Why are observables represented by operators in Hilbert space?

In summary, the motivation behind saying that physical observables can be given by operators is mathematical. Physical observables can be thought of as matrix transformations in Hilbert space, and the values they take are given by their eigenvalues. This is a postulate of QM, and as a postulate, it is not proven. If it is a postulate, what motivates it?
  • #1
Lostinthought
19
0
i have been trying to learn a bit of quantum mechanics,this is some thing that has been bothering me ,
if the states of a system can be expressed as vectors in the Hilbert space,what is the motivation behind saying that physical observables can be given by operators?even then ,how can we say that the values they take are given by their eigenvalues?can this be proved somehow or reasoned out?
 
Physics news on Phys.org
  • #2
This is a postulate of QM, and as a postulate, it is not proven.
 
  • #3
if its a postulate then what motivates it?
 
  • #4
the motivation is purely mathematical. in Hilbert space, you can think of physical observables as matrix transformations. it so happens that a matrix transformation has a set of eigenvalues and eigenfunctions defined for it mathematically. these then corresponde to the possible values of the observable and the set of eigenstates of the system
 
  • #5
Hi Ardie,

Can you give a a physical example of matrix transofrmation or how it may represent a physical process other than observables? Thanks
 
  • #6
for example the hamiltonian in 1 dimension is a 2by1 matrix, that acts on a 1by1 vector (scalar psi) to give the scalar value of observable energy of the system (E)
 
  • #7
the act of transformation by a matrix is simply multiplying the vector by a matrix. you do this when u operate on psi with d/dx or d/dt or multiply etc, all these operations can be encoded into a matrix relation, and hence the analogy to Hilbert space.
 
  • #8
I think the motivation came from the historic development of QM as the matrix mechanics of Heisenberg, Born and Jordan. Namely, the canonical Poisson bracket of two "observables" [itex]f(q, p)[/itex] and [itex]g(q, p)[/itex]:
[tex]
\left\{f, g\right\} \equiv \sum_{k = 1}^{s}{\left(\frac{\partial f}{\partial q_{k}} \, \frac{\partial g}{\partial p_{k}} - \frac{\partial f}{\partial p_{k}} \, \frac{\partial g}{\partial q_{k}}\right)}
[/tex]
"goes over to" a commutator:
[tex]
-\frac{i}{\hbar} \, \left[F, G\right]
[/tex]
The simplest objects that do not commute are matrices. That is why the observables are correspondent to matrices and the new mechanics was called matrix mechanics.

The equation of evolution for a classical observable [itex]f(q, p)[/itex] is given by a Poisson bracket:
[tex]
\frac{d f}{d t} = \left\{f, H\right\}
[/tex]
where [itex]H = H(q, p)[/itex] is the Hamiltonian of the system, should go over to the Heisenberg equation of motion:
[tex]
\frac{d F}{d t} = -\frac{i}{\hbar} \, \left[F, H\right]
[/tex]

Every matrix can be written as:
[tex]
F(t) = U^{-1}(t) \, f \, U(t)
[/tex]
where [itex]f[/itex] is a diagonal matrix with the eigenvalues as the diagonal elements and the time evolution is given by the evolution matrix [itex]U(t)[/itex]. The above equation of motion is then equivalent to:
[tex]
\frac{d U(t)}{d t} = -\frac{i}{\hbar} \, H \, U(t), \ \frac{d U^{-1}(t)}{d t} = \frac{i}{\hbar} \, U^{-1}(t) \, H
[/tex]
 
Last edited:
  • #9
ardie said:
the act of transformation by a matrix is simply multiplying the vector by a matrix. you do this when u operate on psi with d/dx or d/dt or multiply etc, all these operations can be encoded into a matrix relation, and hence the analogy to Hilbert space.

Thank you ardie.
 

1. What is the difference between observables and eigenvalues?

Observables are physical quantities that can be measured in a given system, such as position, momentum, or energy. Eigenvalues, on the other hand, are the possible outcomes of a measurement for a given observable. In other words, observables are the properties of a system that can be observed, while eigenvalues are the specific values that can be measured for those properties.

2. How are observables and eigenvalues related in quantum mechanics?

In quantum mechanics, observables are represented by operators, which are mathematical objects that act on the wavefunction of a system to give the corresponding eigenvalues as outcomes of a measurement. The eigenvalues of an operator represent the possible results of measuring the corresponding observable.

3. Can an observable have multiple eigenvalues?

Yes, an observable can have multiple eigenvalues, depending on the system and the specific observable being measured. For example, the energy observable of a particle in a potential well can have multiple discrete eigenvalues, while the position observable can have a continuous range of eigenvalues.

4. How are observables and eigenvalues used in quantum mechanics?

In quantum mechanics, observables and eigenvalues play a crucial role in describing the behavior of a system. The eigenvalues of an observable represent the possible outcomes of a measurement, and the corresponding eigenvectors represent the states of the system that are associated with those outcomes. The superposition principle allows for the combination of multiple eigenstates, leading to the complex and unique behavior of quantum systems.

5. Are all physical quantities observables and have corresponding eigenvalues?

No, not all physical quantities are observables in quantum mechanics. Only properties that can be measured and have corresponding operators that act on the wavefunction to give eigenvalues are considered observables. Additionally, some observables may have continuous ranges of eigenvalues, while others may only have discrete eigenvalues.

Similar threads

  • Quantum Physics
Replies
24
Views
593
Replies
27
Views
2K
  • Quantum Physics
Replies
2
Views
921
  • Quantum Physics
Replies
7
Views
1K
Replies
3
Views
819
Replies
0
Views
446
Replies
4
Views
2K
Replies
9
Views
1K
  • Quantum Physics
Replies
4
Views
659
  • Quantum Physics
Replies
7
Views
548
Back
Top