How is quantum mechanics consistent with statistical mechanics?
Let's say we have an harmonic oscillator (at Temperature $T$) in a superposition of state 1 and 2:
$$Psi = frac{phi_1+phi_2}{sqrt{2}}$$
where each $phi_i$ has energy $E_i , .$
The probability of finding each the $i$ state would be 50% in this case. However, approaching this problem with statistical mechanics the probability would be proportional to $e^{-E/k_BT}$
What is wrong with my approach?
quantum-mechanics statistical-mechanics
edited 46 mins ago
DanielSank
17.2k45178
asked 2 hours ago
IvanMartinezIvanMartinez
928
add a comment |
Let's say we have an harmonic oscillator (at Temperature $T$) in a superposition of state 1 and 2:
$$Psi = frac{phi_1+phi_2}{sqrt{2}}$$
where each $phi_i$ has energy $E_i , .$
The probability of finding each the $i$ state would be 50% in this case. However, approaching this problem with statistical mechanics the probability would be proportional to $e^{-E/k_BT}$
What is wrong with my approach?
quantum-mechanics statistical-mechanics
edited 46 mins ago
DanielSank
17.2k45178
asked 2 hours ago
IvanMartinezIvanMartinez
928
It does not make sense to speak of a particle in a given state at a certain temperature. If you say your particle is at a temperature T, it can be in any state of energy E with a probability $e^{-E/k_bT}$.
– E. Bellec
2 hours ago
8
The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, because approaching the problem by statistical mechanics, the position distribution of the ball must obey the Boltzmann distribution."
– knzhou
2 hours ago
4
The point is, not everything has to be in thermal equilibrium.
– knzhou
2 hours ago
@knzhou You should make that into an answer
– Aaron Stevens
2 hours ago
add a comment |
Let's say we have an harmonic oscillator (at Temperature $T$) in a superposition of state 1 and 2:
$$Psi = frac{phi_1+phi_2}{sqrt{2}}$$
where each $phi_i$ has energy $E_i , .$
The probability of finding each the $i$ state would be 50% in this case. However, approaching this problem with statistical mechanics the probability would be proportional to $e^{-E/k_BT}$
What is wrong with my approach?
quantum-mechanics statistical-mechanics
edited 46 mins ago
DanielSank
17.2k45178
asked 2 hours ago
IvanMartinezIvanMartinez
928
Let's say we have an harmonic oscillator (at Temperature $T$) in a superposition of state 1 and 2:
$$Psi = frac{phi_1+phi_2}{sqrt{2}}$$
where each $phi_i$ has energy $E_i , .$
The probability of finding each the $i$ state would be 50% in this case. However, approaching this problem with statistical mechanics the probability would be proportional to $e^{-E/k_BT}$
What is wrong with my approach?
quantum-mechanics statistical-mechanics
quantum-mechanics statistical-mechanics
edited 46 mins ago
DanielSank
17.2k45178
asked 2 hours ago
IvanMartinezIvanMartinez
928
edited 46 mins ago
DanielSank
17.2k45178
asked 2 hours ago
IvanMartinezIvanMartinez
928
edited 46 mins ago
DanielSank
17.2k45178
edited 46 mins ago
DanielSank
17.2k45178
edited 46 mins ago
DanielSank
17.2k45178
17.2k45178
asked 2 hours ago
IvanMartinezIvanMartinez
928
asked 2 hours ago
IvanMartinezIvanMartinez
928
asked 2 hours ago
IvanMartinezIvanMartinez
928
928
It does not make sense to speak of a particle in a given state at a certain temperature. If you say your particle is at a temperature T, it can be in any state of energy E with a probability $e^{-E/k_bT}$.
– E. Bellec
2 hours ago
8
The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, because approaching the problem by statistical mechanics, the position distribution of the ball must obey the Boltzmann distribution."
– knzhou
2 hours ago
4
The point is, not everything has to be in thermal equilibrium.
– knzhou
2 hours ago
@knzhou You should make that into an answer
– Aaron Stevens
2 hours ago
add a comment |
It does not make sense to speak of a particle in a given state at a certain temperature. If you say your particle is at a temperature T, it can be in any state of energy E with a probability $e^{-E/k_bT}$.
– E. Bellec
2 hours ago
8
The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, because approaching the problem by statistical mechanics, the position distribution of the ball must obey the Boltzmann distribution."
– knzhou
2 hours ago
4
The point is, not everything has to be in thermal equilibrium.
– knzhou
2 hours ago
@knzhou You should make that into an answer
– Aaron Stevens
2 hours ago
It does not make sense to speak of a particle in a given state at a certain temperature. If you say your particle is at a temperature T, it can be in any state of energy E with a probability $e^{-E/k_bT}$.
– E. Bellec
2 hours ago
It does not make sense to speak of a particle in a given state at a certain temperature. If you say your particle is at a temperature T, it can be in any state of energy E with a probability $e^{-E/k_bT}$.
– E. Bellec
2 hours ago
8
8
The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, because approaching the problem by statistical mechanics, the position distribution of the ball must obey the Boltzmann distribution."
– knzhou
2 hours ago
The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, because approaching the problem by statistical mechanics, the position distribution of the ball must obey the Boltzmann distribution."
– knzhou
2 hours ago
4
4
The point is, not everything has to be in thermal equilibrium.
– knzhou
2 hours ago
The point is, not everything has to be in thermal equilibrium.
– knzhou
2 hours ago
@knzhou You should make that into an answer
– Aaron Stevens
2 hours ago
@knzhou You should make that into an answer
– Aaron Stevens
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
The question doesn't make sense, as not every system has to be in thermal equilibrium.
The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, since by statistical mechanics, the position of the ball must obey the Boltzmann distribution."
answered 2 hours ago
knzhouknzhou
42.8k11117206
add a comment |
You're comparing the wrong things - you must take the appropriate limits when using statistical physics (or statistical mechanics as you called it) to recover quantum and classical results. Also, recall that when we use statistical physics, we must consider the system in an ensemble formalism, i.e. the canonical ensemble. That is, we consider a (statistically) large number of identically prepared copies of our system: a single harmonic oscillator. Additionally, we must use the quantum version of statistical physics, where the ensemble is characterized by the density operator. For a nice intro to this, see Sakurai.
This article should answer you questions regarding the case of a single 1D harmonic oscillator potential viewed statistically. For a given temperature, we use the quantized energy of the oscillator to find the partition function in the canonical ensemble.
Consider the two limits:
1) Classcial: the thermal energy is much greater than the spacing of oscillator energies. Here, one recovers the classical result that the energy is $kT$.
2) Quantum: the thermal energy is much less than the spacing of oscillator energies. Here, one recovers the quantum result that for a given frequency the energy is $frac{1}{2}hbar omega$
Further, using the appropriate density operator, you can recover the probabilities that you sought:
the density operator represented as a matrix in the basis ${phi_1, phi_2 }$, for the system given by pure state $Psi$ is
$$ rho = lvert Psi rangle langle Psi lvert =$$
$begin{bmatrix} frac{1/2} & frac{1/2} \ frac{1/2} & frac{1/2} end{bmatrix}$
answered 53 mins ago
N. SteinleN. Steinle
1,440117
add a comment |
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2 Answers
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2 Answers
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active
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votes
The question doesn't make sense, as not every system has to be in thermal equilibrium.
The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, since by statistical mechanics, the position of the ball must obey the Boltzmann distribution."
answered 2 hours ago
knzhouknzhou
42.8k11117206
add a comment |
The question doesn't make sense, as not every system has to be in thermal equilibrium.
The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, since by statistical mechanics, the position of the ball must obey the Boltzmann distribution."
answered 2 hours ago
knzhouknzhou
42.8k11117206
add a comment |
The question doesn't make sense, as not every system has to be in thermal equilibrium.
The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, since by statistical mechanics, the position of the ball must obey the Boltzmann distribution."
answered 2 hours ago
knzhouknzhou
42.8k11117206
The question doesn't make sense, as not every system has to be in thermal equilibrium.
The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, since by statistical mechanics, the position of the ball must obey the Boltzmann distribution."
answered 2 hours ago
knzhouknzhou
42.8k11117206
edited 28 mins ago
answered 2 hours ago
knzhouknzhou
42.8k11117206
answered 2 hours ago
knzhouknzhou
42.8k11117206
answered 2 hours ago
knzhouknzhou
42.8k11117206
42.8k11117206
add a comment |
add a comment |
You're comparing the wrong things - you must take the appropriate limits when using statistical physics (or statistical mechanics as you called it) to recover quantum and classical results. Also, recall that when we use statistical physics, we must consider the system in an ensemble formalism, i.e. the canonical ensemble. That is, we consider a (statistically) large number of identically prepared copies of our system: a single harmonic oscillator. Additionally, we must use the quantum version of statistical physics, where the ensemble is characterized by the density operator. For a nice intro to this, see Sakurai.
This article should answer you questions regarding the case of a single 1D harmonic oscillator potential viewed statistically. For a given temperature, we use the quantized energy of the oscillator to find the partition function in the canonical ensemble.
Consider the two limits:
1) Classcial: the thermal energy is much greater than the spacing of oscillator energies. Here, one recovers the classical result that the energy is $kT$.
2) Quantum: the thermal energy is much less than the spacing of oscillator energies. Here, one recovers the quantum result that for a given frequency the energy is $frac{1}{2}hbar omega$
Further, using the appropriate density operator, you can recover the probabilities that you sought:
the density operator represented as a matrix in the basis ${phi_1, phi_2 }$, for the system given by pure state $Psi$ is
$$ rho = lvert Psi rangle langle Psi lvert =$$
$begin{bmatrix} frac{1/2} & frac{1/2} \ frac{1/2} & frac{1/2} end{bmatrix}$
answered 53 mins ago
N. SteinleN. Steinle
1,440117
add a comment |
You're comparing the wrong things - you must take the appropriate limits when using statistical physics (or statistical mechanics as you called it) to recover quantum and classical results. Also, recall that when we use statistical physics, we must consider the system in an ensemble formalism, i.e. the canonical ensemble. That is, we consider a (statistically) large number of identically prepared copies of our system: a single harmonic oscillator. Additionally, we must use the quantum version of statistical physics, where the ensemble is characterized by the density operator. For a nice intro to this, see Sakurai.
This article should answer you questions regarding the case of a single 1D harmonic oscillator potential viewed statistically. For a given temperature, we use the quantized energy of the oscillator to find the partition function in the canonical ensemble.
Consider the two limits:
1) Classcial: the thermal energy is much greater than the spacing of oscillator energies. Here, one recovers the classical result that the energy is $kT$.
2) Quantum: the thermal energy is much less than the spacing of oscillator energies. Here, one recovers the quantum result that for a given frequency the energy is $frac{1}{2}hbar omega$
Further, using the appropriate density operator, you can recover the probabilities that you sought:
the density operator represented as a matrix in the basis ${phi_1, phi_2 }$, for the system given by pure state $Psi$ is
$$ rho = lvert Psi rangle langle Psi lvert =$$
$begin{bmatrix} frac{1/2} & frac{1/2} \ frac{1/2} & frac{1/2} end{bmatrix}$
answered 53 mins ago
N. SteinleN. Steinle
1,440117
add a comment |
You're comparing the wrong things - you must take the appropriate limits when using statistical physics (or statistical mechanics as you called it) to recover quantum and classical results. Also, recall that when we use statistical physics, we must consider the system in an ensemble formalism, i.e. the canonical ensemble. That is, we consider a (statistically) large number of identically prepared copies of our system: a single harmonic oscillator. Additionally, we must use the quantum version of statistical physics, where the ensemble is characterized by the density operator. For a nice intro to this, see Sakurai.
This article should answer you questions regarding the case of a single 1D harmonic oscillator potential viewed statistically. For a given temperature, we use the quantized energy of the oscillator to find the partition function in the canonical ensemble.
Consider the two limits:
1) Classcial: the thermal energy is much greater than the spacing of oscillator energies. Here, one recovers the classical result that the energy is $kT$.
2) Quantum: the thermal energy is much less than the spacing of oscillator energies. Here, one recovers the quantum result that for a given frequency the energy is $frac{1}{2}hbar omega$
Further, using the appropriate density operator, you can recover the probabilities that you sought:
the density operator represented as a matrix in the basis ${phi_1, phi_2 }$, for the system given by pure state $Psi$ is
$$ rho = lvert Psi rangle langle Psi lvert =$$
$begin{bmatrix} frac{1/2} & frac{1/2} \ frac{1/2} & frac{1/2} end{bmatrix}$
answered 53 mins ago
N. SteinleN. Steinle
1,440117
You're comparing the wrong things - you must take the appropriate limits when using statistical physics (or statistical mechanics as you called it) to recover quantum and classical results. Also, recall that when we use statistical physics, we must consider the system in an ensemble formalism, i.e. the canonical ensemble. That is, we consider a (statistically) large number of identically prepared copies of our system: a single harmonic oscillator. Additionally, we must use the quantum version of statistical physics, where the ensemble is characterized by the density operator. For a nice intro to this, see Sakurai.
This article should answer you questions regarding the case of a single 1D harmonic oscillator potential viewed statistically. For a given temperature, we use the quantized energy of the oscillator to find the partition function in the canonical ensemble.
Consider the two limits:
1) Classcial: the thermal energy is much greater than the spacing of oscillator energies. Here, one recovers the classical result that the energy is $kT$.
2) Quantum: the thermal energy is much less than the spacing of oscillator energies. Here, one recovers the quantum result that for a given frequency the energy is $frac{1}{2}hbar omega$
Further, using the appropriate density operator, you can recover the probabilities that you sought:
the density operator represented as a matrix in the basis ${phi_1, phi_2 }$, for the system given by pure state $Psi$ is
$$ rho = lvert Psi rangle langle Psi lvert =$$
$begin{bmatrix} frac{1/2} & frac{1/2} \ frac{1/2} & frac{1/2} end{bmatrix}$
answered 53 mins ago
N. SteinleN. Steinle
1,440117
edited 15 mins ago
answered 53 mins ago
N. SteinleN. Steinle
1,440117
answered 53 mins ago
N. SteinleN. Steinle
1,440117
answered 53 mins ago
N. SteinleN. Steinle
1,440117
1,440117
add a comment |
add a comment |
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It does not make sense to speak of a particle in a given state at a certain temperature. If you say your particle is at a temperature T, it can be in any state of energy E with a probability $e^{-E/k_bT}$.
– E. Bellec
2 hours ago
8
The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, because approaching the problem by statistical mechanics, the position distribution of the ball must obey the Boltzmann distribution."
– knzhou
2 hours ago
4
The point is, not everything has to be in thermal equilibrium.
– knzhou
2 hours ago
@knzhou You should make that into an answer
– Aaron Stevens
2 hours ago