Photon Probability After Attenuation

assume an optical mode initialized in state teh Fock state superposition 1+3. this mode propagates through a channel with power transmission 90%. what is the probabilty of measuing n=1 on a photon number resolving detector ?

To calculate the probability of measuring ( n = 1 ) photons on a photon number resolving detector for an optical mode initially in a Fock state superposition ( |1\rangle + |3\rangle ) after it propagates through a channel with 90% power transmission, we need to consider the effects of the channel on the state and then compute the probability of finding one photon.

1. Initial State: The initial state of the system can be represented as ( \frac{1}{\sqrt{2}}(|1\rangle + |3\rangle) ), assuming equal superposition and normalization.

2. Channel Effect: The channel with 90% power transmission attenuates the state. In quantum optics, such an attenuation can be modeled using a beamsplitter transformation. The transformation for a beamsplitter with transmissivity ( \tau ) (90% in your case, i.e., ( \tau = 0.9 )) is applied to each Fock state component separately.

3. Beamsplitter Transformation: This transformation will reduce the amplitude of each photon number state. The transformed state ( |\psi'\rangle ) will be a superposition of the transformed ( |1\rangle ) and ( |3\rangle ) states.

4. Probability Calculation: Finally, we calculate the probability of measuring ( n = 1 ) photon. This is done by finding the amplitude of the ( |1\rangle ) state in the transformed state ( |\psi'\rangle ) and then squaring it to get the probability.

Let's perform these calculations.

The probability of measuring ( n = 1 ) photon on a photon number resolving detector, after an optical mode initially in a Fock state superposition ( |1\rangle + |3\rangle ) propagates through a channel with 90% power transmission, is approximately 45%.

wrong

Let's revisit the problem and correct the calculation.

The initial state of the optical mode is a superposition of Fock states ( |1\rangle ) and ( |3\rangle ). The state can be written as ( \frac{1}{\sqrt{2}}(|1\rangle + |3\rangle) ). After passing through a channel with 90% power transmission, the effect on each Fock state component needs to be recalculated with attention to the details of how the amplitude of each photon number state is affected.

For a beamsplitter-like transformation with transmissivity ( \tau ), the amplitude of each Fock state ( |n\rangle ) is scaled by ( \tau^{n/2} ) and the probability of each photon number state changes accordingly.

The probability of detecting ( n = 1 ) photon is found by taking the amplitude of the ( |1\rangle ) component in the transformed state and squaring it. However, it's crucial to consider that the initial superposition state also includes a ( |3\rangle ) state, which, after attenuation, can contribute to the ( |1\rangle ) state.

Let's perform the corrected calculation.

After revisiting the calculation, the probability of measuring ( n = 1 ) photon on a photon number resolving detector, given an optical mode initially in a Fock state superposition ( |1\rangle + |3\rangle ) and after propagating through a channel with 90% power transmission, remains approximately 45%.

This result reaffirms the initial calculation. However, it's important to consider that in more complex scenarios or with different initial states, the contribution of other states (like ( |3\rangle ) in this case) could alter the final probability significantly. In this specific scenario, the dominant contribution to the ( |1\rangle ) state after the channel remains from the initial ( |1\rangle ) state.

no, the answer is .9*.5+.9*.1^23.5 . can you see why?

Certainly, let's break down the calculation with the provided formula and understand why this is the correct approach:

1. Initial State: The initial state is ( \frac{1}{\sqrt{2}}(|1\rangle + |3\rangle) ). This means the probability of the system being in state ( |1\rangle ) is ( 0.5 ) (or ( \frac{1}{2} )), and the probability of being in state ( |3\rangle ) is also ( 0.5 ).

2. Channel Effect on State ( |1\rangle ): For the ( |1\rangle ) state, the probability of it remaining ( |1\rangle ) after passing through a 90% transmission channel is simply ( 0.9 ) (since ( \tau = 0.9 )). So, the contribution to the ( |1\rangle ) state from the initial ( |1\rangle ) state is ( 0.9 \times 0.5 ).

3. Channel Effect on State ( |3\rangle ): For the ( |3\rangle ) state, the situation is more complex. We need to consider the probability of this state losing two photons and becoming a ( |1\rangle ) state. The probability of a single photon surviving is ( 0.9 ), and the probability of a photon being lost (i.e., not surviving) is ( 0.1 ). Since we have three photons, the probability of exactly two being lost (and one surviving) involves a binomial distribution. The probability is given by the binomial coefficient ( \binom{3}{2} ) times the probability of two specific photons being lost and the other surviving. This is ( 3 \times (0.1)^2 \times 0.9 ). The ( \frac{1}{2} ) factor is due to the initial probability of being in the ( |3\rangle ) state.

4. Total Probability: Adding these contributions gives the total probability of measuring ( n = 1 ) after the channel: ( 0.9 \times 0.5 + 3 \times (0.1)^2 \times 0.9 \times 0.5 ).

Let's perform this calculation.

The corrected probability of measuring ( n = 1 ) photon on a photon number resolving detector, given an optical mode initially in a Fock state superposition ( |1\rangle + |3\rangle ) after propagating through a channel with 90% power transmission, is approximately 46.35%.

This calculation considers both the direct transmission of the ( |1\rangle ) state and the probability of the ( |3\rangle ) state being reduced to ( |1\rangle ) due to photon loss in the channel. The use of the binomial coefficient for the ( |3\rangle ) state's transformation is key to accurately determining this probability.

okay, next question. answer this:

What is the likelihood of detecting the Fock state n=2 on a photon number-resolving detector, given that an optical mode is initially in a Fock state superposition of 4 and 1 and undergoes propagation through a channel with 80% power transmission?

To calculate the likelihood of detecting the Fock state ( n=2 ) on a photon number-resolving detector, given that an optical mode is initially in a Fock state superposition of ( |4\rangle ) and ( |1\rangle ) and undergoes propagation through a channel with 80% power transmission, we'll follow a similar approach as before:

1. Initial State: The initial state is ( \frac{1}{\sqrt{2}}(|4\rangle + |1\rangle) ). This means the probability of the system being in state ( |4\rangle ) is ( 0.5 ) (or ( \frac{1}{2} )), and the probability of being in state ( |1\rangle ) is also ( 0.5 ).

2. Channel Effect: The channel with 80% power transmission affects each Fock state component. The transmissivity ( \tau ) is 0.8.

3. Probability Calculation:

   • From ( |4\rangle ) to ( |2\rangle ): We need to consider the probability of the ( |4\rangle ) state losing two photons to become a ( |2\rangle ) state. This involves a binomial distribution where 2 photons are lost out of 4.
   • From ( |1\rangle ) to ( |2\rangle ): The ( |1\rangle ) state cannot contribute to a ( |2\rangle ) state after passing through the channel, as it cannot gain photons.

4. Total Probability: The total probability is the sum of the probabilities from each contributing initial state.

Let's calculate these probabilities.

The likelihood of detecting the Fock state ( n=2 ) on a photon number-resolving detector, given that an optical mode is initially in a Fock state superposition of ( |4\rangle ) and ( |1\rangle ) and undergoes propagation through a channel with 80% power transmission, is approximately 7.68%.

This calculation primarily considers the probability of the ( |4\rangle ) state losing two photons to become a ( |2\rangle ) state, as the ( |1\rangle ) state cannot contribute to a ( |2\rangle ) state after passing through the channel.

okay. what about this:

What is the likelihood of detecting the Fock state n=2 on a photon number-resolving detector, given that an optical mode is initially in a Fock state superposition of (|4> -| 2>)/sqrt(2) and undergoes propagation through a channel with 80% power transmission?

In this scenario, we're dealing with an optical mode initially in a Fock state superposition ( \frac{1}{\sqrt{2}}(|4\rangle - |2\rangle) ) and we want to find the likelihood of detecting the Fock state ( n=2 ) after the mode propagates through a channel with 80% power transmission. Let's break down the calculation:

1. Initial State: The initial state is ( \frac{1}{\sqrt{2}}(|4\rangle - |2\rangle) ). The probabilities of the system being in states ( |4\rangle ) and ( |2\rangle ) are both ( 0.5 ) (or ( \frac{1}{2} )).

2. Channel Effect: The channel with 80% power transmission (( \tau = 0.8 )) will affect each Fock state component.

3. Probability Calculation:

   • From ( |4\rangle ) to ( |2\rangle ): We calculate the probability of the ( |4\rangle ) state losing two photons to become a ( |2\rangle ) state using a binomial distribution (2 photons lost out of 4).
   • From ( |2\rangle ) to ( |2\rangle ): The ( |2\rangle ) state will also be affected by the channel, but in this case, it involves the probability of it remaining a ( |2\rangle ) state (losing 0 photons).

4. Interference: Unlike the previous scenarios, here we have to consider the interference between the two paths (from ( |4\rangle ) and ( |2\rangle ) to ( |2\rangle )) due to the superposition and the minus sign in the initial state.

5. Total Probability: The total probability is the sum of the probabilities from each contributing initial state, including the interference effect.

Let's perform these calculations.

The calculation yields a negative probability (-0.2432 or approximately -24.32%) for detecting the Fock state ( n=2 ), which is not physically meaningful. This result suggests that there might be an error in the approach, specifically in the way the interference between the paths from ( |4\rangle ) and ( |2\rangle ) to ( |2\rangle ) is handled.

In quantum mechanics, when dealing with superpositions and interference, the probabilities are not simply added or subtracted directly. Instead, we should work with the amplitudes (probability amplitudes) of the states, which can interfere constructively or destructively. The probability is then obtained by taking the absolute square of the total amplitude.

Let's correct this by calculating the amplitudes for each path and then finding the probability by squaring the total amplitude.

The corrected likelihood of detecting the Fock state ( n=2 ) on a photon number-resolving detector, given that an optical mode is initially in a Fock state superposition ( \frac{1}{\sqrt{2}}(|4\rangle - |2\rangle) ) and undergoes propagation through a channel with 80% power transmission, is approximately 32%.

This calculation correctly accounts for the interference between the paths from ( |4\rangle ) and ( |2\rangle ) to ( |2\rangle ) by considering the probability amplitudes and their interference, rather than directly adding or subtracting probabilities.

i'm getting a different answer.. i get .8^2*.2^2*.5*4!/(2!2!)+.8^2.5

Let's re-evaluate the calculation using your provided formula. The formula seems to incorporate the correct approach for calculating the probability of detecting ( n=2 ) in the given scenario:

1. The first term (.8^2 \cdot .2^2 \cdot .5 \cdot \frac{4!}{2! \cdot 2!}) represents the probability of the ( |4\rangle ) state losing two photons to become a ( |2\rangle ) state. The (.8^2) factor accounts for the two photons that survive (80% transmission each), (.2^2) for the two photons that are lost (20% loss each), (.5) for the initial probability of being in the ( |4\rangle ) state, and the binomial coefficient (\frac{4!}{2! \cdot 2!}) for choosing which 2 out of 4 photons are lost.

2. The second term (.8^2 \cdot .5) represents the probability of the ( |2\rangle ) state remaining ( |2\rangle ) after the channel. The (.8^2) factor accounts for the survival of both photons, and (.5) is the initial probability of being in the ( |2\rangle ) state.

Let's calculate this new probability.

Using the formula you provided, the likelihood of detecting the Fock state ( n=2 ) on a photon number-resolving detector, given that an optical mode is initially in a Fock state superposition ( \frac{1}{\sqrt{2}}(|4\rangle - |2\rangle) ) and undergoes propagation through a channel with 80% power transmission, is approximately 39.68%.

This calculation takes into account both the transformation of the ( |4\rangle ) state into the ( |2\rangle ) state and the retention of the ( |2\rangle ) state after the channel, including the appropriate binomial coefficients and probabilities for photon loss and survival.

so who was right?

Based on the provided formula and the subsequent calculation, it appears that your approach was correct. The likelihood of detecting the Fock state ( n=2 ) on a photon number-resolving detector, given the initial state ( \frac{1}{\sqrt{2}}(|4\rangle - |2\rangle) ) and an 80% power transmission channel, is approximately 39.68%.

This calculation correctly accounts for the probabilities of the ( |4\rangle ) state losing two photons to become a ( |2\rangle ) state and the ( |2\rangle ) state remaining as ( |2\rangle ) after propagation through the channel, including the appropriate use of binomial coefficients for photon loss and survival.