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Homework Help: Confused about wave function

  1. Jul 20, 2011 #1

    There is something I can't understand...

    [itex]\psi(0)= \frac {1}{\sqrt 5}|1>+ \frac {2i}{\sqrt 5}|2>

    I need to find [itex]<H> for \rightarrow \psi(t)[/itex]
    [itex]E_n= \hbar \omega (n + \frac {1}{2})[/itex] , [itex]\hat {H} =(a^\dagger a+1/2)[/itex]
    These make me understand that this problem is one dimensional. (+1/2 in the Hemiltonian)
    Finding the time dependance is easy. But now there is a thing I don't understand:

    I was told I could relate to the wave function as a vector:

    [itex] \psi= \frac {1}{\sqrt 5} \begin{pmatrix}
    e^{-it/\hbar}\\ 2ie^{-2it/\hbar}

    \end{pmatrix} [/itex]

    And that means :

    \end{pmatrix} [/itex]


    \end{pmatrix} [/itex]

    But now, when I use a on |1> , for instance- I receive |0>.
    What does that mean?
    I also receive- |3>. according to what I think- I should have defined:
    \\ 0

    , |1>=\begin{pmatrix}
    \\ 0

    Because I have four base vectors, and should have 4 dimensions. But the problem refers to 1 dim...
    What am I not getting here?

    How are dimension related?

    Thank you!
    Last edited: Jul 20, 2011
  2. jcsd
  3. Jul 21, 2011 #2
    You can either write a state as:[itex]\psi = c_0 |0\rangle + c_1 |1\rangle + c_2 |2\rangle + c_3 |3\rangle + \cdots[/itex] or simply denote it by its four coefficient and basically making an implicit assumption about the basis states: [itex]\psi = \begin{pmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ \vdots\end{pmatrix}[/itex]

    This is the same as when you write a spacial vector, you can either write it as [itex]\boldsymbol{b} = b_x \boldsymbol{\hat{x}} + b_y \boldsymbol{\hat{y}} + b_z \boldsymbol{\hat{z}}[/itex] or just by its component [itex]\boldsymbol{b} = (b_x, b_y, b_z)[/itex]

    Presumably, the reason you want to write [itex]\psi[/itex] as a cumbersome column vector is that you are also writing the operators out in matrix form. Try to write out the Hamiltonian [itex]H[/itex]. You will find that it is a diagonal matrix. So when you compute
    [tex]\psi(t) = \exp\left(i\frac{t H}{\hbar}\right) \psi(0)[/tex]
    The exponential of a diagonal matrix is rather trivial to compute and remains diagonal.
  4. Jul 21, 2011 #3
    Thank you!
    But how do I figure out the dimensions?

    What does the length of the vector mean in that context?

    My wave function does not include this vector: |7>. But if I use a^dagger- several times I will get it...

    Since I have this form of a wave function, doesn't it mean there are two coordinate vectors only? (|1> |2> )
    Thank you!

    Another thing I am not getting right:
    when I want to calculate this:

    [itex] <\hat H >=<\psi(t)|\hat H|\psi(t)> [/itex]
    I need to do this using these a, a^dagger

    But the Hemiltonian is
    [itex] \hat H=\hbar \omega (aa^\dagger+1/2) [/itex]

    What should I do about the 1/2? add it as a scalar? use it as a 1/2I?
    Thank you!!
    Last edited: Jul 21, 2011
  5. Jul 21, 2011 #4
    In truth, the eigenbasis for H is infinite dimensional, because it contains [itex]|n\rangle[/itex] for all nonnegative integer n.

    But the "initial state" of your problem only has two non-zero coefficients: one for [itex]|1\rangle[/itex] and one for [itex]|2\rangle[/itex]. So you don't need to include all those other dimensions if the state evolving in time do not mix with other states.

    But this is guaranteed because [itex]|n\rangle[/itex] are eigenstates of the Hamiltonian. They are the called "stationary" state because as the Hamiltonian evolves them in time they do not mix with other states: [itex]|0\rangle[/itex] stays [itex]|0\rangle[/itex], [itex]|1\rangle[/itex] stays [itex]|1\rangle[/itex] etc. The coefficient in front of them can change and find how the coefficient change as function of time is the whole problem!

    [itex]\psi[/itex] is not a stationary state because it is combination of [itex]|1\rangle[/itex] and [itex]|2\rangle[/itex], but each [itex]|1\rangle[/itex] and [itex]|2\rangle[/itex] evolves in a stationary fashion.

    So your problem has always only been two dimensional

    What is the "length" of a vector? You mean like [itex]\boldsymbol{b}\cdot\boldsymbol{b}[/itex]? number of element? The latter is the dimensional of the vector space. The former, in the context of a normalized state, is always 1. Anyway, that was just an analogy that says writing out the coefficient as a column vector is the same as writing in form of a linear combination.

    But the point is that your Hamiltonian contains both [itex]a^\dagger[/itex] and [itex]a[/itex] in equal amount. So you actually never get |7>. Refer to the above comment about stationary states don't mix.


    You have your [itex]a[/itex] and [itex]a^\dagger[/itex] reversed (order is important!!)

    Just act on the state [itex]|0\rangle, |1\rangle, ...[/itex] with [itex]H[/itex] and see what you get. The 1/2 is just a number that just add to your result of [itex]a^\dagger a[/itex].
    Last edited: Jul 21, 2011
  6. Jul 21, 2011 #5
    Thank you!

    The Hemiltonian is given by: [itex] \hat H =\hbar \omega (a^\dagger a +1/2) [/itex]

    Isn't it true, that the 1/2 refers to dimensions of the problem? I have seen such thing for harmonic oscillator...

    If I apply the Hemiltonian on the wave function, because of the orthogonality of the eigenstates, I get:

    [itex] \frac {\hbar \omega} {\sqrt 5} [\sqrt 2 |2> |0> e^{-it/\hbar} +\frac {1} {2} |1> e^{-it/\hbar} +2i\sqrt 6 |3> |1> e^{-2it/\hbar} +i|2> e^{-2it/\hbar} [/itex]

    So, regardless to what you said about the wave function collapsing into |1> and |2>, I should always receive only the 1/2 products... all the others should be zero... am I right?
    But when I think of it now- if what I wrote is right, there is no meaning to the order of the a operators...

    Thank you some much for your explanations...
  7. Jul 21, 2011 #6
    You are confusing the dimension of real space with the dimension of the Hilbert Space.

    Each spacial dimension essentially acts an independent harmonic oscillators. Your problem is a 1D SHO.

    You are confusing acting on a bra on the left side versus acting on a ket on the right side.
    [itex]a[/itex] acting on the right (left) annihilates (creates). [itex]a^\dagger[/itex] acting on the right (left) creates (annihilates).

    [tex]a |n\rangle = \sqrt{n} |n-1\rangle
    \to \langle n| a^\dagger = \langle n-1 | \sqrt{n}[/tex]
    [tex]a^\dagger |n\rangle = \sqrt{n+1} |n+1\rangle
    \to \langle n| a = \langle n+1 | \sqrt{n+1}[/tex]

    Also when you do something like [itex]\langle \psi | H | \psi\rangle[/itex], you should just get a number. Not some convoluted ket times ket like you have up there, which really has no meaning.
    Last edited: Jul 21, 2011
  8. Jul 21, 2011 #7
    Ok... but if I want to calculate [itex] <1|a^\dagger a+1/2|1> [/itex]

    I do it this way:
    [itex] <1|[\sqrt 1 \sqrt 2 |0> |2> +1/2 |1>] [/itex]

    and, when I apply the left part, I then receive:

    [itex] \sqrt 2 <1| |0> |2> + 1/2 <1||1> [/itex]

    The first term vanishes, and only the second remains, giving 1/2I (I is the unit matrix)

    Thank you for your time and patience. You could probably tell- English is not my mother-tongue... so my questions\answers might seem a bit vague...
  9. Jul 21, 2011 #8
    [tex]\langle 1 | a^\dagger a + \frac12 | 1 \rangle
    = \langle 1 | a^\dagger a | 1 \rangle + \frac12 \langle 1 | 1 \rangle[/tex]
    The second term is just 1/2. The first term is:
    [tex]\langle 1 | a^\dagger a | 1 \rangle
    = \langle 1 | a^\dagger | 0 \rangle
    = \langle 1 | 1 \rangle = 1[/tex]

    Notice how there is always 1 bra and 1 ket throughout. Never 2 ket like what you did.
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