Electron two slit experiment with gradual collapse ?

[birulami]

electron two slit experiment with gradual "collapse"?

Just read about the two slit experiment with electrons where you try to find out through which slit the electron went by "shining a light on it" (Feynman lectures on physics I-37-7). I try to summarize the statement and then have a question:

When using high intensity, high frequency light, every electron going through the apparatus can be detected and assigned to one of the two slits. In this case, no interference pattern can be seen.

Now the frequency of the light (not the intensity) is reduced continuously, or in sufficiently small steps for a sufficient number of experiments. Still every electron is detected, but due to the lower frequency, the resolution of the measurement of the electron's position becomes less and less accurate, making an assignment to one of the slits more and more difficult and uncertain. The more the position gets uncertain, the clearer the interference pattern appears.

Now my question: Isn't the disappearance of the interference pattern usually attributed to the "collapse of the wave function"? If yes, does the above describe a "fractional" collapse, or is this a completely different process from what is usually called the "collapse of the wavefunction"?

Harald.

 

[jeblack3]

The disappearance of the interference pattern is due to "decoherence." To try to put it simply, once the electron interacts with the light, you have to consider the joint quantum state of the electron and the light. Without the light, interference requires that possible paths through each hole wind up with the electron in the same place. With the light, interference requires the electron to land in the same place and the scattered light to be in the same state. If the electron going through a different hole always puts the scattered light into a different state, then you have no interference. But if there's some probability amplitude for an electron going through either hole to put the scattered light in the same state, then you only have a reduced interference pattern. That's the rough idea. It could be stated in terms of bras and kets if you would understand and think it would be helpful.

Collapse of the wavefunction has really nothing to do with it. You only need to use the collapse of the wavefunction concept to convert the wavefunction/state vector you get from quantum mechanics into the probability for an observable result.

 

[Sillyboy]

The disappearance of the interference pattern is due to "decoherence." To try to put it simply, once the electron interacts with the light, you have to consider the joint quantum state of the electron and the light. Without the light, interference requires that possible paths through each hole wind up with the electron in the same place. With the light, interference requires the electron to land in the same place and the scattered light to be in the same state. If the electron going through a different hole always puts the scattered light into a different state, then you have no interference. But if there's some probability amplitude for an electron going through either hole to put the scattered light in the same state, then you only have a reduced interference pattern. That's the rough idea. It could be stated in terms of bras and kets if you would understand and think it would be helpful.

Collapse of the wavefunction has really nothing to do with it. You only need to use the collapse of the wavefunction concept to convert the wavefunction/state vector you get from quantum mechanics into the probability for an observable result.

Hello, I am interested in it if you could explain it in terms of Dirac notion. I am not clear about your present explanation. I think your "decoherence" could make my comprehension about this basic principle more clear!

 

[naima]

We have two slits, a source of photons between them, and near each slit, A or B receive the photon.
If a photon is received in A it may have interacted with an electron passing through a or b:
I note this (a(x) T + b(x) F)
T is the amplitude to be received near the slit where the electron passed (and F on the opposite side.
If the photon is received in B: the amplitude is (b(x)T + a(x)F).
Note that the bolded expressions are orthogonal.
If the wavelength is very short T = 1 anf F = 0 the total probability p(x) is
a(x)a*(x)+ b(x)b*(x) (no interference)
The opposite case occurs when F = T = 1/rac(2)
The probability is 1/2 (a(x)+b(x))(a(x)+b(x))* + 1/2 (b(x) + a(x))(b(x) + a(x))*=
(a(x)+b(x))(a(x)+b(x))* (interference)
For intermediate value you may easily find p(x)
We have here to measures: (A ou B) and x so you MAY think there is a wave function of
these variables that have collapsed.

 

[jeblack3]

Okay, bras and kets. This is better because it strips out my hand-waving, and let's you interpret what's going on as you please.

As the electron is passing through either slit A or slit B, it is described by the state vector

$$a|A\rangle + b|B\rangle .$$

Without the light, the electron is detected at some point x on the screen. |A> and |B> evolve into

$$U |A\rangle = \int dx\: f(x) |x\rangle$$
$$U |B\rangle = \int dx\: g(x) |x\rangle ,$$

the final state is

$$\int dx\: (a f(x) + b g(x)) |x\rangle ,$$

and the electron is detected at x with probability

$$P(x) = |a f(x) + b g(x)|^2.$$

Assuming the two waves are of about the same amplitude, we can write

$$a f(x) = z e^{i \theta_A}$$

$$b g(x) = z e^{i \theta_B}$$

and we get

$$P(x) = 2z^2 + z^2 e^{i (\theta_A - \theta_B)} + z^2 e^{-i (\theta_A - \theta_B)} = 2 z^2 (1 + cos(\theta_A - \theta_B)).$$

Now suppose instead there is a photon that scatters off the electron as it is passing through the slit. The state of the photon after scattering is |C> if the electron is at slit A and |D> if the electron is at slit B. The states |C> and |D> need not be orthogonal.

$$U_1 |A\rangle = |A\rangle \times |C\rangle$$
$$U_1 |B\rangle = |B\rangle \times |D\rangle$$

Perhaps the photon is later detected at position y.

$$U_2 |C\rangle = \int dy\: h(y) |y\rangle$$
$$U_2 |D\rangle = \int dy\: k(y) |y\rangle$$

By unitarity, whatever the subsequent evolution of the photon, it must obey

$$\int dy\: |h(y)|^2 = \langle C|U_2^\dagger U_2|C\rangle = \langle C|C\rangle = 1$$

$$\int dy\: |k(y)|^2 = \langle D|U_2^\dagger U_2|D\rangle = \langle D|D\rangle = 1$$

$$\int dy\: h^*(y) k(y) = \langle C|U_2^\dagger U_2|D\rangle = \langle C|D\rangle .$$

The final state is then

$$\int dx\: dy\: (a f(x) h(y) + b g(x) k(y)) |x\rangle \times |y\rangle ,$$

and the probability of electron detection at x and photon detection at y is

$$P(x, y) = |a f(x) h(y) + b g(x) k(y)|^2.$$

But say we're only interested in the electron. The total probability of electron detection at x is

$$P(x) = \int dy\: P(x, y)$$

$$= \int dy\: |a f(x) h(y) + b g(x) k(y)|^2$$

$$= |a f(x)|^2 \left(\int dy\: |h(y)|^2\right) + a^* f^*(x) b g(x) \left(\int dy\: h^*(y) k(y)\right) + b^* g^*(x) a f(x) \left(\int dy\: h(y) k^*(y)\right) + |b g(x)|^2 \left(\int dy\: |k(y)|^2\right)$$

$$= |a f(x)|^2 + a^* f^*(x) b g(x) \langle C|D\rangle + b^* g^*(x) a f(x) \langle C|D\rangle ^* + |b g(x)|^2$$

$$= 2z^2 + z^2 \langle C|D\rangle e^{-i (\theta_A - \theta_B)} + z^2 \langle C|D\rangle ^* e^{i (\theta_A - \theta_B)}$$

$$= 2 z^2 (1 + Re(\langle C|D\rangle ) cos(\theta_A - \theta_B) + Im(\langle C|D\rangle ) sin(\theta_A - \theta_B)).$$

Assuming I didn't make any mistakes above.

A key point is that the total interference pattern for the electron doesn't depend on how or even if we measure the photon, or on any other details about what happens to the photon after it interacts with the electron.

 

[Sillyboy]

Okay, bras and kets. This is better because it strips out my hand-waving, and let's you interpret what's going on as you please.

As the electron is passing through either slit A or slit B, it is described by the state vector

$$a|A\rangle + b|B\rangle .$$

Without the light, the electron is detected at some point x on the screen. |A> and |B> evolve into

$$U |A\rangle = \int dx\: f(x) |x\rangle$$
$$U |B\rangle = \int dx\: g(x) |x\rangle ,$$

the final state is

$$\int dx\: (a f(x) + b g(x)) |x\rangle ,$$

and the electron is detected at x with probability

$$P(x) = |a f(x) + b g(x)|^2.$$

Assuming the two waves are of about the same amplitude, we can write

$$a f(x) = z e^{i \theta_A}$$

$$b g(x) = z e^{i \theta_B}$$

and we get

$$P(x) = 2z^2 + z^2 e^{i (\theta_A - \theta_B)} + z^2 e^{-i (\theta_A - \theta_B)} = 2 z^2 (1 + cos(\theta_A - \theta_B)).$$

Now suppose instead there is a photon that scatters off the electron as it is passing through the slit. The state of the photon after scattering is |C> if the electron is at slit A and |D> if the electron is at slit B. The states |C> and |D> need not be orthogonal.

$$U_1 |A\rangle = |A\rangle \times |C\rangle$$
$$U_1 |B\rangle = |B\rangle \times |D\rangle$$

Perhaps the photon is later detected at position y.

$$U_2 |C\rangle = \int dy\: h(y) |y\rangle$$
$$U_2 |D\rangle = \int dy\: k(y) |y\rangle$$

By unitarity, whatever the subsequent evolution of the photon, it must obey

$$\int dy\: |h(y)|^2 = \langle C|U_2^\dagger U_2|C\rangle = \langle C|C\rangle = 1$$

$$\int dy\: |k(y)|^2 = \langle D|U_2^\dagger U_2|D\rangle = \langle D|D\rangle = 1$$

$$\int dy\: h^*(y) k(y) = \langle C|U_2^\dagger U_2|D\rangle = \langle C|D\rangle .$$

The final state is then

$$\int dx\: dy\: (a f(x) h(y) + b g(x) k(y)) |x\rangle \times |y\rangle ,$$

and the probability of electron detection at x and photon detection at y is

$$P(x, y) = |a f(x) h(y) + b g(x) k(y)|^2.$$

But say we're only interested in the electron. The total probability of electron detection at x is

$$P(x) = \int dy\: P(x, y)$$

$$= \int dy\: |a f(x) h(y) + b g(x) k(y)|^2$$

$$= |a f(x)|^2 \left(\int dy\: |h(y)|^2\right) + a^* f^*(x) b g(x) \left(\int dy\: h^*(y) k(y)\right) + b^* g^*(x) a f(x) \left(\int dy\: h(y) k^*(y)\right) + |b g(x)|^2 \left(\int dy\: |k(y)|^2\right)$$

$$= |a f(x)|^2 + a^* f^*(x) b g(x) \langle C|D\rangle + b^* g^*(x) a f(x) \langle C|D\rangle ^* + |b g(x)|^2$$

$$= 2z^2 + z^2 \langle C|D\rangle e^{-i (\theta_A - \theta_B)} + z^2 \langle C|D\rangle ^* e^{i (\theta_A - \theta_B)}$$

$$= 2 z^2 (1 + Re(\langle C|D\rangle ) cos(\theta_A - \theta_B) + Im(\langle C|D\rangle ) sin(\theta_A - \theta_B)).$$

Assuming I didn't make any mistakes above.

A key point is that the total interference pattern for the electron doesn't depend on how or even if we measure the photon, or on any other details about what happens to the photon after it interacts with the electron.

Ya, I think your intepretation is amazing. You totally explain LZ's problem and you present me a clear view at the double-slit experiment. I think I have a better understanding of this. Thank you, friend!