- #1

KFC

- 488

- 4

a|0> =|0>

or

a|0> = 0 (number)

where a is the destroy operator

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- Thread starter KFC
- Start date

- #1

KFC

- 488

- 4

a|0> =|0>

or

a|0> = 0 (number)

where a is the destroy operator

- #2

Meir Achuz

Science Advisor

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

- #3

ice109

- 1,714

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

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> ?

- #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

KFC

- 488

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

zero ket? Do you mean [tex] a|0\rangle = |0\rangle[/tex] ?

- #6

dx

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- #7

KFC

- 488

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Oh, got it :)

- #8

Pengwuino

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

- #9

Ben Niehoff

Science Advisor

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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

KFC

- 488

- 4

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

- #11

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