Why do we have to use operators in QM?

Why do we have to use operators in QM?
 

tom.stoer

Science Advisor
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Re: Operator

B/c it works. It is a mathematical formalism which successfully describes nature. I don't think that we can explain this based on a deeper reason.
 
1,444
4
Re: Operator

We are also using operators in classical mechanics. For instance when solving differential equations we use sometimes Laplace or Fourier transform. Laplace and Fourier transforms are operators. Every function on the phase space is an operator (multiplication operator) - they commute. Translations and rotations acting on such functions are operators - they do not commute.
 
Re: Operator

B/c it works. It is a mathematical formalism which successfully describes nature. I don't think that we can explain this based on a deeper reason.
What do you mean by "a mathematical formalism" ? :confused:

There are so many mathemaical tools available to physicists. How did physicists know that they had to use operators?

Why/how do operators work anyway?
 
Re: Operator

We are also using operators in classical mechanics. For instance when solving differential equations we use sometimes Laplace or Fourier transform. Laplace and Fourier transforms are operators. Every function on the phase space is an operator (multiplication operator) - they commute. Translations and rotations acting on such functions are operators - they do not commute.
How are Laplace and Fourier transforms operators?

"Every function on the phase space is an operator (multiplication operator) - they commute. Translations and rotations acting on such functions are operators - they do not commute." :
I don't really understand this. I am only a second year physicist. Would you mind explaining this more elaborately?
 
1,444
4
Re: Operator

Fourier transform F applied to a function f gives you another function F(f). The map is linear. So, you gave a linear operator. By the Plancherel's theorem this is a unitary operator. Moreover its square is the inversion:

(F^2 f)(x) = f(-x)

Very interesting operator. But I am not sure if being a second year physicist you have already met with Fourier's transform?
 
Re: Operator

Fourier transform F applied to a function f gives you another function F(f). The map is linear. So, you gave a linear operator. By the Plancherel's theorem this is a unitary operator. Moreover its square is the inversion:

(F^2 f)(x) = f(-x)

Very interesting operator. But I am not sure if being a second year physicist you have already met with Fourier's transform?
Yes I have met Fourier transforms.

But I don't understand why it is a linear opeartor.
 

HallsofIvy

Science Advisor
Homework Helper
41,728
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Re: Operator

Fourier transforms are linear operators because integration is linear.

You probably learned in Calculus I that [itex]\int af(x)+ bg(x)dx= a\int f(x)dx+ b\int g(x)dx[/itex] where a and b are constants.
 
Re: Operator

Thanks!

I'm still wondering if anyone would take on the challenge of answering my other question? :cool:
 
1,341
3
Re: Operator

You use operators everywhere in physics not just in quantum mechanics.

Every time you take a derivative or an integral of some function you've used an operator.

In physics generally operators can be thought of like this:

Suppose you are walking down the street. There are many parameters that can be used to describe this. Position, velocity, momentum, acceleration, etc...

All of these quantities can be collectively grouped together and interconnected in quantum they are grouped together by the wave function for a given system. These quantities can be seen by using certain operators on the wave function.

It's not just in QM that we use operators. The only thing that changes is that you have to take extra caution with the operators you do use in QM. That's really the motivation for learning a little bit of abstract algebra in QM.

Asking why we use operators is like asking why we use a derivative of position to describe velocity.
 
Re: Operator

All of these quantities can be collectively grouped together and interconnected in quantum they are grouped together by the wave function for a given system. These quantities can be seen by using certain operators on the wave function.
Would you please elaborate?

It's not just in QM that we use operators. The only thing that changes is that you have to take extra caution with the operators you do use in QM. That's really the motivation for learning a little bit of abstract algebra in QM.
I don't really get the last two sentences.
 
1,341
3
Re: Operator

In physics we talk about stuff happening within a system. But before we can talk about of the stuff in our system behaves have a way to describe our system.

In quantum mechanics the way to describe our system is with the wave function.

In classical mechanics you can describe the system using the sum of the forces or the Lagrangian, etc...

Once we have a general description of the system we can observe any of the quantities you would be interested in using operators, to operate on our system.
 

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