- #1

- 488

- 4

a|0> =|0>

or

a|0> = 0 (number)

where a is the destroy operator

- Thread starter KFC
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- #1

- 488

- 4

a|0> =|0>

or

a|0> = 0 (number)

where a is the destroy operator

- #2

Meir Achuz

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It is the number zero. If a were the identity operator, you would get |0> back.

- #3

- 1,707

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you sure about that? i know that the books say zero but those operators aren't hermitian anyway hence unphysical so what does it matter what it does to |0> ?It is the number zero. If a were the identity operator, you would get |0> back.

- #4

dx

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Neither. [tex] a|0\rangle = [/tex] the zero ket.

EDIT: The zero ket and [tex] |0\rangle [/tex] are different things.

EDIT: The zero ket and [tex] |0\rangle [/tex] are different things.

Last edited:

- #5

- #6

dx

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- #7

- #8

Pengwuino

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Or better yet, a|0> = null vector.

- #9

Ben Niehoff

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Yes. The thing to remember is that all physical states must be normalizable. So

<0|0> = 1

<0|0> = 1

- #10

- 488

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[tex]\hat{N}|0\rangle = 0|0\rangle = 0 [/tex] (number) ?

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