# Normalised states

1. Mar 1, 2010

### pnaj

I've been following the lectures given by Leonard Susskind on Quantum Entanglement. In Lecture 6, he describes a simplified version of the two-slit experiment so that we can use finite dimensional spaces to describe the system (which is all we've learnt so far).

He also uses what he calls the 'Time Evolution Postulate' which states: if |A> -> (evolves to) |A'> and |B> -> |B'>, then any superposition of the two states must evolve linearly as a|A> + b|B> -> a|A'> + b|B'>.

My question is: assuming that both |A> and |B> are both unit vectors, which until this point in the course was necessary for a state to describe a real-world system, shouldn't we now normalise this state for it to make sense?

It does seem to matter, even though Prof. Susskind suggests that it doesn't in this case.

In the lecture |A> represents the state in which the particle goes through slit A, |B> - particle goes through B. Then the particle lands up on one of N 'detectors'.

He represents this by letting |A> and |B> evolve to a superposition of the N detectors ...

|A> -> |A'> = a1|1> + ... + aN|N>
Then with only slit A open: prob(A, m) = <m|A'> = am*am (am* conjugate to am)

|B> -> b1|1> + ... + bN|N>
With only slit B open: prob(B, m) = <m|B'> = bm*bm

Classical result would be prob(A, B, m) = prob(A, m) + prob(B, m) = am*am + bm*bm

But, by the 'Time Evolution Postulate'

|A> + |B> -> (a1 + b1)|1> + ... + (aN + bN)|N>

So quantum result would be prob(A, B, m) = (am + bm)*(am + bm) = (am*am + bm*bm) + (am*bm + ambm*)

It seems to me that normalising |A> + |B> would give a different probability, so surely it does matter.

Any help would be mucch appreciated.

Paul

2. Mar 1, 2010

### Fredrik

Staff Emeritus
The time evolution operator $$U(t)=e^{-iHt}$$ is unitary, so it doesn't change the normalization of the state it acts on. We have $\|Ux\|=\|x\|[/tex] for all state vectors x. 3. Mar 1, 2010 ### pnaj Hi Fredrik, Thanks for answering - I think I understand. But shouldn't the superposition of |A> and |B> be normalised in the first place (before we apply the time evolution operator). P. 4. Mar 1, 2010 ### Fredrik Staff Emeritus It seems that I didn't read your question carefully enough. You weren't really asking what I thought. I think you're right. We should be working with the normalized state [itex]\frac{1}{\sqrt 2}\big(|A\rangle+|B\rangle\big)$ instead.

You have a couple of minor mistakes in your post. The middle part of prob(A, m) = <m|A'> = am*am should really be |<m|A'>|2, and you did the same thing with prob(B,m).

5. Mar 1, 2010

### pnaj

Hi Fredrik

Ah - maybe this is where the confusion lies.

To calculate the probabilities I took the average value of the projection operator for m, which is a definition given in the course ...
$P(A, m) = \big(\sum \langle k|a_k^* \big) |m\rangle \langle m| \big(\sum a_n |n \rangle \big) = a_m^* a_m$

I think this is the right idea but I can't see where my calculation is wrong.

Thanks for help with this.
P.

Last edited: Mar 1, 2010
6. Mar 1, 2010

### Fredrik

Staff Emeritus
I don't think your calculation is wrong. You just typed the middle part wrong when you wrote prob(A, m) = <m|A'> = am*am. The right-hand side is still correct (i.e. equal to the left-hand side).

7. Mar 1, 2010

### pnaj

Hi once again,

Yes - just noticed that - now I see why you said 'minor' mistake now ... it should just have read ...$P(A, m) = |\langle m|A'\rangle|^2 = a_m^* a_m$.

But I still am having trouble following the logic in Prof. Susskind's lecture.

So, we have:
$P(A, m) = a_m^* a_m$.
$P(B, m) = b_m^* b_m$.

Classical result:
$P(A, B, m) = a_m^* a_m + b_m^* b_m$.

Quantum result (using the normalised state this time):
$P(A, B, m) = \frac{1}{2}(a_m^* a_m + b_m^* b_m + a_m^* b_m + a_m b_m^*)$.

In the lecture there's an example where the 0th 'detector' is located symmetrically wrt the two slits implying that then, $a_0 = b_0$ due to the symmetry, which means that, (with the normalized composite state) $P(A, B, m) = \frac{1}{2}(a_0^* a_0 + a_0^* a_0 + a_0^* a_0 + a_0^* a_0) = 2(a_0^* a_0)$ ... which is the same as the classical result.

Prof Susskind does not normalize, so his result is $P(A, B, m) = 4(a_0^* a_0)$ ... twice the classical result.

I can't tell whether this is due to the simplification or not - or whether it's just an straightforward error in his logic.

P.

Last edited: Mar 1, 2010
8. Mar 1, 2010

### xepma

Well, the statement that $P(A, m) = |\langle m|A\rangle|^2 = a_m^* a_m$ is already wrong if you're not working with normalized states. You should divide by the norm of |A> as well.

9. Mar 1, 2010

### pnaj

Hi xepma,

We are assuming that $|A\rangle$ is normalized.

P.