Confused about wavefunctions and kets

[dyn]

Roughly, a ket can be represented as a column vector. So let's say there are only 2 possible positions. Also let us choose to represent the ket |x=1> as the column vector [1 0]^T, and the ket |x=2> as the column vector [0 1]^T, ie. we choose as basis vectors states of definite position. An arbitary ket is then |ψ>=ψ(1)|x=1>+ψ(2)|x=2>, or equivalently as the wavefunction [ψ(1) ψ(2)]^T, or equivalently the wavefunction ψ(x) where x is an index that runs from 1 to 2.

However, x actually is not discrete with only 2 values, it runs continuously. So if we use basis vectors with a definite position, then the ket |ψ> is an infinite dimensional column vector. An element of this column vector ψ(x) is the probability amplitude that a particle will be found at location x.

The above is rigourously incorrect, because there are sonme subtleties for infinite dimensional spaces, but the idea is roughly ok. Take a look at the explanations in http://physics.mq.edu.au/~jcresser/Phys304/Handouts/QuantumPhysicsNotes.pdf (chapters 8-10).

Thanks. You said an element of the column vector ψ(x). Did you mean an element of the ket |ψ> ? But you then relate it to location x. I thought kets are independent of basis ? So why would it be location x and not momentum p or some other basis ?

 

[atyy]

Thanks. You said an element of the column vector ψ(x). Did you mean an element of the ket |ψ> ?

Before you represent a ket as a column vector, you must always choose a basis. In the above the choice of basis means we choose |x=1> to be the column vector [1 0]^T, and |x=2> to be the column vector [0 1]^T.

Then the ket |ψ> will be the column vector which can be written [ψ(1) ψ(2)]^T, or for short ψ(x) which is an element of the column vector [ψ(1) ψ(2)]^T.

But you then relate it to location x. I thought kets are independent of basis ? So why would it be location x and not momentum p or some other basis ?

Yes, because I chose at the start a basis in which a state with a definite position |x=1> is the column vector [1 0]^T, and the state with definite position |x=2> is the column vector [0 1]^T. This is why the column vector [ψ(1) ψ(2)]^T is also written as [ψ(x=1) ψ(x=2)]^T, or for short ψ(x) is an element of that column vector.

If at the start I had chosen to represent the state of definite momentum as the basis, eg. choose |p=1> as the column vector [1 0]^T, then the elements of the column vector representing the ket |ψ> would be ψ(p).

 

[dyn]

Some things are clearer now but as for the rest ; my head is spinning more and more. I just want to thank everyone who has persevered with me on this thread.