# 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.