An example of state determination (Ballentine Problem 8.4)

  • #1
EE18
112
13
Homework Statement
See attached image.
Relevant Equations
See below.
I've given the question as an image as some of the formatting was difficult for me in the small window given:
Screen Shot 2023-08-08 at 9.00.09 PM.png

My work is below. I got (a), but cannot get (b):

(a) It was a theorem proved in the text that any measurement on one subsystem will always be fully determined by the reduced state operator of the corresponding subsystem. That is, any measurement of an observable on the composite system represented by an operator of the form (take subsystem 1 WLOG) ##R^{(1) }\otimes I^{(2)}## can be predicted only in terms of the reduced state operator ##\rho^{(1)}## (see the bottom of 217). Thus we begin answering this problem by computing the two reduced state operators. First we need ##\rho##, which can be found (since this is a pure state initially given in ket representation) as (note we use the shorthand that, for example, ##\ket{+-} \equiv \ket{+} \otimes \ket{-}##)
$$\rho = \ket{\psi_c}\bra{\psi_c} = \frac{1}{2}\left(\ket{+-}\bra{+-} +c\ket{-+}\bra{+-} +c^*\ket{+-}\bra{-+} +\ket{-+}\bra{-+}\right),$$
where we have used ##|c|^2 = 1##.
Then we immediately find (tracing over the relevant subsystem and using the definition of the inner product on a tensor product space)
$$\rho^{(1)} = \frac{1}{2}\left(\ket{+}\bra{+} + \ket{-}\bra{-}\right) = \rho^{(2)}.$$
The reduced state operators are independent of ##c##, and it's thus immediately clear that no measurements on only one subsystem can distinguish between composite states with different values of ##c## (since any such measurement would use the reduced state operator which is independent of ##c##).

(b) Obviously we'll need to make use of "cross terms". Consider measuring 1 along the ##x## and 2 along the ##y##. We can see from the structure of the Pauli matrices that in this ##\sigma_z## basis we have
$$\sigma_x = \ket{-}\bra{+} + \ket{+}\bra{-}$$
$$\sigma_y = i\ket{-}\bra{+} -i\ket{+}\bra{-}$$
Then using the linearity of the tensor product we get to
$$\sigma_x^{(1)} \otimes\sigma_y^{(2)} = (\ket{-}\bra{+} + \ket{+}\bra{-}) \otimes (i\ket{-}\bra{+} -i\ket{+}\bra{-}) = i\ket{--}\bra{++} +i\ket{+-}\bra{-+} -i\ket{-+}\bra{+-} -i\ket{++}\bra{--},$$
where we have used our shorthand notation. Next, we compute
$$\rho \sigma_x^{(1)} \otimes\sigma_y^{(2)} = \frac{1}{2}\left(i\ket{+-}\bra{-+} + ci\ket{-+}\bra{-+} -c^*i\ket{+-}\bra{+-} -i\ket{-+}\bra{+-}\right)$$
Finally, we have
$$\langle\sigma_x^{(1)} \otimes\sigma_y^{(2)}\rangle = Tr({\rho \sigma_x^{(1)} \otimes\sigma_y^{(2)}}) = \frac{1}{2}(i + ci -c^*i - i) = \frac{1}{2}(ci + (ci)^*) = \Re(ci) =-\Im(c).$$

But I can't figure out another measurement to get me the real part of ##c##. I tried ##\sigma_y^{(1)} \otimes\sigma_x^{(2)}## as well as ##\sigma_z^{(1)} \otimes\sigma_x^{(2)}## but the former gave me the imaginary part again and the latter gave me zero IIRC. Can anyone supply a hint for what to check, or a better way to proceed here?
 
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  • #2
I guess you should try measuring arbitrary directions, i.e., the observable ##\vec{n_1} \cdot \hat{\vec{\sigma}}^{(1)} \otimes \vec{n}_2 \cdot \hat{\vec{\sigma}}^{(2)}## with arbitrary unit vectors ##\vec{n}_1## and ##\vec{n}_2##.
 
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