Hermitian Operators: Finding Psi(p) from Psi(x)

In summary, the conversation discusses the concept of eigenfunctions and eigenvalues in quantum mechanics, specifically in relation to the momentum operator. Eigenvalues are important numbers as they represent the possible values of observables. Eigenfunctions are used as basis functions in expansion and the eigenvalue is the probability amplitude for that specific eigenfunction. These mathematical methods are well known and crucial in understanding quantum mechanics.
  • #1
khemist
248
0
I recently thought of this, please excuse me if it is way off the mark!

If I act on a state with a hermitian operator, am I able to find the psi(p) (momentum), where I had psi(x) (position) before (and wise versa)? Or does the operator do what it appears to do, and that is find the derivative of a particular state (if it does this, what does a derivative of a state mean?)?

Thanks
 
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  • #2
If we solve the eigenvalue equation for the momentum operator we get the momentum eigenvalues and momentum eigenfunctions. The eigenvalues of momentum are the possible values obtained when we measure momentum. If we are measuring momentum, then the wavefunction (obtained from the experimental configuration) must be expanded in terms of momentum eigenfunctions. The momentum eigenfunctions are the basis functions in this expansion and the expansion coefficients are the probability amplitudes.

Operating on any wavefunction with the momentum operator has no special significance, unless that wavefunction is a momentum eigenfunction. Operating on a momentum eigenfunction gives the momentum eigenvalue corresponding to that eigenfunction.
Best wishes
 
  • #3
The eigenfunction/ value equation being [tex]\hat{A}[/tex]x = [tex]\lambda[/tex]x

In this case, x is the eigenfunction, and lambda is the eigenvalue. Now I assume there is a method to find these functions, rather than simply plugging in particular ones and hoping that when we apply the momentum operator to it that it yields an eigenvalue. Now if that function is in fact an eigenfunction, we are able to use that function as a basis vector for the vector space (is it an arbitrary space, or a particular space?), and the scalar lambda will be the probability amplitude for the particular eigenfunction.

Does this make sense and sound reasonably correct? Thanks for the help!
 
  • #4
khemist said:
The eigenfunction/ value equation being [tex]\hat{A}[/tex]x = [tex]\lambda[/tex]x

In this case, x is the eigenfunction, and lambda is the eigenvalue. Now I assume there is a method to find these functions, rather than simply plugging in particular ones and hoping that when we apply the momentum operator to it that it yields an eigenvalue. Now if that function is in fact an eigenfunction, we are able to use that function as a basis vector for the vector space (is it an arbitrary space, or a particular space?), and the scalar lambda will be the probability amplitude for the particular eigenfunction.

Does this make sense and sound reasonably correct? Thanks for the help!

For the momentum operator [tex]\hat p_x = - i\hbar \partial /\partial x[/tex], the eigenvalue equation is [tex] - i\hbar \partial \varphi /\partial x = \lambda \varphi [/tex]. You must solve this differential equation subject to the appropriate boundary conditions for eigenfunctions [tex]\varphi [/tex] and eigenvalues [tex]\lambda [/tex]. [tex]\lambda [/tex] is not the probability amplitude. If we measure momentum for arbitrary state function
[tex]\psi [/tex], the probability amplitude is [tex]\left\langle {\varphi }
\mathrel{\left | {\vphantom {\varphi \psi }}
\right. \kern-\nulldelimiterspace}
{\psi } \right\rangle = 1/\sqrt {2\pi \hbar } \int\limits_{ - \infty }^\infty {\varphi ^* \psi dx} [/tex]. You may recognize this as the Fourier Transform of [tex]\psi [/tex]. The mathematical methods are well known.
 
  • #5
So is the eigenvalue not a necessarily important number, rather the important part is the fact that there are eigenvalues. The eigenfunction represents the basis for a particular space that one is solving for.
 
  • #6
khemist said:
So is the eigenvalue not a necessarily important number, rather the important part is the fact that there are eigenvalues

Eigenvalues (or generally spectral values) of operators describing observables are very important numbers, they are the only possible values of the observables, if one measures these physical properties/quantities.

From the mathematical perspective, not choosing properly the operators accounting for the description of the observables in the mathematical formalism will generally not give you any eigenvalue.
 

1. What is a Hermitian operator?

A Hermitian operator is a mathematical operator used in quantum mechanics to describe the properties of a physical system. It is a linear operator that is equal to its own conjugate transpose, which means that its eigenvalues are all real. In other words, it is a type of operator that preserves the probability of finding a particle in a certain state.

2. How do you find Psi(p) from Psi(x) using a Hermitian operator?

To find Psi(p) from Psi(x), you can use the Fourier transform, which is a mathematical operation that converts a function in one domain (such as position) to a function in another domain (such as momentum). In this case, the Hermitian operator is used as a multiplier in the Fourier transform equation. By plugging in the function Psi(x) and solving the equation, you can find the function Psi(p) in momentum space.

3. What is the relationship between the Hermitian operator and the uncertainty principle?

The Hermitian operator and the uncertainty principle are closely related. The uncertainty principle states that it is impossible to know both the exact position and momentum of a particle at the same time. This is because the Hermitian operator, which represents the measurement of momentum, and the position operator do not commute. In other words, the order in which you measure these two properties affects the outcome, and there will always be some uncertainty in one of the measurements.

4. Can a Hermitian operator have complex eigenvalues?

No, a Hermitian operator must have real eigenvalues. This is because the definition of a Hermitian operator requires that it is equal to its own conjugate transpose. Complex eigenvalues would violate this condition. Additionally, in quantum mechanics, the eigenvalues of a Hermitian operator represent the possible outcomes of a measurement, and these outcomes must be real numbers.

5. How is the Hermitian property of an operator related to the conservation of probability?

The Hermitian property of an operator is directly related to the conservation of probability in quantum mechanics. This is because the Hermitian property ensures that the operator is self-adjoint, meaning that it will always return real eigenvalues. In quantum mechanics, the probability of finding a particle in a certain state is given by the square of the magnitude of the wave function, so it is important that the operator used to describe the system preserves this property.

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