Shouldn't the operator be applied to both the wf anD its modulus?

In summary, the two formulas presented are different ways of calculating the expectation value of a random variable in quantum mechanics. The first formula follows from the standard definition of expectation value, while the second formula is not linear and may not make as much mathematical sense. There may be a deeper reason why the first formula works, possibly related to the postulates of quantum mechanics.
  • #1
jshrager
Gold Member
24
1
Why is it:

$$ \langle A \rangle = \int dV ~ \psi^* (\hat{A} \psi) $$

As opposed to:

$$ \langle A \rangle = \int dV ~ (\hat{A} \psi^*) (\hat{A} \psi) $$

The op can substantially change the wf, so it would seem to make more mathematical sense (at least from a linear algebra pov) to operate on both the wf and its mod, otherwise, as we do it now (first expr) you're getting a very odd product. Is this is a property of the way that wfs are formulated? Are they a special case where this isn't required? Or maybe it's a magical property of hermetian operators? Or maybe it just works and I shouldn't think too hard about it.
 
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  • #2
1) it works
2) the second formula is not linear
3) There is probably some deeper reason why this works
 
  • #3
If you start with the standard definition of the expectation value of a random variable <A> = Ʃi piai -where the ais are the possible outcomes and pi is the probability to obtain the i-th- your first formula follows from the QM postulates about these quantities. I don't know if this is satifying to you. ;-)
 
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Related to Shouldn't the operator be applied to both the wf anD its modulus?

1. What is the purpose of applying the operator to both the wave function and its modulus?

The operator is applied to both the wave function and its modulus in order to obtain a complete description of the physical system. The wave function represents the amplitude of the particle's probability distribution, while the modulus represents the probability of finding the particle at a specific location. By applying the operator to both, we can determine the average value of a physical quantity such as position or momentum.

2. Can the operator be applied to just the wave function or modulus alone?

No, the operator must be applied to both the wave function and its modulus to obtain accurate results. Applying it to only one of these quantities would not provide a complete description of the physical system and could lead to incorrect conclusions about the system's behavior.

3. How does applying the operator to both the wave function and its modulus affect the results?

Applying the operator to both the wave function and its modulus allows us to obtain a more accurate and complete description of the system. The results will reflect both the particle's amplitude and probability, providing a more nuanced understanding of its behavior.

4. Is it necessary to apply the operator to both the wave function and its modulus every time?

Yes, it is necessary to apply the operator to both the wave function and its modulus every time we want to obtain a complete description of the system. This is a fundamental aspect of quantum mechanics and cannot be skipped or simplified without sacrificing the accuracy of the results.

5. Are there any exceptions where the operator does not need to be applied to both the wave function and its modulus?

No, there are no exceptions where the operator does not need to be applied to both the wave function and its modulus. This is a fundamental principle in quantum mechanics and applies to all physical systems and scenarios.

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