# Quantum Mechanics basiscs (confused about wavefunctions)

1. Jan 6, 2010

### Hipp0

Hello, we started Quantum Mechanics last semester, and somehow I manged to do most of the homework during that semester, but now I'm trying to revise it again, and I can't seem to understand the very basics of it, in particular about wavefunctions. Please read this carefully, because you might not understand where the confusion is (I tried Physics chat room, but no-one seemed to understand what i was saying, just saying that I was wrong, and all of it is complete nonsense)

At first I thought that there is like a general wavefunction, that is something abstract that changes with time(like e.g. we know what rotation does, but the rotation matrix looks differently in different bases) so i thought a wave function is like a function of infinitely many variables, e.g. x-positing, p-momentum, spin, energy... and we need to "project" it on one of those bases <x|psi(t)> = psi (x, t) (so this step is like representing a rotation with a matrix in a basis). And then we are working in one of those bases (position), I can do other stuff now, like find probabilities amplitudes that it's in some state q using : <q | psi (x,t)> and do other find expected values etc... But after reading Griffiths introduction to quantum mechanics for a bit, I realized that my logic is probably wrong. So I started thinking again.

As I understand now, a general wavefunction is like a vector space of infinitely many square integrable functions (and their linear combinations and stuff are obviously in that vector space as well), so it's just like a database of infinitely many functions, which are all functions of x and t (in 1 dimension). And so now, we just take this wavefunction and take scalar product with state vectors that we are interested in straight away, <q | psi(x,t)> and we'll get the probability amplitude.
Is that correct? I'm a bit confused because states are vectors, can you give some examples of states? can a state be a function (e.g. q is some function, of what variables?) then obviously <q (...) | psi (x,t)> = integral q*psi(x,t)dx or something.

p.s. At our lectures, we were actually never told what a wavefunction is, our lecturer thought it made more sense to start with Probability Amplitudes and assume that everyone knows what a wavefunction means.

2. Jan 6, 2010

### Hipp0

my 1st understanding was that a wavefunction is actually a function of infinitely many spectrums or something, like a function of position, momentum, energy etc... and then we have to "project" it on one of those bases to get it in e.g. position basis psi(x,t) or momentum basis to get psi(p, t). But from Griffith I understood that a wavefunction is always a function of (x,t) .

Because I heard people many times, saying that the wavefunction IN POSITION basis is blah blah blah.. so in my understanding there must be a general wavefunction (i.e. before expressing it in position basis) and therefore it can be expressed in other bases like momentum bases . So I'm confused again now.

If a wave function in position basis is psi(x,t) what is it before "projecting" onto this basis? And what other bases are available to "project" it on? this is the root of my confusion

Last edited: Jan 6, 2010
3. Jan 6, 2010

### Mentz114

You're overcomplicating this. Forget about projecting into bases for a moment and just accept that the wave function is a function of x, t and has a complex value. This is the Hilbert space vector. The wave function assigns a value to every element of this vector.
Any vector space has any number of bases, which may be orthogonal or not, so this position vector can be written as a linear combination of an infinite number of basis vectors. But this is actually not very useful.

More useful is that the Fourier transform of the wave equation is the momentum space wave function $\phi(k, t)$. So we can talk about the same wave function either as representing positions or momenta.

Maybe this is what you remember as 'projecting into momentum space'.

The other element of QM is the operators. You'll no doubt remember that they can be represented as matrices that act on the state vectors by multiplication. The energy of the system can be represented as an operator that can be pre and post multipled by a state vector to give a scalar value that is the expectation ( average ) energy that will be measured. The measured value will always be an eigenvalue of the energy operator (H).

If we take an eigenvector of H and outmultiply it by itself, we get a matrix of the same form as an operator. If you multiply a statevector by this operator, it gets projected against this eigenvector.

4. Jan 6, 2010

### peteratcam

I'm going to try to be very precise and I hope I cover the confusion you have.

In both classical and quantum physics, the aim of the game is to describe how a system evolves in time according to the equations of motion. The idea is that we describe a system at a particular instant by its 'state': in classical physics, this is the position and momentum of all the constituents; in quantum physics, it is a 'complete set of probability amplitudes', or the 'ket' for the system, or a thing often called psi - in quantum physics, the state of a system is an abstract object, compared to in classical physics.

The equations of motion describe the evolution of the state: in classical physics, these are Newton's/Hamilton's equations. In quantum physics, it is the schroedinger equation. Given the 'state' at time t0, the Schroedinger equation uniquely determines the state at a later time t1. But what is the mathematical object which we use to describe the 'state' of a quantum mechanical system?
In the mathematical formalism, the state of a quantum mechanical system is an *element* of a complex vector space, which physicists loosely call 'the Hilbert space for the problem'. The dimension of the vector space depends on the degrees of freedom in the system being studied - a 'two state' system has just a 2-dimensional hilbert space, whereas a single particle system has an infinite-dimensional Hilbert space. There is *no* time dependence of the Hilbert space, or its elements; the state of the system does evolve in time though. You can think of the state psi(t) as a time dependent choice of vector from the hilbert space.

Specifically for a single particle problem, the hilbert space has a natural interpretation as the space of square integrable complex functions of x, so in this case the state of the system is some function of x, which changes with time. I prefer to think of 'wavefunction' to refer specifically to the components of the system's 'state' vector in the position basis, but some authors call the abstract thing the wavefunction also.

Strictly, your notation is a bit off:
You can either write $$|\psi(t)\rangle$$, the abstract state of the system, or write $$\psi(x,t) = \langle x | \psi(t) \rangle$$, the wavefunction of the system (which is the projection of the state on the position basis.
Also, you should be careful about what you mean by 'get the probability amplitude?'. *If* <q| has the interpretation as being an eigenstate of an hermitian operator Q with eigenvalue of q, *then* <q | psi(t)> has the interpretation of being the probability amplitude that a measurement of Q at time t will give the value q. But if the state <q| does not have that interpretation, I don't think I'd call it a probability amplitude.

Anything called a 'state' is an element of the Hilbert space discussed above. A careful notation might reserve 'state' for the state of the system, but 'possible state' for all other elements of the Hilbert space. When considering Hermitian operators on Hilbert space, the operators have eigenstates, or eigenvectors.

It is almost impossible to give an example of a vector without choosing a basis. So I'll choose the position basis for a single particle problem. A possible state is then any element of the space of square integrable functions, eg a gaussian, or a gaussian multiplied by a sine wave, etc.

Are you at Oxford?

You should have these equations in mind:
$$\psi(x,t) = \langle x | \psi(t)\rangle$$
$$|\psi(t)\rangle = \int dx \psi(x,t)|x\rangle$$
where <x| is a position eigenstate: a possible state where the particle is definitely located at x. As you can see from the second equation, the wavefunction can be thought of as the components of the abstract state psi(t) in the position basis. (Compare with a normal finite dimensional expansion). You can project onto any basis of the Hilbert space, but people don't normally call what you get a 'wavefunction'. A few basis's are very common:
the energy basis (project onto eigenfunctions of Hamiltonian)
the momentum basis (project onto momentum eigenstates)
the position basis (project onto position eigenstates - the most common)

5. Jan 7, 2010

### Hipp0

Hi, thanks for your replies,especially to peteratcam .I think I get what a wavefunction and a state is for a continuous case of a single particle, but a bit confused about the discreet case of two-state system. Here are my thoughts:
http://img697.imageshack.us/img697/8175/72630741.jpg [Broken]
I think an example of a two-state system can be spin, e.g. the two basis elements that are in that two dimensional Hilbert space are UP or DOWN, their linear combinations are in that Hilbert space as well. So a general wavefunction in this case can be expressed as a linear combination of |UP> or |DOWN> so |psi (t) > = alpha |UP> + beta |DOWN> but in single particle case we couldn't really express |psi (t)> in mathematical form, we had to choose |x> basis to do it.

Is what I'm saying so far correct?
(sorry about writing a lot, I just have to understand how one thing follows from another, otherwise I don't get the whole picture)
p.s. peteratcam, how did you know I'm at Oxford? :D

Last edited by a moderator: May 4, 2017
6. Jan 7, 2010

### Hipp0

Well actually, now that I tried this discrete case, can I actually express a general state |psi(t)> as $$\left|\Psi\left(t\right)\right\rangle=\sum\alpha\left|\phi\left(t\right)\right\rangle$$ for a single particle system. That's an infinite linear combination of infinitely many orthonormal square integrable functions phi(t) (e.g. bases of that infinite-dimensional Hilbert space, because it would have infinitely many of them, right?)
Thanks

7. Jan 7, 2010

### peteratcam

Yes, you can always expand a vector in that way, thats the point of a basis.

This is basically right, but I'm going to pick you up on details again. *A* choice of basis is |UP> and |DOWN>, but it is not *the* basis.

5 examples of expanding a vector as a linear combination of basis vectors:
1. $$\mathbf v = v_1 \mathbf i+v_2 \mathbf j + v_3 \mathbf k$$
2. $$|a\rangle = \sum_{i=1}^n a_i |e_i\rangle$$
3. $$|\phi\rangle = \sum_{i=\uparrow,\downarrow} a_i |i\rangle$$
4. $$|\psi\rangle = \int dx \psi(x) |x\rangle$$
5. $$|\psi\rangle = \sum_{i=0}^{\infty} c_i |E_i\rangle$$
The 3rd example is appropriate to a two state system.
The 4th and 5th example is appropriate to an infinite state system (eg, particle on the real line).
Example 4, I've chosen the position basis, which is continuous. In example 5, I'm expanding in a discrete (but infinite) basis, eg the energy basis.

The emphasis on probability amplitudes is very James Binney and I'm tutoring 2nd yr QM this year.

8. Jan 7, 2010

### Hipp0

That's excellent! it's all starting to make sense now. I'll definitely write these down in my summary notes.
By the way, did you have a chance to read my thoughts on a particular example (handwritten picture?) Are they ok-ish for that particular example? here is the link again:
http://img697.imageshack.us/img697/8175/72630741.jpg [Broken] If handwriting is unreadable then it's fine, no need to reply.

Oh yeah, that's our lecturer :) are you at Oxford as well?

Last edited by a moderator: May 4, 2017
9. Jan 7, 2010

### peteratcam

I read the handwritten stuff, all seems basically fine, and it will refine itself as you do more QM.
Regarding the last question in the handwritten stuff - 'position basis' is only appropriate to the quantum mechanics of particles (a position degree of freedom). But quantum mechanics has a wider validity to other degrees of freedom, such as spin degrees of freedom. So there isn't really a natural position basis for a two-state system describing spin. In fact in a mathematical sense any hermitian matrix/operator defines a quantum mechanical problem if you call the hermitian matrix/operator the Hamiltonian - the system you get won't in general correspond to any bit of real physics though.

Also, I trust you realise that wavefunctions are in general complex, and basically impossible to sketch unless they happen to be real. A popular choice for sketching is instead to plot the probability density, which is $$|\psi(x)|^2$$ - your sketches look like you are sketching a probability density.

Yes, I'm a 3rd year DPhil student in theoretical physics, also a college tutor.

10. Jan 7, 2010

### Hipp0

We are half-way through the course, so I bet it will refine itself before the exams :)

So, bases that we can choose to express $$\left|\Psi\left(t\right)\right\rangle$$ are really physical quantities that can change for that system, so when we have a particle in space, e.g. accelerated by a linear particle accelerator, we can represent it's position along a line, but alternatively, we can represent it's velocity (momentum) along that line and choose momentum basis to express the same thing. Because it's moving in 1D, other position dimensions do not change, therefore they can't be bases for this problem. As I understand, it's the same analogy with 2-state spin system, the system may have some position, but it doesn't change, only spin "state" describes that system, and position is irrelevant here as y,z-dimensions before. Is this correct?

Oh yeah, I realize that, I'm writing summary notes atm, "digging" a bit deeper in what we've done during the 1st term, because it helps me to remember when I rewrite stuff, and I like to use a lot of simplifications and diagrams to help me visualize and understand things, like you mentioned above, or saying that unitary matrix is like rotation in complex space and hermitian matrix is like a symmetric matrix in complex space, etc. But I'm aware of why it's not exactly true.

Nice, you are probably not my tutor though :) I'm actually surprised that you have time to help other students on this forum and not tired of teaching people yet :D

And thanks a lot for your help! it cleared my confusion completely!

11. Jan 7, 2010

### peteratcam

I think so, although I didn't quite follow.

You sound like the perfect student! Seeing as you like pictures, is the following in your notes yet:
The Schroedinger equation is
$$i\hbar\frac{d}{dt}|\psi(t)\rangle = \hat H|\psi(t)\rangle$$
A formal solution is:
$$|\psi(t)\rangle = U(t,t_0)|\psi(t_0)\rangle$$ where the time-evolution operator is $$\hat U(t,t_0) = \exp(-i\hat H (t-t_0)/\hbar)$$ and where t0 is a time when the state of the system is known.

The time evolution operator is unitary, because H is hermitian. Unitary transformations preserve length (like rotations). So the picture of quantum dynamics is that the state-vector traces a path in a vector space a bit like a dot on the surface of (complex) hypersphere. The fact that time-evolution is unitary is intimately associated with the conservation of probability.

I'm far too easily distractable, so basically I don't have time, I should be doing my research! Perhaps I should send your college an invoice for my time!

12. Jan 7, 2010

### Hipp0

Oh right, I actually just drew a picture in my notes randomly: a dotted curve and a few arrows touching that curve, saying that: Schrodinger equation kinda "picks" a state-vector "path" from Hilbert space(it tells which arrow is next after another, where arrows represent functions |psi(t1)>,| psi (t2)>...). But obviously I haven't look that far yet :) so I guess, I need to correct my picture to look like a rotation really of a vector.
They wouldn't like it :D

13. Jan 8, 2010

### Hipp0

Sorry, I'm a bit confused about solving Schrodinger equation now. Here is my attempt:

http://img691.imageshack.us/img691/9438/72723108.jpg [Broken]

I used separation of variables and the part that leads to Schrodinger Time Independent equation is ok, but the other part, which involves functions of time is not. Because when we integrate that part (on the left hand side of the page, at the bottom), I have to add like a constant, but because it's a partial differential equation, instead of constant I added a function of the other variable (x). Is this correct? or am I confusing this method with something else?
But on other websites like http://scienceworld.wolfram.com/physics/SchroedingerEquation.html
they say the solution to time-dependent part is only the exponent. How did they arrive at that result?

And also
I'm a bit confused about the notation. All psi here are only functions of time, does it mean you are not really using solutions of Schrodinger's equation, but more the fact that physical properties of state vectors have to stay unchanged, and so Unitary operator does not change their properties. Because I thought , solutions to Schrodinger Equations should have a function of x (I guess if you express the Hamiltonian in x-basis, although maybe there is a way to do it generally).
Thanks

P.S. Now after a bit of thinking, I guess I was wrong to add that f(x) part, because it was an ordinary differential equation, not a partial. But still there should be a constant, how do they know if the constant is 1? because it's psi(x,t) that should be normalized, not necessary c*exp(... t ) or psi(x)

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14. Jan 8, 2010

### Feldoh

In classical mechanics we describe a particular system by classifying thinks such as energies, positions, times, etc..

The way that I understand it in quantum mechanics a wave function is a description of a particular system. The wave function contains everything about our system. So if you like it must be independent of a basis since regardless of the basis you should get the same measurements (same descriptions of our system this is described by the wave function).

15. Jan 8, 2010

### peteratcam

The whole point of separation of variables is that the T bit *only* depends on t. By assumption, you only consider such solutions. Fortunately, the general solution in this case can be written as a sum of separable solutions. Another way of looking at it is to think that the 'integration' constant should be fixed by the boundary conditions, and the boundary conditions at t=0 are that the T part is 1.

The most general way to write the TDSE is in the operator notation I wrote. In the position basis, it becomes the more familiar TDSE which you wrote, where there is a del^2 etc. For a two state system for example, the Schrodinger equation (with a particular choice of hamiltonian) looks like:
$$i\hbar \frac{d}{dt}\begin{pmatrix}u_1(t)\\ u_2(t)\end{pmatrix} = \begin{pmatrix}4 & 3i\\-3i &-4\end{pmatrix}\begin{pmatrix}u_1(t)\\u_2(t)\end{pmatrix}$$
No x dependence obviously.
The TDSE is only a partial differential equation in the position basis, at heart, it is an ODE for a long vector.
Notice that the formal solution I gave has a matrix exponential in the definition of the time-evolution operator and calculating that is just as hard as solving the TDSE by diagonalisation.