# properties

• Jul 8th 2009, 12:43 PM
Sampras
properties
Show the following:

• $|0 \rangle$ 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?
• Jul 8th 2009, 09:53 PM
siclar
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 $|W\rangle=|W\rangle+|0\rangle=|W\rangle+|V\rangle+ |-V\rangle=|0\rangle+|-V\rangle=|-V\rangle$