Explicit demonstration of a measurement interaction

[EE18]

Homework Statement

See below.

Relevant Equations

See below.

$$\newcommand{\bra}[1]{\left \langle #1 \right \rvert}
\newcommand{\braxket}[3]{\left \langle #1 \middle \rvert #2 \middle \rvert #3 \right \rangle}
\newcommand{\ket}[1]{\left \rvert #1 \right \rangle}
\newcommand{\expec}[1]{\langle #1 \rangle}$$

Ballentine asks us the question at the end of this post. I am unclear on how to proceed because of the exponential of a tensor product operator.

My work:

We note from the outset that ##c## is unitless, as is obvious on dimensional grounds.

Suppose we have some initial, unentangled state (we assume pure states). The initial state of the object is some superposition of position states, of course.
$$\ket{\Psi_0} = \ket{\psi_0} \otimes \ket{\alpha},$$
where ##\ket{\alpha} = \int dx \, \alpha(x) \ket{x}## denotes a preparatory state of the apparatus and ##\ket{\psi_0} = \int dx\, \psi_0(x) \ket{x}## is the initial state of the object. We note that the time-development operator is given by
$$U(t) = e^{-i\int dt\, H(t)/\hbar} = e^{-icQ^{(1)}P^{(2)}\int dt\, \delta(t)/\hbar} = e^{-icQ^{(1)}P^{(2)}/\hbar}.$$
We now consider the measurement interaction:
$$\ket{\Psi_0} = \ket{\psi_0} \otimes \ket{\alpha} \to \ket{\Psi_f} = e^{-icQ^{(1)}P^{(2)}/\hbar}\ket{\psi_0} \otimes \ket{\alpha} = \int dx \int dx' \, \psi_0(x) \alpha(x') \left[ e^{-icQ^{(1)}P^{(2)}/\hbar} \ket{x} \otimes \ket{x'} \right]$$
$$ = \int dx \int dx' \, \psi_0(x) \alpha(x') \sum_{n = 0}^\infty \frac{1}{n!} \left[(-icQ^{(1)}P^{(2)}/\hbar)^n \ket{x} \otimes \ket{x'} \right]$$
$$= \int dx \int dx' \, \psi_0(x) \alpha(x') \sum_{n = 0}^\infty \frac{1}{n!} \left(\frac{-ic}{\hbar} \right)^n \left((Q^{(1)})^n \ket{x}\right) \otimes \left((P^{(2)})^n \ket{x'}\right) $$
$$= \int dx \int dx' \, \psi_0(x) \alpha(x') \sum_{n = 0}^\infty \frac{1}{n!}\left(\frac{-icx}{\hbar} \right)^n \ket{x} \otimes \left((P^{(2)})^n \ket{x'}\right) $$
$$\stackrel{(1)}{=} \int dx \int dx' \, \psi_0(x) \alpha(x') \ket{x} \otimes \left(\sum_{n = 0}^\infty \frac{1}{n!}\left(\frac{-icx}{\hbar}P^{(2)} \right)^n \ket{x'}\right)$$
$$ = \int dx \int dx' \, \psi_0(x) \alpha(x') \ket{x} \otimes \left(e^{\frac{-icx}{\hbar}P^{(2)}} \ket{x'}\right)$$
$$ = \int dx\, \psi_0(x) \ket{x}\otimes \left(\int dx' \, e^{\frac{-icx}{\hbar}P^{(2)}} \ket{x'}\alpha(x')\right)$$
$$ = \int dx\, \psi_0(x) \ket{x}\otimes \left(\int dx' \, e^{\frac{-icx}{\hbar}P^{(2)}} \ket{x'}\bra{x'}\ket{\alpha}\right)$$
$$\int dx\, \psi_0(x) \ket{x}\otimes \left( e^{\frac{-icx}{\hbar}P^{(2)}}\ket{\alpha}\right)$$
$$ \equiv \int dx\, \psi_0(x) \ket{x}\otimes \left( T^{(2)}(cx)\ket{\alpha}\right),$$
where in (1) we use the linearity of the tensor product and where in the last equality we have identified the translation operator.

Now let's consider computing ##\expec{Q^{(1)}Q^{(2)}}## on the post-interaction state (this expectation value is related to the correlation coefficient and has been the proxy which Ballentine uses for correlation). We obtain
$$\expec{Q^{(1)}Q^{(2)}} = \left[ \int dx'\, \psi^*_0(x') \bra{x'}\otimes \left( \bra{\alpha}T^{(2)}(-cx')\right)\right]Q^{(1)}Q^{(2)} \left[ \int dx\, \psi_0(x) \ket{x}\otimes \left( T^{(2)}(cx)\ket{\alpha}\right)\right]$$
$$ \stackrel{(1)}{=} \int dx \, x|\psi^*_0(x)|^2 \bra{\alpha}T^{(2)}(-cx)Q^{(2)}T^{(2)}(cx)\ket{\alpha}$$
where in (1) we've used the inner product definition on a tensor product space and ##\braxket{x'}{Q^{(1)}}{x} = x\delta(x-x')##.

But this doesn't seem to be what Ballentine wants in the end. What does he mean by the "value of ##Q^{(2)}##? I also can't see where to go past where I've gotten to. If anyone can help out I'd greatly appreciate it.

 

[vanhees71]

I think, it's easier to keep the time finite. The interaction operator is simple since the operator at different times commute. Thus the time-evolution operator of states in the interaction picture simply is (with ##t_0<0## to avoid trouble with the ##\delta## distribution)
$$\hat{C}(t,t_0)=exp \left (-\mathrm{i} \int_{0}^t \mathrm{d} t' \hat{H}_I(t')/\hbar \right ) = \exp[-\mathrm{i} c \hat{Q}^{(1)} \hat{P}^{(2)}/\hbar \Theta(t)].$$
Low you can set ##t>0## and just evaluate
$$|q^{(1)},q^{(2)},t \rangle=\exp(-\mathrm{i} c \hat{Q}^{(1)} \hat{P}^{(2)}/\hbar) |q^{(1)},q^{(2)},0 \rangle,$$
and then use the result to calculate the "asymptotic free state"
$$|\Psi' \rangle =\int_{q^{(1)},q^{(2)}} \mathrm{d} q^{(1)} \mathrm{d} q^{(2)} \exp(-\mathrm{i} c \hat{Q}^{(1)} \hat{P}^{(2)}/\hbar) |q^{(1)},q^{(2)},0 \rangle \langle q^{(1)},q^{(2)},0|\Psi_0 \rangle.$$

 

[EE18]

I think, it's easier to keep the time finite. The interaction operator is simple since the operator at different times commute. Thus the time-evolution operator of states in the interaction picture simply is (with ##t_0<0## to avoid trouble with the ##\delta## distribution)
$$\hat{C}(t,t_0)=exp \left (-\mathrm{i} \int_{0}^t \mathrm{d} t' \hat{H}_I(t')/\hbar \right ) = \exp[-\mathrm{i} c \hat{Q}^{(1)} \hat{P}^{(2)}/\hbar \Theta(t)].$$
Low you can set ##t>0## and just evaluate
$$|q^{(1)},q^{(2)},t \rangle=\exp(-\mathrm{i} c \hat{Q}^{(1)} \hat{P}^{(2)}/\hbar) |q^{(1)},q^{(2)},0 \rangle,$$
and then use the result to calculate the "asymptotic free state"
$$|\Psi' \rangle =\int_{q^{(1)},q^{(2)}} \mathrm{d} q^{(1)} \mathrm{d} q^{(2)} \exp(-\mathrm{i} c \hat{Q}^{(1)} \hat{P}^{(2)}/\hbar) |q^{(1)},q^{(2)},0 \rangle \langle q^{(1)},q^{(2)},0|\Psi_0 \rangle.$$

Perhaps I don't follow, but isn't your last line more or less what I've given in my last line? Would you be able to comment on how ##Q^2## now provides a measurement of the initial state of the object's position ##Q^2## beforehand?

 

[vanhees71]

May be, I'm not sure. What should come out is an entangled state between the particle and the detector such that a measurement of ##Q^{(2)}## (reading of the pointer position) leads to a measurement of ##Q^{(1)}## (position of the particle). So you should calculate the state after the interaction and discuss.