Help with QM state vectors

In summary, A harmonic oscillator system is initially in state n=2 for time t<0. At time t>0, a perturbation occurs and the system is given by V(t) = sqrt(3/4)*h_bar*omega* (|2><1| + |1><2|). The task is to compute the probability amplitudes, however, the individual is unsure of how to work with the outer products of two state vectors and how mixing works. They ask for hints or links for help with this. They also inquire about the use of LaTeX in the forums and receive instructions on how to use it. Additionally, they ask for clarification on manipulating algebra with state vectors. They are provided with a useful calculation
  • #1
Jumbuck
2
0
For my homework, I have a problem in which a (harmonic oscillator) system is prepared in state n=2 for t<0.

For time t>0, there is a perturbation given by

V(t) = sqrt(3/4)*h_bar*omega* (|2><1| + |1><2|)

After this I need to compute the probability amplitudes. However, my background is in engineering, so I'm unsure how to work with these outer products of two state vectors, or even how this mixing works. If anyone has any hints or links on how to work with these, I would appreciate it very much.

Also, for future reference, do these forums automatically generate LaTeX, or do you import the LaTeX equations I've seen in other posts?
 
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  • #2
In these forums, you can get the $ ... $ environment by using the [ itex ] ... [ /itex] tags. And you can get the \[ ... \] environment with the [ tex ] ... [ /tex ] tags. (Remove the spaces to use those tags) (note the direction of the slash)
 
  • #3
As to doing the algebra, just manipulate it formally. If you were faced with the product of |1><2| with |2>, that's given by |1><2|2> = |1>. Just remember that the distributive rules work (i.e. (A+B)C = AC + BC), but the commutative rule only works for scalars (i.e. for most S and T: [itex]ST \neq TS[/itex], but rS = Sr)
 
  • #4
Hurkyl said:
As to doing the algebra, just manipulate it formally. If you were faced with the product of |1><2| with |2>, that's given by |1><2|2> = |1>. Just remember that the distributive rules work (i.e. (A+B)C = AC + BC), but the commutative rule only works for scalars (i.e. for most S and T: [itex]ST \neq TS[/itex], but rS = Sr)

Thanks!

Since these are state vectors, would

(|2><1| + |1><2|) * |2>) = |2><1|2> + |1><2|2> = |1> ?

I believe these state vectors are orthogonal, so the <1|2> term is 0, but my textbooks isn't very clear.
 
  • #5
Here's a useful little calculation: suppose that v and w are eigenstates of a hermetian operator T, with different associated eigenvalues. Then, compute:

[tex]\langle v | T | w \rangle[/tex]

and

[tex]\langle w | T | v \rangle[/tex]

Since these two expressions are complex conjugates of each other, it tells you something about [itex]\langle v | w \rangle = \langle w | v \rangle^*[/itex].
 

1. What are quantum state vectors?

Quantum state vectors, also known as wave functions, are mathematical representations of a quantum system that describe its physical properties and behavior.

2. How are quantum state vectors used in quantum mechanics?

In quantum mechanics, state vectors are used to calculate the probabilities of different outcomes of a measurement on a quantum system. They also help to determine the time evolution of a quantum system.

3. What is the difference between a pure state and a mixed state in quantum mechanics?

A pure state is represented by a single quantum state vector, while a mixed state is a combination of multiple quantum state vectors with different probabilities. Pure states are deterministic, while mixed states have inherent uncertainty.

4. Can quantum state vectors be observed or measured?

No, quantum state vectors cannot be directly observed or measured. They are mathematical representations of a quantum system and do not have a physical existence.

5. How do quantum state vectors relate to the uncertainty principle?

Quantum state vectors are subject to the uncertainty principle, which states that it is impossible to know both the position and momentum of a particle with absolute certainty. This is because the state vector of a particle can only be described by a probability distribution in space and time.

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