Are all selfadjoint operators in quantum mechanics bounded?

In summary: The spectrum of an operator is defined for operators that are bounded. However, there is an extended spectrum for unbounded operators. However, this spectrum may be empty.
  • #1
hitche
3
0
Hi, I'd like to know if the following statement is true:
Let [itex] \hat{A}, \hat{B} [/itex] be operators for any two observables [itex]A, B[/itex]. Then [itex] \langle \hat{A} \rangle_{\psi} = \langle \hat{B} \rangle_{\psi} \forall \psi [/itex] implies [itex] \hat{A} = \hat{B} [/itex].
Here, [itex] \langle \hat{A} \rangle_{\psi} = \int_\mathbb{R^3} d^3x \psi^* \hat{A} \psi [/itex] is standard definition of expectation value of an operator [itex]\hat{A}[/itex].
If this doesn't hold, could you provide any counterexample? Thank you!
 
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  • #2
Any two operators with odd parity would give zero for any wave function.
 
  • #3
yes - because the expected value of A-B will be zero for all psi, then by taking psi to be A-B's eigenvector, you reached the conclusion that A-B has only zero eigenvalues, which means A-B must be a zero operator.
 
  • #4
Of course, not. The general solution of [itex]\langle\psi, A\psi \rangle= \langle\psi, B\psi\rangle [/itex] is that either one is an extension of the other. Equality follows iff both operators are bounded. As a counterexample would be any symmetric operator which is not self-adjoint.
 
  • #5
cattlecattle said:
yes - because the expected value of A-B will be zero for all psi, then by taking psi to be A-B's eigenvector, you reached the conclusion that A-B has only zero eigenvalues, which means A-B must be a zero operator.

Not that's wrong. If either A or B is unbounded then clearly your argument won't apply.
 
  • #6
WannabeNewton said:
Not that's wrong. If either A or B is unbounded then clearly your argument won't apply.

sorry, I failed to see which step was not applicable due to unboundedness
 
  • #7
cattlecattle said:
sorry, I failed to see which step was not applicable due to unboundedness

I'm new to the theory, but I think the reason is that spectrum of operator is defined for bounded operators. There exists also some extended definition of spectrum for unbounded operators, but it can be empty (unlike the spectrum of a bounded operator).
BUT I read that in quantum mechanics every observable is represented by selfadjoint linear operator and according to another resource every selfadjoint everywhere defined operator is bounded. On the other hand bounded self adjoint operator may have no eigenvalue. So.. I don't know!
 

FAQ: Are all selfadjoint operators in quantum mechanics bounded?

What is an expectation value?

An expectation value is a mathematical concept used in quantum mechanics to represent the average value of a measurable quantity for a given quantum state.

How is the expectation value calculated?

The expectation value is calculated by taking the sum of the product of each possible value of the quantity and its corresponding probability, for all possible values of the quantity.

What is the significance of the expectation value in quantum mechanics?

The expectation value represents the most probable outcome of a measurement for a given quantum state. It also helps to determine the uncertainty or spread of the measurements.

What is the difference between expectation value and actual measurement?

The expectation value is a theoretical prediction based on the probabilities of different outcomes, while the actual measurement is the result obtained when measuring the quantity on a specific quantum system.

How is the concept of expectation value applied in real-world scenarios?

The concept of expectation value is applied in many areas of quantum mechanics, such as in the calculation of energy levels and transition probabilities in atoms and molecules, and in the prediction of particle behavior in quantum systems.

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