Eigenvalues of operator in dirac not* (measurement outcomes)

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 2K views
12x4
Messages
28
Reaction score
0

Homework Statement


A measurement is described by the operator:

|0⟩⟨1| + |1⟩⟨0|

where, |0⟩ and |1⟩ represent orthonormal states.

What are the possible measurement outcomes?

Homework Equations


[/B]
Eigenvalue Equation: A|Ψ> = a|Ψ>

The Attempt at a Solution



Apologies for the basic question but just very unsure of myself when it comes to this stuff. I have had a go and come up with a solution but I'm not sure if its right so any help would be much appreciated.

We're told that:
A = |0⟩⟨1| + |1⟩⟨0|

Can I then assume something like: Ψ = α|1> + β|0>?

using this I've then solved the eigenvalue equation, AΨ=aΨ, and found:

α|0> + β|1> = aα|1> + aβ|0>

giving:

α=aβ & β = aα

thus,

β=a2β

a = (+-) 1

hence, my eigenvalues are -1 and 1.

and these are the possible outcomes?
 
Physics news on Phys.org
You don't assume that [itex]|\psi>=\alpha |1>+ \beta |2>[/itex], there is a reason for that. The completeness relation gives,

[itex]I=|1><1|+|2><2|[/itex] [look up completeness relation if you don't know about it.]

which means, [itex]|\psi>=I|\psi>=|1><1|\psi>+|2><2|\psi>[/itex]

or, [itex]|\psi>=\alpha |1>+ \beta |2>[/itex],

where [itex]\alpha=<1|\psi>[/itex]

and [itex]\beta=<2|\psi>[/itex] are c-number.

Except that there are no more problem with your work.
 
12x4 said:

Homework Statement


A measurement is described by the operator:

|0⟩⟨1| + |1⟩⟨0|

where, |0⟩ and |1⟩ represent orthonormal states.

What are the possible measurement outcomes?

Homework Equations


[/B]
Eigenvalue Equation: A|Ψ> = a|Ψ>

The Attempt at a Solution



Apologies for the basic question but just very unsure of myself when it comes to this stuff. I have had a go and come up with a solution but I'm not sure if its right so any help would be much appreciated.

We're told that:
A = |0⟩⟨1| + |1⟩⟨0|

Can I then assume something like: Ψ = α|1> + β|0>?

using this I've then solved the eigenvalue equation, AΨ=aΨ, and found:

α|0> + β|1> = aα|1> + aβ|0>

giving:

α=aβ & β = aα

thus,

β=a2β

a = (+-) 1

hence, my eigenvalues are -1 and 1.

and these are the possible outcomes?

Yes, that looks just fine to me. The possible measurements of an experiment are the eigenvalues of the operator.
 
Last edited: