How Do You Normalize a Quantum State in Quantum Mechanics?

[quixi]

Homework Statement

Homework Equations

The Attempt at a Solution

OK I have

$$\left| \psi \right> = \alpha \left| 000 \right> + \beta \left| 001 \right> + ... + \theta \left| 111 \right>$$

which I need to normalise.

I know that

$$\left| \psi \right>^{*} = \left< \psi \right|$$

and so have derived expression

$$\left< \psi \right| = \alpha^{*} \left| 000 \right> + \beta^{*} \left| 001 \right> + ... + \theta^{*} \left| 111 \right>$$

which means each have 8 terms.. so now doing

$$\left< \psi | \psi \right> = \alpha \alpha^{*} \left< 000 | 000 \right> + \alpha \beta^{*} \left< 000 | 001 \right> + ...$$

this multiplication will result in 64 terms! I don't know if that's right?!

then I know $\left< 00|00 \right> = 1$ and $\left< 00|11 \right> = 0$ which I assume I can apply to $\left< 000|000 \right> = 1$ and $\left< 00|11 \right> = 0$ ?

basically then end up with

$$\left< \psi | \psi \right> = \alpha \alpha^{*} + \beta \beta^{*} + ..$$

I think that's along the right lines now? Its just the daunting 8x8=64 multiplication that I was unsure on!

 

[Chopin]

Yes, that is correct--there will be 64 terms in the multiplication. However, as you've noted, 56 of them will be 0, so they can be immediately discarded. The only ones that remain are the products of a state with its own conjugate, so whenever you see such a multiplication, you can immediately go straight to your last step, without even bothering to write down all of the intermediate work.

Things like this are the reason that equations in QM use a lot of ...'s.

 

[quixi]

OK good, thought it must be right but seemed a lot of maths, glad I'm not going mad!

The only ones that remain are the products of a state with its own conjugate

I'm a bit confused over which ones will remain (but I do see it will be only a few, most cancel out).

So for example if I had:

$$|\psi> = 3|000> + 7i|001>$$ +...

$$<\psi| = 3|000> - 7i|001>$$ +...

$$<\psi|\psi> = 9<000|000> -21i<000|001> + 21i<000|001> + 49<001|001> +...$$

then the only terms there that remain will be the first (since [itex]\alpha = 3|000> = \alpha^{*}[/tex]) i.e. where the complex conjugate is the same as the normal.<br /> <br /> .. is that correct?<br /> <br /> Perhaps that's not a great example with knowing <000|000> = 1 but hopefully it still makes sense the understanding. Just want to make sure I'll be ignoring the correct ones![/itex]

 

[Chopin]

It's not that the coefficients are the same as their conjugates, its whether one state times the other is nonzero or not.

I'm going to change your notation a bit to make it easier to demonstrate. Say that instead of states like $$|000\rangle$$, $$|101\rangle$$, etc. we have a set of n states $$|\Psi_0\rangle$$, $$|\Psi_1\rangle$$, etc. In general, a state will be some linear combination of these states: $$\Psi = a_0|\Psi_0\rangle + a_1|\Psi_1\rangle + ...$$. We can write this compactly as $$\Psi = \sum_{i=0}^n{a_i|\Psi_i\rangle}$$. Now, let's try to find the norm of this state.

$$\langle\Psi|\Psi\rangle = (\sum_i{a_i^*\langle\Psi_i|})(\sum_j{a_j|\Psi_j\rangle})$$
$$= \sum_i\sum_j{(a_i^*\langle\Psi_i|)(a_j|\Psi_j\rangle)}$$
$$= \sum_i\sum_j{a_i^*a_j\langle\Psi_i|\Psi_j\rangle}$$

This is just a general way of saying what you had before. You can see that if you have $$n$$ states, there will be $$n^2$$ terms in this expansion, because of the double summation. In general, all of those terms could be nonzero, in which case you have a whole lot of terms on your hands. For a system with 1,000 states, for instance, you're going to have to keep track of 1,000,000 terms. indeed!

However, in many cases, the set of states you're working with are of a special kind, called an orthonormal set. This means that for any two terms, $$\langle\Psi_i|$$ and $$|\Psi_j\rangle$$, their product $$\langle\Psi_i|\Psi_j\rangle$$ is 1 if i=j, and 0 if it isn't. We can write this compactly using a symbol called the Kronecker Delta, $$\langle\Psi_i|\Psi_j\rangle = \delta^i_j$$. This symbol is simply defined to be 1 if i=j, and 0 if not. If you have a set of orthonormal states, things get a lot easier. Let's substitute this into our equation and see what happens:

$$\langle\Psi|\Psi\rangle = \sum_i\sum_j{a_i^*a_j\langle\Psi_i|\Psi_j\rangle} = \sum_i\sum_j{a_i^*a_j\delta^i_j$$.

Now we made all the bras and kets go away, and it's just an equation with numbers. Furthermore, let's examine a few terms of this sum. If i=7 and j=25, then we have $$a_{7}^*a_{25}\delta^{7}_{25}$$. However, $$\delta^{7}_{25} = 0$$, so we know this term will vanish. If i=10 and j=10, then we have $$a_{10}^*a_{10}\delta_{10}^{10}$$. But $$\delta_{10}^{10}=1$$, so this is just $$a_{10}^*a_{10}$$. In general, of the $$n^2$$ terms in our expansion, only the $$n$$ terms with equal indices will actually matter. All the other ones (called the cross terms) vanish. So really, we could write our double sum as a single sum:

$$\sum_i\sum_j{a_i^*a_j\delta^i_j = \sum_i{a_i^*a_i}$$

Which is a lot easier to solve. Whenever you see a delta in an equation, it will eventually allow you to do this--it eats a summation, and substitutes the variable on that summation with the other variable on the delta. This is an enormous help because it cuts down the number of terms dramatically. This is a very common operation in quantum mechanics, so it's important to become familiar with it if you aren't already.

So if your states are orthonormal, their product will be a delta function, and you can do the above trick. If they're not, then all of the terms will contribute, and you've got a big mess. So it pays to have an orthonormal set of base states. Fortunately, most of the sets of states you come across are orthonormal (an electron spin-up state is 100% equal to an electron spin-up state, and 0% equal to an electron spin-down state, for instance) so you can often use this fact to make your life a lot easier.