Basic question about QM Operators

In summary, the book is discussing the concept of measuring the energy of an electron in a hydrogen atom multiple times and the resulting distribution of energy values. It explains that the absolute square of the state-coefficients represents the probability of obtaining a particular energy value after measurement, and that the average energy can be calculated using this concept. The book also introduces the use of equations (1) and (2) to obtain the average energy, but the reasoning behind this step may not be entirely clear.
  • #1
maverick280857
1,789
4
Hello,

Just out of interest and curiosity, I am reading a book on QM which introduces QM from a highly practical point of view, minus the formalism, assuming basic knowledge of complex numbers, algebra, calculus and physics. I will probably do QM formally much later in college, but while going through the book, I came across some operations and I would like to confirm one such operation as it is not explicitly described in the book.

Suppose we consider an experiment in which the energy of an electron in a hydrogen atom is measured N times. We denote the total energy in the i-th state by [itex]E_{i}[/itex] so that

[tex]E_{i} = <i|\hat{H}|i>[/tex] (1)

where [itex]\hat{H}[/itex] is the Hamiltonian operator. If [itex]C_{i}[/itex] for integers i, are complex numbers, then we since the system will be in some general state [itex]|\psi>[/itex], we can write it as a superposition of the eigenstates, i.e. as

[tex]|\psi> = C_{1}|1> + C_{2}|2> + ... + C_{i}|i>[/tex] (2)

If [itex]n_{i}[/itex] is the number of times the system's energy is measured to be [itex]E_{i}[/itex] then, [itex]\sum_{i}n_{i} = N[/itex].

First issue The book says that for large enough N, we must have

[tex]\frac{n_{i}}{N} = |C_{i}|^2[/tex] (3)

As I understand, the phrase "for large enough N" has either got something to do with statistics or with the fact that the measurement of energy forces the system to jump to one of the base states. Is this correct? Either way, is it correct to say that [itex]|C_{i}|^2[/itex] is a kind of weight which represents the fractional occurence of a particular energy [itex]E_{i}[/itex] in the distribution as the number of observations increases?

Now, the average energy is given by

[tex]E_{avg} = \frac{\sum_{i}n_{i}E_{i}}{\sum_{i}n_{i}} = \frac{\sum_{i}n_{i}E_{i}}{N}[/tex] (4)

If N is large, we use equation (3) and write,

[tex]E_{avg} = \sum_{i}|C_{i}|^{2}E_{i}[/tex]

Second Issue

Now, the book says that we use equation (1) and obtain,

[tex]E_{avg} = <\psi|\hat{H}|\psi>[/tex] (5)

This is the step I have trouble with. I tried to reason as follows. I know that

[tex]<j|\hat{O}|i> = \int \psi^{*}_{j}O(x)\psi_{i}(x)dx[/tex]
and
[tex]<j|\hat{H}|i> = \int \psi^{*}_{i}H\psi_{i}(x)dx[/tex]

(where O is a dipole operator which causes the electron to go from state [itex]|i>[/itex] to state [itex]|j>[/itex]), because the book states these as more practical versions of the above relationships. So, I figured that we could write [itex]|C_{i}|^2[/itex] as

[tex]|C_{i}|^2 = C_{i}^{*}C_{i}[/tex]

and then use the definitions (2) and the conjugate properties of bra and ket states to get equation (5). I am not sure if this is correct because the trick (if at all it is valid) seems obvious from the integrals and from the commutativtity of the product of 2 complex numbers.

I hope you will not mind mistakes or inadequacies in my reasoning/analysis too much as I have had virtually zero grounding in the development of QM in a rigorous mathematical fashion and I am pursuing it purely out of interest. I have done total derivatives, integrals, vectors, complex numbers, algebra and classical mechanics, electrodynamics minus tremendous partial derivatives (so no Lagrangian, Hamiltonian mechanics from the ab-initio).

Thanks and cheers
Vivek
(PS--This is not homework, so I figured this is the right place for the post.)
 
Last edited:
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  • #2
You have the correct idea for your first issue, the absolute square of the state-coefficients [tex]|C_{i}|^2[/tex] are interpreted as probability that state i is obtained after measurement and ni/N is just a statement that the law of large numbers works in probability theory eg. [tex]n_i/N[/tex] tends to the probability of [tex]n_i[/tex] when N goes to infinity.

Edit: my post got all f*cked up cause some latex issue that displayed the wrong things.. so I've edited away my strange interpretations :smile: if anyone happened to notice em.
 
Last edited:
  • #3
maverick280857 said:
Hello,

Just out of interest and curiosity, I am reading a book on QM which introduces QM from a highly practical point of view, minus the formalism, assuming basic knowledge of complex numbers, algebra, calculus and physics. I will probably do QM formally much later in college, but while going through the book, I came across some operations and I would like to confirm one such operation as it is not explicitly described in the book.

Suppose we consider an experiment in which the energy of an electron in a hydrogen atom is measured N times. We denote the total energy in the i-th state by [itex]E_{i}[/itex] so that

[tex]E_{i} = <i|\hat{H}|i>[/tex] (1)

where [itex]\hat{H}[/itex] is the Hamiltonian operator. If [itex]C_{i}[/itex] for integers i, are complex numbers, then we since the system will be in some general state [itex]|\psi>[/itex], we can write it as a superposition of the eigenstates, i.e. as

[tex]|\psi> = C_{1}|1> + C_{2}|2> + ... + C_{i}|i>[/tex] (2)

If [itex]n_{i}[/itex] is the number of times the system's energy is measured to be [itex]E_{i}[/itex] then, [itex]\sum_{i}n_{i} = N[/itex].

First issue The book says that for large enough N, we must have

[tex]\frac{n_{i}}{N} = |C_{i}|^2[/tex] (3)

As I understand, the phrase "for large enough N" has either got something to do with statistics or with the fact that the measurement of energy forces the system to jump to one of the base states. Is this correct? Either way, is it correct to say that [itex]|C_{i}|^2[/itex] is a kind of weight which represents the fractional occurence of a particular energy [itex]E_{i}[/itex] in the distribution as the number of observations increases?
Yes, it is the probability of measuring the energy at E_i

Now, the average energy is given by

[tex]E_{avg} = \frac{\sum_{i}n_{i}E_{i}}{\sum_{i}n_{i}} = \frac{\sum_{i}n_{i}E_{i}}{N}[/tex] (4)

If N is large, we use equation (3) and write,

[tex]E_{avg} = \sum_{i}|C_{i}|^{2}E_{i}[/tex]

Second Issue

Now, the book says that we use equation (1) and obtain,

[tex]E_{avg} = <\psi|\hat{H}|\psi>[/tex] (5)

This is the step I have trouble with. I tried to reason as follows. I know that

[tex]<j|\hat{O}|i> = \int \psi^{*}_{j}O(x)\psi_{i}(x)dx[/tex]
and
[tex]<j|\hat{H}|i> = \int \psi^{*}_{i}H\psi_{i}(x)dx[/tex]

(where O is a dipole operator which causes the electron to go from state [itex]|i>[/itex] to state [itex]|j>[/itex]), because the book states these as more practical versions of the above relationships. So, I figured that we could write [itex]|C_{i}|^2[/itex] as

[tex]|C_{i}|^2 = C_{i}^{*}C_{i}[/tex]

and then use the definitions (2) and the conjugate properties of bra and ket states to get equation (5). I am not sure if this is correct because the trick (if at all it is valid) seems obvious from the integrals and from the commutativtity of the product of 2 complex numbers.


[tex]E_{avg} = \sum_{i}|C_{i}|^{2}E_{i} = \sum_{i}|C_{i}|^{2} <i|\hat{H}|i> [/tex]

But [tex]\sum_{i}C_i|i> = |\psi> [/tex] and the dual gives [tex]\sum_{i}<i|C_i^* = <\psi| [/tex]

So plugging in gives the required result. Is this what you're saying ?
 
  • #4
What books are you using?There are not too many books out there dealing with virtual statistical ensembles...

Daniel.
 
  • #5
maverick280857 said:
Now, the book says that we use equation (1) and obtain,

[tex]E_{avg} = <\psi|\hat{H}|\psi>[/tex] (5)

That's right. Just write [itex]|\psi \rangle = \sum_i C_i|i\rangle[/itex] and use the orthonormality of the eigenfunctions: [itex]\langle i|j\rangle = \delta_{ij}[/itex].
 
  • #6
Hi everyone

Thanks very much for the clarification. Yes Gokul, that's what I meant.

Cheers
Vivek
 

What are quantum operators?

Quantum operators are mathematical representations of physical observables in quantum mechanics. They are used to describe the behavior and properties of quantum systems.

What is the role of quantum operators in quantum mechanics?

Quantum operators are essential tools in quantum mechanics, as they allow us to calculate the probabilities of different outcomes when measuring physical quantities and to predict the behavior of quantum systems.

How do quantum operators differ from classical operators?

Quantum operators differ from classical operators in that they do not commute, meaning their order matters in calculations. They also have complex eigenvalues, and their eigenfunctions are not necessarily mutually orthogonal.

What is the uncertainty principle in relation to quantum operators?

The uncertainty principle states that certain pairs of physical quantities, such as position and momentum, cannot be known simultaneously with arbitrary precision. This is due to the non-commutativity of quantum operators.

How are quantum operators used in practical applications?

Quantum operators are used in many practical applications, including quantum computing, quantum cryptography, and quantum sensing. They are also used in the development of new technologies, such as quantum sensors and advanced materials.

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