• Empty
Quantum mechanics question. Why is the derivative of the expectation value of position the expectation of the velocity? I don't think these two things are exactly equal.
  • Stacey Warren - Expert brainly.com
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  • jamiebookeater
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  • Empty
Specifically, in math terms what I am saying is I don't think these two things are equal. \[\frac{d \langle x \rangle}{dt} \ne \langle \frac{dx}{dt} \rangle \]
  • anonymous
Off the top of my head, they should be equivalent. For a single observable, I can't think of any time the two quantities wouldn't be the same (definitely doesn't mean that it doesn't exist. Quantum mechanics can be very...subtle). The issue you are alluding to usually arises when we try to calculate products of observables (including the squares of observables). To take this a bit further, let's look at the expectation value of x. A wave function is a probability distribution of various things. We can calculate the expectation value of a particle by doing: \[\langle x\rangle = \int\limits \Psi^*(\vec{x},t)x \Psi (\vec{x},t)dx\] Now, if we calculate the expectation value of dx/dt, we see that \[\langle\frac{dx}{dt}\rangle = \left\langle\int\limits\Psi^*\frac{dx}{dt}\Psi\right\rangle = \left\langle\frac{d(\int\limits\Psi^* x\Psi dx)}{dt}\right\rangle =\frac{d(\int\limits \Psi^*\langle x \rangle\Psi dx)}{dt}=\frac{d\langle x\rangle}{dt}\] But, if you tried this with a product of observables (or a square of an observable), it wouldn't work. That is: \[\langle AB\rangle \neq \langle A\rangle\langle B\rangle\] \[\langle x^2\rangle \neq \langle x\rangle\langle x\rangle\]

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