Confused about wave function

1. Jul 20, 2011

noamriemer

Hello!

There is something I can't understand...

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

I need to find $<H> for \rightarrow \psi(t)$
For
$E_n= \hbar \omega (n + \frac {1}{2})$ , $\hat {H} =(a^\dagger a+1/2)$
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:

$\psi= \frac {1}{\sqrt 5} \begin{pmatrix} e^{-it/\hbar}\\ 2ie^{-2it/\hbar} \end{pmatrix}$

And that means :
$|1>=\begin{pmatrix} 1\\0 \end{pmatrix}$

$|2>=\begin{pmatrix} 0\\1 \end{pmatrix}$

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:
$|0>=\begin{pmatrix} 1\\0 \\ 0 \\0 \end{pmatrix}$
, |1>=\begin{pmatrix}
0\\1
\\ 0
\\0

\end{pmatrix}[/itex]
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. Jul 21, 2011

mathfeel

You can either write a state as:$\psi = c_0 |0\rangle + c_1 |1\rangle + c_2 |2\rangle + c_3 |3\rangle + \cdots$ or simply denote it by its four coefficient and basically making an implicit assumption about the basis states: $\psi = \begin{pmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ \vdots\end{pmatrix}$

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

Presumably, the reason you want to write $\psi$ as a cumbersome column vector is that you are also writing the operators out in matrix form. Try to write out the Hamiltonian $H$. You will find that it is a diagonal matrix. So when you compute
$$\psi(t) = \exp\left(i\frac{t H}{\hbar}\right) \psi(0)$$
The exponential of a diagonal matrix is rather trivial to compute and remains diagonal.

3. Jul 21, 2011

noamriemer

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:

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

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

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
4. Jul 21, 2011

mathfeel

In truth, the eigenbasis for H is infinite dimensional, because it contains $|n\rangle$ for all nonnegative integer n.

But the "initial state" of your problem only has two non-zero coefficients: one for $|1\rangle$ and one for $|2\rangle$. 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 $|n\rangle$ 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: $|0\rangle$ stays $|0\rangle$, $|1\rangle$ stays $|1\rangle$ etc. The coefficient in front of them can change and find how the coefficient change as function of time is the whole problem!

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

So your problem has always only been two dimensional

What is the "length" of a vector? You mean like $\boldsymbol{b}\cdot\boldsymbol{b}$? 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 $a^\dagger$ and $a$ in equal amount. So you actually never get |7>. Refer to the above comment about stationary states don't mix.

Yes!

You have your $a$ and $a^\dagger$ reversed (order is important!!)

Just act on the state $|0\rangle, |1\rangle, ...$ with $H$ and see what you get. The 1/2 is just a number that just add to your result of $a^\dagger a$.

Last edited: Jul 21, 2011
5. Jul 21, 2011

noamriemer

Thank you!

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

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:

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

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...

6. Jul 21, 2011

mathfeel

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.
$a$ acting on the right (left) annihilates (creates). $a^\dagger$ acting on the right (left) creates (annihilates).

$$a |n\rangle = \sqrt{n} |n-1\rangle \to \langle n| a^\dagger = \langle n-1 | \sqrt{n}$$
$$a^\dagger |n\rangle = \sqrt{n+1} |n+1\rangle \to \langle n| a = \langle n+1 | \sqrt{n+1}$$

Also when you do something like $\langle \psi | H | \psi\rangle$, 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
7. Jul 21, 2011

noamriemer

Ok... but if I want to calculate $<1|a^\dagger a+1/2|1>$

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

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

$\sqrt 2 <1| |0> |2> + 1/2 <1||1>$

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...

8. Jul 21, 2011

mathfeel

$$\langle 1 | a^\dagger a + \frac12 | 1 \rangle = \langle 1 | a^\dagger a | 1 \rangle + \frac12 \langle 1 | 1 \rangle$$
The second term is just 1/2. The first term is:
$$\langle 1 | a^\dagger a | 1 \rangle = \langle 1 | a^\dagger | 0 \rangle = \langle 1 | 1 \rangle = 1$$

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