How is quantum mechanics consistent with statistical mechanics?












6















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







share|cite|improve this question

























  • 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
















6















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







share|cite|improve this question

























  • 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














6












6








6


1






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







share|cite|improve this question
















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 46 mins ago









DanielSank

17.2k45178




17.2k45178










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



















  • 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










2 Answers
2






active

oldest

votes


















10














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."






share|cite|improve this answer

































    1














    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



    $$ rho = lvert Psi rangle langle Psi lvert =$$
    $begin{bmatrix} frac{1/2} & frac{1/2} \ frac{1/2} & frac{1/2} end{bmatrix}$






    share|cite|improve this answer

























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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      10














      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."






      share|cite|improve this answer






























        10














        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."






        share|cite|improve this answer




























          10












          10








          10







          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."






          share|cite|improve this answer















          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."







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 28 mins ago

























          answered 2 hours ago









          knzhouknzhou

          42.8k11117206




          42.8k11117206























              1














              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



              $$ rho = lvert Psi rangle langle Psi lvert =$$
              $begin{bmatrix} frac{1/2} & frac{1/2} \ frac{1/2} & frac{1/2} end{bmatrix}$






              share|cite|improve this answer






























                1














                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



                $$ rho = lvert Psi rangle langle Psi lvert =$$
                $begin{bmatrix} frac{1/2} & frac{1/2} \ frac{1/2} & frac{1/2} end{bmatrix}$






                share|cite|improve this answer




























                  1












                  1








                  1







                  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



                  $$ rho = lvert Psi rangle langle Psi lvert =$$
                  $begin{bmatrix} frac{1/2} & frac{1/2} \ frac{1/2} & frac{1/2} end{bmatrix}$






                  share|cite|improve this answer















                  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



                  $$ rho = lvert Psi rangle langle Psi lvert =$$
                  $begin{bmatrix} frac{1/2} & frac{1/2} \ frac{1/2} & frac{1/2} end{bmatrix}$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 15 mins ago

























                  answered 53 mins ago









                  N. SteinleN. Steinle

                  1,440117




                  1,440117






























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