We will inspect the time evolution of an expectation value of an operator Ô and then apply the result to the position operator. (A) Consider the expectation value of Ô for quantum state represented by the wavefunction (z): (0) = / ®(z, t)Ô ®(z, t)dz We denote the expectation value by angle brackets, i.e., (-..). Use the time-dependent Schrödinger equation: ih(z,t) = Ĥ®(x, t) and ih (z, t) = Ĥ®°(1, t) where the second identity stems from the self-adjointness of the hamiltonian operator. Prove that the following identity holds for an expectation value (o) of an operator Ô : d To prove this, consider taking the time derivative of the expectation value integral with wavefunction of an arbitrary quantum state &(x). Note: the partial derivative means we consider the explicit dependence of Ô on time - we have not encountered such operators in practice. (B) In the class, we only consider special cases in which the operators are time-independent, = 0. This holds for all the operator we are considering, e.g., position â for which đf/at = 0. Use the above expression to prove that the classical relation between momentum and the time derivative of position holds even in quantum mechanics, i.e., i.e., prove that: d m(x) = (p)

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We will inspect the time evolution of an expectation value of an operator Ô and then apply
the result to the position operator.
(A) Consider the expectation value of Ô for quantum state represented by the wavefunction
(x):
(0) = / 0*(z, t)Ô +(z, t)dæ
We denote the expectation value by angle brackets, i.e., (.). Use the time-dependent
Schrödinger equation:
iħ(x, t) = Ĥ&(x, t)
– iħ*(x, t) = Ĥ¢*(x,t)
and
at
where the second identity stems from the self-adjointness of the hamiltonian operator.
Prove that the following identity holds for an expectation value (o) of an operator Ö :
(0)
dt
%3D
iħ
at
To prove this, consider taking the time derivative of the expectation value integral
with wavefunction of an arbitrary quantum state (x). Note: the partial derivative
means we consider the explicit dependence of Ô on time – we have not encountered
such operators in practice.
(B) In the class, we only consider special cases in which the operators are time-independent,
i.e., () = 0. This holds for all the operator we are considering, e.g., position ât for
which dể/ôt = 0. Use the above expression to prove that the classical relation between
momentum and the time derivative of position holds even in quantum mechanics, i.e.,
prove that:
m (a) = (p)
dt
(C) In the previous question, you encountered the cannonical commutation relation:
• What is the cannonical commutation relation?
• What does it mean for two observables if two associated operators do not commute?
• How does the lack of commutativity impact quantum measurements?

Expert Solution

Step 1 Part (a)

$T h e o p e r a t o r b e < \hat{O} > N o w, e q u a t i o n o f t h e m o t i o n \frac{d}{d t} <\hat{O}> = \frac{d}{d t} \{<ϕ |\hat{O}| ϕ>\} N o w, d i f f e r e n t i a t e \frac{d}{d t} ⟨ \hat{O} ⟩ = ⟨ \frac{\partial ϕ}{\partial t} | \hat{O} | ϕ ⟩ + ⟨ ϕ | \frac{\partial \hat{O}}{\partial t} | ϕ ⟩ + ⟨ ϕ | \hat{O} | \frac{\partial ϕ}{\partial t} ⟩ - - - (A)$

$T h e S c h r o d i n g e r t i m e - i n d e p e n d e n t E q u a t i o n i h = \frac{\partial ϕ}{\partial t} > = \hat{H} | ϕ > \frac{\partial ϕ}{\partial t} > = \frac{\hat{H}}{i h} | ϕ > H e r e, s t h e c o n j u g a t e c o m p l e x \frac{\partial ϕ}{\partial t} > = \frac{i \hat{H}}{h} | ϕ > f r o m e q . (1) \frac{d}{d t} <\hat{O}> = \frac{i}{h} ⟨ ϕ | \hat{H} \hat{O} | ϕ ⟩ + ⟨ ϕ | \frac{\partial \hat{O}}{\partial t} | ϕ ⟩ + \frac{1}{i h} ⟨ ϕ | \hat{O} \hat{H} | ϕ ⟩ \frac{d}{d t} <\hat{O}> = \frac{- 1}{i h} ⟨ ϕ | \hat{H} \hat{O} | ϕ ⟩ + ⟨ ϕ | \frac{\partial \hat{O}}{\partial t} | ϕ ⟩ + \frac{1}{i h} ⟨ ϕ | \hat{O} \hat{H} | ϕ ⟩ h e r e, i^{2} = - 1$

$N o w, t a k e c o m m o n \frac{1}{i h} \frac{d}{d t} <\hat{O}> = \frac{1}{i h} [- ⟨ ϕ | \hat{H} \hat{O} | ϕ ⟩] + [<ϕ | \hat{O} \hat{H} | ϕ>] + ⟨ ϕ | \frac{\partial \hat{O}}{\partial t} | ϕ ⟩$

$\frac{d}{d t} <\hat{O}> = \frac{1}{i h} [<ϕ | \hat{O}, \hat{H} | ϕ>] + ⟨ ϕ | \frac{\partial \hat{O}}{\partial t} | ϕ ⟩ U s i n g c o m m u t a t e$

$\frac{d}{d t} <\hat{O}> = \frac{[<\hat{O}, \hat{H}>]}{i h} + <\frac{\partial \hat{O}}{\partial t}> \frac{d}{d t} ⟨ \hat{O} ⟩ = (\frac{1}{i h}) ⟨ \hat{O}, \hat{H} ⟩ + ⟨ \frac{\partial \hat{O}}{\partial t} ⟩$

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