Expectation value of non-physical observable

In summary, the conversation discusses the concept of physical observables in quantum mechanics and how they must yield real expectation values. The question is posed about a specific calculation involving a first derivative and potential, which is not considered a physical observable. The answer is that the linear operators used to describe quantum observables must be self-adjoint and have a purely real spectrum, leading to real expectation values. However, valid observables can be constructed by getting the operator ordering right. The combination of position and momentum, represented by the operator D, is self-adjoint and can be interpreted as the product of the two physical observables.
  • #1
Visceral
59
0

Homework Statement


This may be incredibly obvious, but I just need to check. Of course we all know that physical observables must yield real expectation values. What if you tried to calculate, say, <xV(d/dx)>, where x is the position, d/dx is a first derivative, and V is the potential? This isn't an observable...so do we automatically know the answer is zero(or non-real)?

Homework Equations





The Attempt at a Solution

 
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  • #2
Nope.
 
  • #3
vela said:
Nope.

So after searching around I agree with your answer, but I am not sure why this is so. For example, <xp> is not zero but has a complex component. xp isn't really considered a physical observable though...is it? Is the only restriction on for a real expectation value that the quantity be a "physically meaningful" observable such as momentum, energy, etc..?
 
  • #4
The basic requirement is for the linear operators which describe quantum observables to be self-adjoint. xp isn't. Self-adjoint operators have a spectral decomposition with a purely real spectrum, so that every expectation value on a state in their domain is real.
 
  • #5
However, the combination

[tex]D = \frac{XP+PX}{2},[/tex]

which looks the same classically, is self adjoint. One can often construct valid observables by getting the operator ordering right.
 
  • #6
Oxvillian said:
However, the combination

[tex]D = \frac{XP+PX}{2},[/tex]

which looks the same classically, is self adjoint. One can often construct valid observables by getting the operator ordering right.

So since this is self adjoint...then how to we interpret this physically?
 
  • #7
Simply as the product of position and momentum, I would think. We could in principle design a clever experiment to measure D.
 

1. What is the expectation value of a non-physical observable?

The expectation value of a non-physical observable is a statistical measure of the average value that would be obtained from a large number of measurements of that observable in a given quantum state. It is a mathematical concept used in quantum mechanics to describe the expected behavior of a system.

2. How is the expectation value of a non-physical observable calculated?

The expectation value of a non-physical observable is calculated by taking the inner product of the state vector with the observable operator. This is then multiplied by the complex conjugate of the state vector and integrated over all possible values of the observable. The result is a single number that represents the average value of the observable in that state.

3. Can the expectation value of a non-physical observable be negative?

Yes, the expectation value of a non-physical observable can be negative. In quantum mechanics, the expectation value is a probability-weighted average, so it is possible for the average value to be negative if the probabilities for different outcomes are appropriately weighted.

4. How does the uncertainty principle affect the expectation value of a non-physical observable?

The uncertainty principle states that it is impossible to know with certainty both the position and momentum of a particle at the same time. This principle also applies to other non-physical observables. Therefore, the expectation value of a non-physical observable may have a larger uncertainty due to the inherent uncertainty in the quantum system.

5. What is the significance of the expectation value of a non-physical observable in quantum mechanics?

The expectation value of a non-physical observable is a fundamental concept in quantum mechanics and plays a crucial role in predicting the behavior of a system. It allows scientists to make probabilistic predictions about the outcomes of measurements and provides a way to compare different quantum states. It is also used in the formulation of important principles such as the conservation of energy and the uncertainty principle.

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