# Homework Help: Show that the dot product is linear: Bra-ket notation

1. Feb 4, 2014

### lausco

1. The problem statement, all variables and given/known data
Show that the dot product in two-dimensional space is linear:
<u|(|v> + |w>) = <u|v> + <u|w>

3. The attempt at a solution
I feel like I'm missing some grasp of the concept here ...
I would think to just distribute the <u| and be done in that one step,
but I'm being asked to prove this.
Is there a reason the <u| can't simply be distributed?

2. Feb 4, 2014

### collinsmark

I think you're being asked to demonstrate this longhand, in order to prove it.

In other words, recall (for two dimensional space),

$$| v \rangle = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}, \ \ | w \rangle = \begin{pmatrix} w_1 \\ w_2 \end{pmatrix}, \ \ \langle u | = (u_1^*, u_2^*)$$
and then work things out longhand, using more conventional methods, to eventually show that it does distribute.

3. Feb 4, 2014

### lausco

I know that <u|v> = the length of |u> times the projection of |v> along |u> . . . Are the conjugates related the the projection?

4. Feb 4, 2014

### collinsmark

I think so, yes. I'm not the best person to be explaining math, so don't rely on me for a graceful explanation of this. But yes, I think the conjugates ultimately comes down to some sort of generalization of projections.

$| A \rangle$ and $\langle A |$ represent conjugate transposes of one another. Of course if you deal only with real numbers, you don't need to worry about the conjugate. But in general, when dealing the complex numbers, the conjugate is necessary.

Perhaps it's easiest to demonstrate the motivation of this with a special case of taking the inner product of a vector with itself. In other words, let's examine $\langle A | A \rangle$.

Let's further simplify this to 1 dimension for now. Suppose we have a simple, one-dimensional vector $| A \rangle = \left( 3 + i4 \right)$. Suppose our goal is to find the magnitude squared of this complex, one dimensional vector. In complete agreement with the length of A times the projection of A onto itself, we can find the length squared of A by finding $\langle A | A \rangle$. In this example, the answer is (3 - i4)(3 + i4) = 25. Notice I found |A|2 by multiplying the complex conjugate of A by A. In other words, |A|2 = A*A for this one dimensional case. Notice that since we were trying to find the magnitude squared of a vector, the answer will always be real, even if A is complex. We couldn't do that without the complex conjugate.

We can move on to larger dimensional spaces by saying that in general, $\langle A |$ is the conjugate transpose of $| A \rangle$.