Understanding the M-Theory Equation: Wavefunction and Constant Explained

[dimension10]

In the following equation:

Γ^i [X^i ,λ]ψ

Does psi refer to the wavefunction? If not what does it refer to?

Also, is i the square root of negative 1 or is it a constant?

Thanks in advance.

 

[mitchell porter]

(Continued from https://www.physicsforums.com/showthread.php?t=62567#2"...)

Does psi refer to the wavefunction? If not what does it refer to?

Yes, in Urs Schreiber's equation, psi is the wavefunction. But you should probably retain the http://en.wikipedia.org/wiki/Bra-ket_notation" , |psi>, in order to emphasize that the other factors are operators.

This is what happens when you "quantize" a theory. The classical theory will be described by a differential equation, and the algebraic objects in the equation are just numbers ("c-numbers", c for classical), describing field strengths and so on. But in the quantum theory, you have a "wavefunction" over the entire "space" of classical configurations, which determines probabilities according to the Born rule, and you now also have an operator for each classical quantity.

Operators take a wavefunction and turn it into another wavefunction. Ultimately we use a combination of these operators to determine how the wavefunction changes over time (more on this below). But if we just focus on a single, simple operator, like "position" (x). If you have a wavefunction which says that x definitely has some value, then the x operator applied to that wavefunction just multiplies it by that value. So $\hat{x}$|x=1> = 1 . |x=1>, $\hat{x}$|x=2> = 2 . |x=2>, and so on. The other rule to how it works is that it's linear, thus $\hat{x}$(|x=1>+|x=2>) = 1.|x=1> + 2.|x=2>.

This may seem a little mindless and meaningless, but in fact this notation then allows you to write down a formula for something like "the expected value of x", for any wavefunction, not just the "eigenstates" or eigenfunctions I was just writing about, like |x=1>, for which x=1 with probability 100%. But for a superposition like |psi> = |x=1>+|x=2>, the expected value of x is the weighted average of all the possibilities, so here it's 1.5 (equally weighted between 1 and 2), and you can get this back from <psi|$\hat{x}$|psi>, where <psi| means that you are projecting onto the |psi> part of the wavefunction resulting from $\hat{x}$|psi>...

You really need to read up on the basics of quantum mechanics in order to understand this properly - something you should do and must do if you want to understand M-theory as mathematics, and not just qualitatively. Anyway, we started out with a classical theory, and I mentioned that when you move to quantum mechanics, there will be an operator for each of the basic classical quantities occurring in the classical equation of motion. It turns out that if you algebraically combine the quantum operators in the same way that the classical quantities are combined, in the classical equation of motion, you get one big operator which describes how the wavefunction develops with time: the Hamiltonian operator H that appears in the Schrödinger equation!

The Hamiltonian version of classical mechanics which comes closest to quantum mechanics (that is, the version of classical mechanics which makes the transition from classical to quantum most natural) is the http://en.wikipedia.org/wiki/Hamilt...elationship_to_the_Schr.C3.B6dinger_equation", which has a strong resemblance to the "hidden variables" version of quantum mechanics called Bohmian mechanics. The more usual path from classical to quantum - the path that Feynman used - is the principle of least action. Here, you specify initial and final conditions, and you use the Euler-Lagrange equation to minimize/maximize/extremize the action, and this will determine a classical trajectory. In Feynman's sum over histories, you calculate the probability for a transition from the initial condition to the final condition by summing e^iS for every possible path, where S is the action, usually calculated from a function called the Lagrangian, or Lagrangian density for fields.

But Urs's "M-theory equation" is a Schrödinger equation, so it's using a Hamiltonian for M-theory (M-theory on a particular background space-time, as I emphasized last time), rather than a Lagrangian. The important thing to understand is that this is a quantum equation of motion, made up of operators acting on a probability wavefunction; it is not a classical equation, in which definite physical quantities evolve deterministically over time. You could write down a sort of "classical M(atrix) theory" version of this equation, in which the fields and matrices were ordinary classical quantities rather than operators. The solutions to the classical equations of motion do have a relationship to the corresponding quantum theory - what happens in the quantum theory can often be understood as a fluctuation around a particular classical solution. But you need the framework of operators and wavefunctions to see the full quantum theory at work.

Also, is i the square root of negative 1 or is it a constant?

Neither of these - here, "i" is just a label running from 1 to 9. There are 9 Xs and 9 Gammas. The term is actually a sum, but http://en.wikipedia.org/wiki/Einstein_notation" is being used, according to which, if an index like "i" occurs twice in a product, you should interpret it as a sum over the possible values of i.

By the way, if you want to see the original appearance of this expression (but in Lagrangian form), see equation 4.1 of http://arxiv.org/abs/hep-th/9610043.

Also, I really recommend that you look at Barton Zwiebach's textbook on string theory. That starts with classical electromagnetism, and takes you right through to string theory, introducing quantum mechanics on the way.

 

[dimension10]

(Continued from https://www.physicsforums.com/showthread.php?t=62567#2"...)
Yes, in Urs Schreiber's equation, psi is the wavefunction. But you should probably retain the http://en.wikipedia.org/wiki/Bra-ket_notation" , |psi>, in order to emphasize that the other factors are operators.

This is what happens when you "quantize" a theory. The classical theory will be described by a differential equation, and the algebraic objects in the equation are just numbers ("c-numbers", c for classical), describing field strengths and so on. But in the quantum theory, you have a "wavefunction" over the entire "space" of classical configurations, which determines probabilities according to the Born rule, and you now also have an operator for each classical quantity.

Operators take a wavefunction and turn it into another wavefunction. Ultimately we use a combination of these operators to determine how the wavefunction changes over time (more on this below). But if we just focus on a single, simple operator, like "position" (x). If you have a wavefunction which says that x definitely has some value, then the x operator applied to that wavefunction just multiplies it by that value. So $\hat{x}$|x=1> = 1 . |x=1>, $\hat{x}$|x=2> = 2 . |x=2>, and so on. The other rule to how it works is that it's linear, thus $\hat{x}$(|x=1>+|x=2>) = 1.|x=1> + 2.|x=2>.

This may seem a little mindless and meaningless, but in fact this notation then allows you to write down a formula for something like "the expected value of x", for any wavefunction, not just the "eigenstates" or eigenfunctions I was just writing about, like |x=1>, for which x=1 with probability 100%. But for a superposition like |psi> = |x=1>+|x=2>, the expected value of x is the weighted average of all the possibilities, so here it's 1.5 (equally weighted between 1 and 2), and you can get this back from <psi|$\hat{x}$|psi>, where <psi| means that you are projecting onto the |psi> part of the wavefunction resulting from $\hat{x}$|psi>...

You really need to read up on the basics of quantum mechanics in order to understand this properly - something you should do and must do if you want to understand M-theory as mathematics, and not just qualitatively. Anyway, we started out with a classical theory, and I mentioned that when you move to quantum mechanics, there will be an operator for each of the basic classical quantities occurring in the classical equation of motion. It turns out that if you algebraically combine the quantum operators in the same way that the classical quantities are combined, in the classical equation of motion, you get one big operator which describes how the wavefunction develops with time: the Hamiltonian operator H that appears in the Schrödinger equation!

The Hamiltonian version of classical mechanics which comes closest to quantum mechanics (that is, the version of classical mechanics which makes the transition from classical to quantum most natural) is the http://en.wikipedia.org/wiki/Hamilt...elationship_to_the_Schr.C3.B6dinger_equation", which has a strong resemblance to the "hidden variables" version of quantum mechanics called Bohmian mechanics. The more usual path from classical to quantum - the path that Feynman used - is the principle of least action. Here, you specify initial and final conditions, and you use the Euler-Lagrange equation to minimize/maximize/extremize the action, and this will determine a classical trajectory. In Feynman's sum over histories, you calculate the probability for a transition from the initial condition to the final condition by summing e^iS for every possible path, where S is the action, usually calculated from a function called the Lagrangian, or Lagrangian density for fields.

But Urs's "M-theory equation" is a Schrödinger equation, so it's using a Hamiltonian for M-theory (M-theory on a particular background space-time, as I emphasized last time), rather than a Lagrangian. The important thing to understand is that this is a quantum equation of motion, made up of operators acting on a probability wavefunction; it is not a classical equation, in which definite physical quantities evolve deterministically over time. You could write down a sort of "classical M(atrix) theory" version of this equation, in which the fields and matrices were ordinary classical quantities rather than operators. The solutions to the classical equations of motion do have a relationship to the corresponding quantum theory - what happens in the quantum theory can often be understood as a fluctuation around a particular classical solution. But you need the framework of operators and wavefunctions to see the full quantum theory at work.
Neither of these - here, "i" is just a label running from 1 to 9. There are 9 Xs and 9 Gammas. The term is actually a sum, but http://en.wikipedia.org/wiki/Einstein_notation" is being used, according to which, if an index like "i" occurs twice in a product, you should interpret it as a sum over the possible values of i.

By the way, if you want to see the original appearance of this expression (but in Lagrangian form), see equation 4.1 of http://arxiv.org/abs/hep-th/9610043.

Also, I really recommend that you look at Barton Zwiebach's textbook on string theory. That starts with classical electromagnetism, and takes you right through to string theory, introducing quantum mechanics on the way.

Thanks a lot. I guess I was getting confused over "i" and the superscript.

 

[mitchell porter]

http://arxiv.org/abs/1106.1179" have a proposal which, if it works out, might be the biggest development in Matrix Theory since Seiberg 1997. Matrix models of M-theory compactified on a manifold of 5 or less dimensions "work" because the gravitational modes decouple (like in AdS/CFT), leaving a lower-dimensional gauge theory which is equivalent to the full higher-dimensional theory. But matrix models for M-theory compactified on 6 or more dimensions don't work out because gravity is still present. The new idea is to use a new type of noncommutative geometry for the compact dimensions. I don't understand what this means from the perspective of gravitational decoupling, but Banks is one of the inventors of Matrix Theory, so this has to be taken seriously...

 

[dimension10]

http://arxiv.org/abs/1106.1179" have a proposal which, if it works out, might be the biggest development in Matrix Theory since Seiberg 1997. Matrix models of M-theory compactified on a manifold of 5 or less dimensions "work" because the gravitational modes decouple (like in AdS/CFT), leaving a lower-dimensional gauge theory which is equivalent to the full higher-dimensional theory. But matrix models for M-theory compactified on 6 or more dimensions don't work out because gravity is still present. The new idea is to use a new type of noncommutative geometry for the compact dimensions. I don't understand what this means from the perspective of gravitational decoupling, but Banks is one of the inventors of Matrix Theory, so this has to be taken seriously...

Thanks. Anyway, is there any advantage of using the Lagrangian form instead of the Hamiltonian form?

 

[mitchell porter]

Anyway, is there any advantage of using the Lagrangian form instead of the Hamiltonian form?

Only in context. They both get used. See how http://www.people.fas.harvard.edu/~xiyin/matrix.pdf" jump back and forth.

 

[dimension10]

Thanks.