇⋅∇ or sometimes Δ but since it must act on something, it isn't something that should be thought of as existing by itself (unlike a number, vector or function). This is simplified, of course. Anyway ∇²(F) for 1 dimension is ∂²F/∂x² (assuming x is the dimension) while for 3 dimensions it is ∂²F/∂x² + ∂²F/∂y² + ∂²F/∂z² (assuming cartesian coordinates (dimensions) x,y,& z). Functions of the form e^x have derivatives of every order, so they are a "natural" for use with differential equations. The function e^ix is, by definition, cos(x) + i*sin(x) (and you probably know that sin & cos also have derivatives of all orders...). Also the imaginary constant i here can pretty much be treated as any other number (with the obvious point that i*i = -1), so shouldn't throw your understanding. Since either ∑An*sin(x) where An is a set of constants with n ranging from 1 to ∞, or ∑Bn*cos(x) can (in the limit) exactly equal any function, F(x). The utility of expressing any function in terms of its (fourier series) basis should be apparent. Note that the notation can be confusing since it is abbreviated short-hand. For instance in the above sums, "x" is meant to be short-hand to imply "for the variable x in its range of validity, for any x..." So, a lot is implied and lecturers try to start introducing that to the student gently. I've never seen an advanced text (or paper) write out in full what exactly the authors mean, they all use shorthand, so get used to it. And get used to expanding their equations into a form meaningful to you. I suppose I should also note that classical wave equations usually were differentials with respect to spatial dimensions (coordinates) and time, but in relativistic QM time and space are mixed so ∇² may range from 1 to 4 (or from 0 to 3)... As already answered, using the form e^ix for wave functions makes working with them much much easier and you can do it without losing the universality of application to any wave equation...they're general purpose. Plus, for the simple cases (infinite potential well, harmonic oscillator, etc.) they make things neat and tidy...and easy!
 
  • #8
ogg said:
BACKGROUND FYI
Wave equations were important in physics long before the invention of quantum mechanics. A wave equation is a function which satisfies the second order differential equation ∂²F/∂t² = c²∇²F. Probably, you haven't seen the ∇² ( ) operator before. An operator acts on something (a number or a vector or a function or...) You're familiar with the addition "operator" +, and the cosine operator cos( ) and probably the derivative operator d( ). The Laplacian operator ∇² (del squared) is also symbolized as ∇⋅∇ or sometimes Δ but since it must act on something, it isn't something that should be thought of as existing by itself (unlike a number, vector or function). This is simplified, of course. Anyway ∇²(F) for 1 dimension is ∂²F/∂x² (assuming x is the dimension) while for 3 dimensions it is ∂²F/∂x² + ∂²F/∂y² + ∂²F/∂z² (assuming cartesian coordinates (dimensions) x,y,& z). Functions of the form e^x have derivatives of every order, so they are a "natural" for use with differential equations. The function e^ix is, by definition, cos(x) + i*sin(x) (and you probably know that sin & cos also have derivatives of all orders...). Also the imaginary constant i here can pretty much be treated as any other number (with the obvious point that i*i = -1), so shouldn't throw your understanding. Since either ∑An*sin(x) where An is a set of constants with n ranging from 1 to ∞, or ∑Bn*cos(x) can (in the limit) exactly equal any function, F(x). The utility of expressing any function in terms of its (fourier series) basis should be apparent. Note that the notation can be confusing since it is abbreviated short-hand. For instance in the above sums, "x" is meant to be short-hand to imply "for the variable x in its range of validity, for any x..." So, a lot is implied and lecturers try to start introducing that to the student gently. I've never seen an advanced text (or paper) write out in full what exactly the authors mean, they all use shorthand, so get used to it. And get used to expanding their equations into a form meaningful to you. I suppose I should also note that classical wave equations usually were differentials with respect to spatial dimensions (coordinates) and time, but in relativistic QM time and space are mixed so ∇² may range from 1 to 4 (or from 0 to 3)... As already answered, using the form e^ix for wave functions makes working with them much much easier and you can do it without losing the universality of application to any wave equation...they're general purpose. Plus, for the simple cases (infinite potential well, harmonic oscillator, etc.) they make things neat and tidy...and easy!
I actually have experience with the Laplacian opperator but I generally think of it as div(grad(F)) and not del squared. That is interesting though. I have worked with it in classical mechanics but not in quantum mechanics.
 
  • #9
EDIT: Sorry a mistyping, in the last sentence in post #2 I should have said basis function, rather than eigenfunction.
 
  • #10
blue_leaf77 said:
A wavefunction is any functions satisfying the Schroedinger equation. The specific form you have there is an eigenfunction of a free particle (##V(x)=C<E##).
... and, very important, a wave function, representing a pure quantum state, must be square integrable, i.e., the integral
$$\|\psi \|^2=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\psi(\vec{x})|^2$$
must exist. In order to represent a probability amplitude one must be able to normalized the wavefunction such that ##\|\psi \|=1##, and then ##|\psi(\vec{x})|^2## is the probability distribution of the particle's position.

The plane-wave functions are generalized functions (or distributions) in the sense of functional analysis. They cannot represent states, but you can represent wave functions in terms of these generalized functions. In the case of the momentum eigenstates (I set ##\hbar=1## for convenience)
$$u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi)^{3/2}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}).$$
You can represent any position-space wave function in terms of the corresponding momentum-space wave function. Since the momentum eigenstates are represented by the exponential function that's just Fourier transforming between the two representations:
$$\psi(\vec{x})=\int_{\mathbb{R}} \mathrm{d}^3 \vec{p} \tilde{\psi}(\vec{p}) u_{\vec{p}}(\vec{x})$$
and the other way around
$$\tilde{\psi}(\vec{x})=\int_{\mathbb{R}} \mathrm{d}^3 \vec{x} \psi(\vec{x}) u_{\vec{p}}^*(\vec{x}).$$
The momentum eigenstates are of course not square integrable, but they are "renormalized to a Dirac ##\delta## distribution" in the sense that
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}_1}^*(\vec{x}) u_{\vec{p}_2}(\vec{x})=\delta^{(3)}(\vec{p}_1-\vec{p}_2).$$
 
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  • #11
vanhees71 said:
The plane-wave functions are generalized functions (or distributions) in the sense of functional analysis. They cannot represent states, but you can represent wave functions in terms of these generalized functions.

:smile::smile::smile::smile::smile::smile::smile::smile::smile::smile::smile:

To the OP the full resolution to this requires what are called Rigged Hilbert spaces that was developed by some of the greatest mathematicians of the middle 20th century eg Schwatz (distribution theory), Grothendieck (Nuclear Spaces) and Gelfland (Gelfland Triple). It will take a lot of study to come to grips with that.

But as a start I highly recommend the following:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

Thanks
Bill
 
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1. What is the concept of "form of wavefunctions" in quantum mechanics?

The form of wavefunctions refers to the mathematical function that describes the behavior of a quantum system. In quantum mechanics, particles are described as waves, and the wavefunction is a mathematical representation of the probability of finding a particle at a given position. It contains information about the position, momentum, and energy of the particle.

2. What does the term "e^ikx" represent in the wavefunction?

The term "e^ikx" represents the phase factor of the wavefunction. It is a complex number that describes the oscillatory behavior of the wavefunction. The "i" represents the imaginary unit, and "k" represents the wavenumber, which is related to the momentum of the particle.

3. How does the form of wavefunctions change when exploring outside of e^ikx?

When exploring outside of e^ikx, the form of wavefunctions can take on different mathematical functions, such as trigonometric functions, exponential functions, or a combination of both. These functions depend on the potential energy of the system and can be solved using the Schrödinger equation.

4. What is the significance of exploring outside of e^ikx in quantum mechanics?

Exploring outside of e^ikx allows us to study a wider range of quantum systems that cannot be described by the simple plane wave function. It also helps us understand the effects of different potentials on the behavior of particles and provides insight into the behavior of complex systems.

5. Can the form of wavefunctions be experimentally observed?

No, the form of wavefunctions cannot be directly observed as they represent the probability of finding a particle at a given position. However, the predictions of the wavefunction can be experimentally tested and verified through various quantum experiments, such as the double-slit experiment or the Stern-Gerlach experiment.

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