# Scaling of a quantum state

1. Oct 24, 2012

### Lindsayyyy

Hi everyone,

I'm new to quantum mechanics, so bear with me

1. The problem statement, all variables and given/known data

I'm not sure if scaling is the right word here, but my problem is about the absolut value of a quantum mechanics state to be one. I have the state $$| \phi>$$ which is a linear combination of the states $$|+>$$ and $$|->$$. The first task is about scaling my phi.

2. Relevant equations

.

3. The attempt at a solution

I know $$| \phi> = \lambda_1 |+> + \lambda_2 |->$$ whereas the lambdas are complex numbers.

Afterwards I used the definition of the scalar product to get the norm.
$$<\phi|\phi> = (\lambda_{1}' <+| + \lambda_{2}'<-|)(\lambda_1 |+> + \lambda_2 |->)$$

lambda' is the complex conjugated
If I solve this I get an expression like : sqrt(a+b) wheres my a equals the realpart of my lambdas1 and b equals the real part of my lambdas2

Is that solution right or totally wrong?

2. Oct 24, 2012

### TSny

I'm not sure what you mean when you say "If I solve this I get an expression like : sqrt(a+b) " What are you saying sqrt(a+b) should equal?

Anyway, I think the imaginary parts of each λ will be just as important as the real parts.

Note $(\lambda_{1}' <+| + \lambda_{2}'<-|)(\lambda_1 |+> + \lambda_2 |->)$ will expand to

$\lambda_{1}'\lambda_{1} <+| +> \;+ \;\lambda_{1}'\lambda_{2}<+|-> \;+ \; \lambda_{2}'\lambda_{1}<-|+> \;+ \;\lambda_{2}'\lambda_{2}<-|->$

Try to simplify this.

3. Oct 24, 2012

### Lindsayyyy

first of all: thanks for the help

Is my attempt at a solution even right? I'm uncertain if that's the right approach.

Yes, I had the expression you mentioned already, where I think I can say the following:

<+|+> and <-|-> equals 1 and <+|-> equals zero

And while I'm writing this down. I realised I forgot to say in my beginning post that + and - are orthonormal quantum states :/. Sorry about that.

edit: I got to the point where I have the followng left:
$$<\phi|\phi> = \lambda_{1}'\lambda_{1} +\lambda_{2}'\lambda_{2}$$

And a complex number times its conjugated version gives me a real part only

4. Oct 24, 2012

### TSny

That looks good. Yes, a complex number times it's conjugate gives a real number in the sense that if z = a + ib then z'z = a^2 + b^2, which includes both the real and imaginary parts of z. (I thought you might have been saying that only the real part of z would contribute to z'z.)

When you talk about "scaling" $|\phi>$, I am guessing you are talking about what is usually called "normalizing" $|\phi>$. So, you want to find a number, $A$, such that if you multiply $|\phi>$ by $A$, the resultant state vector has a norm of 1. That is, if you let $|\psi> = A|\phi>$ then $<\psi|\psi> = 1$.

5. Oct 25, 2012

### Lindsayyyy

yes, normalizing is the word I'm looking for. I'm not a native speaker and my translation program didn't give me "to normalize" :) .

So, my attempt is right?

the normalized vector would then be:

$$\frac {1}{\lambda_1' \lambda_1 \lambda_2' \lambda_2} \mid \phi \rangle$$

Is that correct?

Furthermore I want to calculate the possibility of |+>. I've done it with

$$p_n = \frac {|a_n|^2} {\sum |a_n|^2}$$

where my a_n's are the coefficients.
So finally I'd get

$$p_+=\frac {|\lambda_1|^2}{|\lambda_1|^2+|\lambda_2|^2}$$

is that correct?

Thanks for the help

6. Oct 25, 2012

### TSny

That's not quite correct. You want to find $A$ such that $A|\phi>$ is normalized. Which means

$<\phi|A'A|\phi> = A'A<\phi|\phi> = 1$. So, $A'A = \frac{1}{<\phi|\phi>}$ You are free to choose $A$ to be a real number, so you can choose $A^2= \frac{1}{<\phi|\phi>}$
That's correct.

7. Oct 25, 2012

### Lindsayyyy

Hm I don't understand that.

Let's take a look at "normal" vectors. When I want to normalize a vector its:

$$\vec v' = \frac {1}{\sqrt{a²+b²+c²}} \cdot \vec v$$

doesn't that also work now for quantum states?

like in my example

$$\mid \phi \rangle' = \frac {1}{\sqrt{\lambda_1' \lambda_1 + \lambda_2' \lambda_2}} \cdot \mid \phi \rangle$$

8. Oct 25, 2012

### TSny

Yes, that's correct. Good.