- #1
woodyhouse
- 1
- 0
Firstly hello, this is the first time I have posted here (although I have used the site to find info in the past). My query is best illustrated, I think, with an example. Suppose we have some physical system with corresponding state vector
[tex]
\left| \psi \right> = a \left| 0 \right> + b \left| 1 \right> + c \left| 2 \right> + d \left| 3 \right> \in \mathbb C^4
[/tex]
and some physical quantity represented by the operator
[tex]
\hat E = E_0 \left| 0 \right> \! \left< 0 \right| + E_1 \left| 1 \right> \! \left< 1 \right|
+ E_2 \left| 2 \right> \! \left< 2 \right| + E_3 \left| 3 \right> \! \left< 3 \right|.
[/tex]
First suppose that we have that [itex] E_0 = E_1 = E [/itex]. Then the 2-dimensional subspace spanned by [itex] \left| 0 \right>[/itex] and [itex] \left| 1 \right> [/itex] is an eigenspace of [itex]\hat E [/itex] and any measurement with outcome [itex] E [/itex] will leave us with the projection (up to normalization) of [itex] \left| \psi \right> [/itex] onto this subspace. Expressed as a density matrix, the final state is
[tex]
\rho = \mathcal N \; \big(
\left | a \right|^2 \left| 0 \right> \! \left< 0 \right| +
\left | b \right|^2 \left| 1 \right> \! \left< 1 \right| +
ab^* \left| 0 \right> \! \left< 1 \right| +
a^*b \left| 1 \right> \! \left< 0 \right|.
\big)
[/tex]
Now consider the following: we have some experiment that is not accurate enough to distinguish [itex] E_0 [/itex] and [itex] E_1 = E_0 + \epsilon[/itex] but that can distinguish all others (for instance [itex] E_0 [/itex] and [itex] E_1 [/itex] may correspond to very close spectral lines compared to [itex] E_3 [/itex] and [itex] E_4 [/itex]). We perform a measurement, the outcome of which is [itex] E_0 \pm 10\epsilon[/itex]. Then we could argue that the state must have collapsed to either [itex] \left| 0 \right> [/itex] or [itex] \left| 1 \right> [/itex] with probabilities [itex] \left| a \right|^2 [/itex] and [itex] \left| b \right|^2 [/itex]respectively. According to the lack-of-knowledge interpretation of density matrices, the corresponding state (as a density matrix) after measurement is
[tex]
\rho' = \mathcal N '\; \big(
\left | a \right|^2 \left| 0 \right> \! \left< 0 \right| +
\left | b \right|^2 \left| 1 \right> \! \left< 1 \right|
\big)
[/tex]
where [itex] \mathcal N [/itex] is a normalizing factor.
The point of this is that (provided my reasoning holds) [itex] \rho [/itex] and [itex] \rho' [/itex] are physically distinct states. But when do we distinguish between the two scenarios? For instance if we have 2 degenerate levels that we know can be split with a magnetic field, do we always have to assume the presence of a magnetic field too weak to measure, or do we assume there is no magnetic field at all? Do we have to distinguish between `true' degeneracy and degeneracy relating to experimental inaccuracy?I had a look in various literature and over previous posts in this forum and haven't been able to find an answer to this; I apologize if my search was not sufficiently thorough or if I am missing something obvious.
[tex]
\left| \psi \right> = a \left| 0 \right> + b \left| 1 \right> + c \left| 2 \right> + d \left| 3 \right> \in \mathbb C^4
[/tex]
and some physical quantity represented by the operator
[tex]
\hat E = E_0 \left| 0 \right> \! \left< 0 \right| + E_1 \left| 1 \right> \! \left< 1 \right|
+ E_2 \left| 2 \right> \! \left< 2 \right| + E_3 \left| 3 \right> \! \left< 3 \right|.
[/tex]
First suppose that we have that [itex] E_0 = E_1 = E [/itex]. Then the 2-dimensional subspace spanned by [itex] \left| 0 \right>[/itex] and [itex] \left| 1 \right> [/itex] is an eigenspace of [itex]\hat E [/itex] and any measurement with outcome [itex] E [/itex] will leave us with the projection (up to normalization) of [itex] \left| \psi \right> [/itex] onto this subspace. Expressed as a density matrix, the final state is
[tex]
\rho = \mathcal N \; \big(
\left | a \right|^2 \left| 0 \right> \! \left< 0 \right| +
\left | b \right|^2 \left| 1 \right> \! \left< 1 \right| +
ab^* \left| 0 \right> \! \left< 1 \right| +
a^*b \left| 1 \right> \! \left< 0 \right|.
\big)
[/tex]
Now consider the following: we have some experiment that is not accurate enough to distinguish [itex] E_0 [/itex] and [itex] E_1 = E_0 + \epsilon[/itex] but that can distinguish all others (for instance [itex] E_0 [/itex] and [itex] E_1 [/itex] may correspond to very close spectral lines compared to [itex] E_3 [/itex] and [itex] E_4 [/itex]). We perform a measurement, the outcome of which is [itex] E_0 \pm 10\epsilon[/itex]. Then we could argue that the state must have collapsed to either [itex] \left| 0 \right> [/itex] or [itex] \left| 1 \right> [/itex] with probabilities [itex] \left| a \right|^2 [/itex] and [itex] \left| b \right|^2 [/itex]respectively. According to the lack-of-knowledge interpretation of density matrices, the corresponding state (as a density matrix) after measurement is
[tex]
\rho' = \mathcal N '\; \big(
\left | a \right|^2 \left| 0 \right> \! \left< 0 \right| +
\left | b \right|^2 \left| 1 \right> \! \left< 1 \right|
\big)
[/tex]
where [itex] \mathcal N [/itex] is a normalizing factor.
The point of this is that (provided my reasoning holds) [itex] \rho [/itex] and [itex] \rho' [/itex] are physically distinct states. But when do we distinguish between the two scenarios? For instance if we have 2 degenerate levels that we know can be split with a magnetic field, do we always have to assume the presence of a magnetic field too weak to measure, or do we assume there is no magnetic field at all? Do we have to distinguish between `true' degeneracy and degeneracy relating to experimental inaccuracy?I had a look in various literature and over previous posts in this forum and haven't been able to find an answer to this; I apologize if my search was not sufficiently thorough or if I am missing something obvious.
Last edited: