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July 8th 2009, 11:43 AM #1
Sampras

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May 2009

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properties
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$|0 \rangle$ is unique
$0|V \rangle = |0 \rangle$
$|-V \rangle = -|V \rangle$
$|-V \rangle$ is the unique additive inverse of $|V \rangle$

So there exists a null vector $|0 \rangle$ obeying $|V \rangle + |0 \rangle = |V \rangle$ . Suppose there is another vector $|0' \rangle$ such that $|V \rangle + |0' \rangle = |V \rangle$ . Then $|0 \rangle + |0' \rangle = (|V \rangle + |V \rangle)-(|V \rangle + |V \rangle) = |0 \rangle$ . Hence $|0 \rangle$ is unique.

We know that $|0 \rangle = (0+1)|V \rangle + |-V \rangle$ . This is equaled to $0|V \rangle + |V \rangle + |-V \rangle = 0|V \rangle$ .

We know that $|V \rangle + (-|V \rangle) = 0|V \rangle = |0 \rangle$ . Then by associativity, $|V \rangle + (|-V \rangle) = 0|V \rangle = |0 \rangle$ . Hence $|-V \rangle = -|V \rangle$ .

Suppose there is some $|W \rangle$ such that $|V \rangle + |W \rangle = |0 \rangle$ . Then $|V \rangle + |W \rangle = |V \rangle + |-V \rangle$ . So $|W \rangle = |-V \rangle$ since $|0 \rangle$ is unique.

Is this correct?
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July 8th 2009, 08:53 PM #2

siclar

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Jul 2008

Posts

81

Your ideas are mostly correct, but can be simplified to be more elegant.

If $|0\rangle$ and $|0'\rangle$ are both identity elements, then $|0\rangle=|0\rangle+|0'\rangle=|0'\rangle$ , where the left equality comes from $0'$ being the identity, and the right equality comes from $0$ being the identity.

$0|V\rangle=(1-1)|V\rangle=|V\rangle-|V\rangle=|0\rangle$

In this next one I assume $|-V\rangle$ is the inverse element of $|V\rangle$ . Then $|-V\rangle=|-V\rangle+ 0|V\rangle=|-V\rangle+ (1-1)|V\rangle=|-V\rangle+ |V\rangle-|V\rangle=|0\rangle-|V\rangle=-|V\rangle$

Suppose $|V\rangle+|W\rangle=|0\rangle$ . Then $|W\rangle=|W\rangle+|0\rangle=|W\rangle+|V\rangle+ |-V\rangle=|0\rangle+|-V\rangle=|-V\rangle$

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