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Math Help - properties

  1. #1
    Senior Member Sampras's Avatar
    Joined
    May 2009
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    301

    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?
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  2. #2
    Member
    Joined
    Jul 2008
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    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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