1. Dec 1, 2008

### joker_900

OK so I am looking for a definitive answer. I am very confused and don't know why! Any help would be much appreciated, I will try and convey the source of my confusion below.

Firstly what exactly is psi(x) which crops up everywhere? Does it mean the probability amplitude of measuring the particle at x, or is it "the state of the system", or are these the same things?

I ask because I have seen the dirac ket psi a lot, e.g.
H|psi>=ih (d/dt)|psi>

But I also see: <x|psi> = psi(x) which is described as the probability amplitude of measuring the particle at x when the system has state |psi>.

However I also see
H psi(x) = ih (d/dt)psi(x)

Then I see
psi(x) = 2^(-0.5) (phi1(x) + phi2(x))
Here the phi are described as "energy states", and they are eigenfunctions of the TISE. So here I guess psi(x) is the state of the system: are the 2^(-0.5) the amplitudes of measuring E1 and E2, or is phi(x) part of the amplitude?

2. Dec 1, 2008

### Tac-Tics

The psi in schrodinger's equation is a function of time and position to amplitudes (thus, the partial derivative on the one side).

Even though it is a complex-valued function, in the math, it's treated as a vector. If you have two functions f and g, (f+g)(x) = f(x) + g(x) and for a scalar a, you have (af)(x) = a f(x), and the zero of the space is the constant 0 function f(x) = 0.

When you see <x|psi>, you are taking the dot product of two vectors. The dot product between two functions f and g is $$<f|g> = f \cdot g = \int f(x) g(x) dx$$. And the norm (the length) of f would be the square root of <f|f>

So, say you want to know the amplitude of a particle at some position x0 along a single axis. To do this, you take the dot product of the state the particle is in and the state representing position x0. So psi is your particle's state, and x is a dirac delta function-like thingy which is infinite at x0 and zero everywhere else. The dot product is then $$<x|\Psi> = \int x \Psi = \int \delta_{x0} \Psi = \Psi$$. Or something like that!

3. Dec 1, 2008

### dx

$$|\psi>$$ is essentially the state of the system. Consider the case of a particle. In classical physics, the state of the particle at any given time is just two real numbers representing position and velocity, but in quantum physics a more elaborate object is needed. Say you have a quantum particle in a box. At a time t, you made an accurate observation of its position. At that instant, the observation essentially puts the particle in what is called an eigenstate of the position operator. After that, if you make no further observations, the state will evolve according to Schrodinger's equation, and will determine the probabilities of various outcomes of experiments at any given time. For example, if the state at a time t' is $$|\psi>$$, then the probability of finding the particle at a position x is $$|<x|\psi>|^2$$ where $$|x>$$ is the position eigenstate corresponding to the position x. $$<x|\psi>$$ is usually denoted $$\psi(x)$$ and is called the wave function of the particle.

4. Dec 1, 2008

### joker_900

Awesome thanks to both of you, I understand a little bit more now. One more thing, I know you can write the state $$|\psi>$$ as a sum of well-defined energy states multiplied by their respective amplitudes. If you were to multiply both sides by <x|, on the left we would have $$\psi(x)$$ and on the right the sum of: the amplitude of measuring the system having energy Ei multiplied by the amplitude of measuring the particle at x when the system definitely has an energy Ei. Is this right?

And does the first poster mean that $$\psi(x)$$ = $$|\psi>$$?

Thanks!

5. Dec 1, 2008

### Fredrik

Staff Emeritus
That last equation in #2 is a bit strange. I think he means

$$\langle x|\psi\rangle=\int dx' \delta(x'-x)\psi(x')=\psi(x)$$

but I'm not crazy about that notation either. I wouldn't write the "position eigenfunction with eigenvalue x" as "x". This "x" is the function $x'\mapsto\delta(x'-x)$.

Note that he's not using bra-ket notation. He's just writing the scalar product of two functions f and g as <f|g>.

6. Dec 2, 2008

### dx

Yes, that's right.

No, that's not correct. The difference between $$|\psi>$$ and $$\psi(x)$$ is essentially the same as the difference between a vector $$v$$ and its components in some basis, $$(v_1, v_2, v_3)$$. $$\psi(x)$$ tells you the components of $$|\psi>$$ in the position basis.

7. Dec 2, 2008

### Tac-Tics

Sitting in bed last night, I realized I made an error in this post, and it's been bugging the hell out of me! =-P

In the equation

you actually need to take the complex congugate:

$$<f|g> = f \cdot g = \int f(x)^* g(x) dx$$

8. Dec 2, 2008

### joker_900

Excellent, thanks. I have one more question, and that's it I promise!

If you can write

$$|\psi>$$ = Sum{ai |Ei>}

where ai is the amplitude of a measurement yielding the value E1 (which is an eigenvalue corresponding to the eigenstate |Ei>). Then what is the probability of the measurement yielding either E1 or E2? Is it:

|a1|^2 + |a2|^2
or
|a1 + a2|^2

?

Thanks

9. Dec 2, 2008

### dx

It's $$|a_1|^2 + |a_2|^2$$. (this has nothing to do with quantum mechanics; It comes from probability theory.)

Last edited: Dec 2, 2008
10. Dec 2, 2008

### joker_900

But in the two slit experiment, if you have an amplitude A1(x) that the particle will go through slit 1 and hit x, and A2(x), then the probability of it hitting x is given by

|A1(x) + A2(x)|^2 which gives an interference term. What is the difference here?

Thanks

11. Dec 3, 2008

### dx

First let me explain why it is $|a_1|^2 + |a_2|^2$ for getting either $E_1$ or $E_2$. If the state of a system is $|\psi> = a_1|E_1> + a_2|E_2> + ...$, then when you measure the energy of the system, the probability of getting $E_i$ is $|a_i|^2$. Since the different energies $E_i$ are mutually exclusive outcomes of the experiment, it follows that the probability of getting either $E_1$ or $E_2$ is just the sum of their respective probabilities, i.e. $|a_1|^2 + |a_2|^2$. The same logic is being used when you conclude that the probability of getting a 5 or a 6 when you roll a die is (1/6) + (1/6), because they are mutually exclusive outcomes.

Now, the double-slit experiment. The amplitudes that you are talking about when you say "the amplitude of going to x through hole 1" is actually not the same kind of amplitude as the $a_i$ above. This is roughly the picture of the double-slit experiment that is given by standard quantum mechanics: The wave function of the electron is like a blob of high probability in a certain region, which tapers off as you move away from that central region. To observe the interference pattern, the uncertainty in position must be larger than the distance between the two slits, i.e. the approximate size of the blob is larger than the distance between the two slits. Now imagine this blob moving towards the slits. As it moves, it also gets larger, according to the Schrodinger equation. What happens when it reaches the slits? It splits into two, and behind the slits you see two smaller blobs coming out of the two slits. These blobs again diffuse and get larger as they move towards the screen, and after a little distance, overlap. This is where the interference comes from. Remember that even though the wave function has been split spatially, it still represents the single electron, and is a single state $|\psi>$. Let $P$ be a point on the screen. The amplitude of finding the electron there is just $<P|\psi>$, where the $|P>$ is an eigenstate of position corresponding to the position P. But we've just seen that we can think of the wave function as being made up of two parts coming from each slit, $|\psi_1>$ and $|\psi_2>$. So the amplitude $|<P|\psi>|$ can be written as $<P| (|\psi_1> +|\psi_2>) = <P|\psi_1> + <P|\psi_2>$. Therefore, the probability is $|<P|\psi_1> + <P|\psi_2>|^2$. In the technical sense, the amplitude is $<P|\psi>$, but you can think of it as being the sum of two "amplitudes" $<P|\psi_1>$ and $<P|\psi_2>$, and you can think of them as "the amplitude of the particle going to P through hole 1" and "the amplitude of the particle going to P through hole 2", but remember that there's a slight difference between the two uses.

Last edited: Dec 3, 2008
12. Dec 3, 2008

### Fredrik

Staff Emeritus
This old post of mine might be (somewhat) helpful too:

13. Dec 3, 2008

### shaun_o_kane

Don't really agree with this. The interaction with the slit is a measurement - the slit puts constraints on the location of the photon. It's possible to view the interference pattern as the result of Heisenberg's Uncertainty Principle - constraining the location of the particle as it goes through the slits increases the spread of it's momentum.

The photon changes "state" as a result of going through the screen (the waveform spreads out) and being absorbed by the screen.

The idea that superimpositions are destoyed as a result of decoherence is dependent on particular interpretations of QM, so I'm not so sure it's "basic" QM. What is "macroscopically distinguishable" is again very dependent on particular interpretations of QM.

IMHO, I'd advice to a newbee to understanding the the math models of the waveform - don't bother with decoherence etc until that is understood.