What is the meaning of a ket in quantum mechanics?

[kweierstrass]

What is meant when writing a "one-ket" like this |1> ?

 

[Academic]

That usually means the principle quantum number is 1.

 

[tom.stoer]

It depends on the system; for the one-dim. harmonic oscillator it means the first excited state |1> above the ground state |0>.

 

[kweierstrass]

Maybe I should have meantioned the whole context, I now realize that it means the eigenket coresponding to the eigenvalue 1 in my specific problem. Thanks anyway

 

[skynelson]

My understanding is that it is often an abstract way of referring to the n=1 basis vector. This is similar to what someone wrote above.

The basis vectors must be implicitly defined, and then |n> refers to the nth basis vector. Unfortunately I am not clear on exactly where and how the |n> basis vectors are typically defined. I often see the notation |1> without it being clear what the ACTUAL eigenfunctions being referred to are. In that type of situation, I guess it is simply an abstraction; and index referring to a real function.

 

[DevilsAvocado]

I think this explains it:

http://en.wikipedia.org/wiki/Bra-ket_notation#Most_common_use:_Quantum_mechanics"

In quantum mechanics, the state of a physical system is identified with a ray in a complex separable Hilbert space, $$\mathcal{H}$$, or, equivalently, by a point in the projective Hilbert space of the system. Each vector in the ray is called a "ket" and written as $$|\psi\rangle$$, which would be read as "ket psi". (The $$\psi\!$$ can be replaced by any symbols, letters, numbers, or even words—whatever serves as a convenient label for the ket.) The ket can be viewed as a column vector and (given a basis for the Hilbert space) written out in components,

$$|\psi\rangle = [ c_0 \; c_1 \; c_2 \; \dots ] ^T,$$

when the considered Hilbert space is finite-dimensional. In infinite-dimensional spaces there are infinitely many components and the ket may be written in complex function notation, by prepending it with a bra (see below). For example,

$$\langle x|\psi\rangle = \psi(x)\ = c e^{- ikx}.$$

(:uhh: ... is it because English is not my native language ... but "infinite-dimensional spaces" and "prepending bra" makes me think of something "completely different" ... )