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

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Homework Statement


Show that the dot product in two-dimensional space is linear:
<u|(|v> + |w>) = <u|v> + <u|w>

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?
 

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  • #2
collinsmark
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Homework Statement


Show that the dot product in two-dimensional space is linear:
<u|(|v> + |w>) = <u|v> + <u|w>

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

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

[tex] | 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^*) [/tex]
and then work things out longhand, using more conventional methods, to eventually show that it does distribute.
 
  • #3
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I know that <u|v> = the length of |u> times the projection of |v> along |u> . . . Are the conjugates related the the projection?
 
  • #4
collinsmark
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I know that <u|v> = the length of |u> times the projection of |v> along |u> . . . Are the conjugates related the the projection?
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. :redface: But yes, I think the conjugates ultimately comes down to some sort of generalization of projections.

[itex] | A \rangle [/itex] and [itex] \langle A | [/itex] represent conjugate transposes of one another. Of course if you deal only with real numbers, you don't need to worry about the conjugate. :smile: 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 [itex] \langle A | A \rangle [/itex].

Let's further simplify this to 1 dimension for now. Suppose we have a simple, one-dimensional vector [itex] | A \rangle = \left( 3 + i4 \right) [/itex]. 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 [itex] \langle A | A \rangle [/itex]. 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, [itex] \langle A | [/itex] is the conjugate transpose of [itex] | A \rangle [/itex].
 

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