Novice Guide to Understanding Bra-ket Notation

[JDude13]

I am new to qm and very new to bra-ket notation.
If you, as a physicist, saw this:
|\phi&gt;=\Sigma(\sqrt{\Lambda_n}|x=x_n&gt;)<br />
what would you understand about the system it is describing?

 

[maverick_starstrider]

You shouldn't feel bad that notation is a little bit obtuse. Basically \vert \phi \rangle is a wavefunction of some particle. You are then choosing to represent it in the x eigenbasis. Essentially then all the information in \vert \phi \rangle is then represented as the probability \Lambda_n of the system being in the state given by \vert x_n \rangle. Since

\vert \langle x_0 \vert \phi \rangle \vert^2 = \Lambda_0

would be the probability of it being in state 0.

 

[JesseM]

To me that notation is a little weird, I don't think it's quite right to represent the probability amplitude as \sqrt{\Lambda_n}, since the amplitude is a complex number while the probability is a positive real...of course people often say that the probability is equal to the amplitude squared (JDude13 might take a look at [post=3250764]this post[/post] of mine for a quick explanation of this), but what they really mean is that the probability is equal to the amplitude multiplied by its own complex conjugate. And when you use vector decomposition to represent a quantum state vector as a weighted sum of eigenstates of some observable like position or momentum, the "weights" attached to each amplitude have to be complex amplitudes.

Since the amplitude associated with a given eigenvector \vert x_0 \rangle is just \langle x_0 \vert \phi \rangle, I think a better notation would be \vert \phi \rangle =\Sigma(\langle x_n \vert \phi \rangle \vert x_n \rangle)

 

[JDude13]

Isnt that a mathematically null statement? Like saying 2=2.

 

[JesseM]

Isnt that a mathematically null statement? Like saying 2=2.

I don't follow, aren't all mathematical statements inevitable given whatever axioms you're using? I don't think it's trivially obvious that any state vector can be expressed as a weighted sum of the eigenvectors of one or more observables.

 

[CondMattGuy]

Strictly speaking, | \phi \rangle isn't a "wavefunction." It's a state or vector in an abstract vector space. The quantity \langle \vec{r} | \phi \rangle is the position-space wavefunction \phi ( \vec{r} ) you're thinking of.