What Is the Position Operator if Momentum Is Given by a Specific Operator?

In summary, operators in quantum mechanics are mathematical objects that act on quantum states to produce new states, and are essential for describing the behavior of particles at the quantum level. They perform mathematical operations on the wave function to create new states and are central to the mathematical formalism of quantum mechanics. Hermitian operators are self-adjoint and have real eigenvalues, while non-Hermitian operators do not have these properties and are used to describe systems with dissipation or gain. Operators represent physical observables in quantum mechanics and their actions on quantum states produce eigenvalues, allowing us to make predictions about the behavior of particles.
  • #1
kemiisto
2
0

Homework Statement


Find the operator for position [tex]x[/tex] if the operator for momentum p is taken to be [tex]\left(\hbar/2m\right)^{1/2}\left(A + B\right)[/tex], with [tex]\left[A,B\right] = 1[/tex] and all other commutators zero.

Homework Equations


Canonical commutation relation
[tex]\left [ \hat{ x }, \hat{ p } \right ] = \hat{x} \hat{p} - \hat{p} \hat{x} = i \hbar[/tex]

The Attempt at a Solution


Using [tex]c = \left(\hbar/2m\right)^{1/2}[/tex]
[tex]\hat{x} \hat{p} f - \hat{p} \hat{x} f = i \hbar[/tex]
[tex]\hat{x} c \left(\hat{A} + \hat{B}\right) f - c \left(\hat{A} + \hat{B}\right) \hat{x} f = i \hbar[/tex]
[tex]c \hat{x} \left(\hat{A} + \hat{B}\right) f - c \left(\hat{A} + \hat{B}\right) \hat{x} f = i \hbar[/tex]
[tex]\hat{x} \hat{A} f + \hat{x} \hat{B} f - \hat{A} \hat{x} f - \hat{B} \hat{x} f = i \hbar / c[/tex]
[tex]\hat{x} \hat{A} f - c \hat{A} \hat{x} f + \hat{x} \hat{B} f - \hat{B} \hat{x} f = i \hbar / c[/tex]
[tex]\left [ \hat{x}, \hat{A} \right ] + \left [ \hat{x}, \hat{B} \right ] = i \hbar / c[/tex]
"all other commutators zero"
[tex]0 + 0 = i \hbar / c[/tex]

:confused:

Problem 1.2 from http://www.oup.com/uk/orc/bin/9780199274987/" .
 
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  • #2
If [A,B]=1, then what will be the commutator of A with A+B? Can you go from there?

P.S. Your "f" on the left hand side is unnecessary.
 
  • #3
If [A, B] = 1 then [A, A + B] = 1

[A, A + B] = A(A + B) - (A + B)A = AA + AB - AA - BA = AB - BA = [A, B] = 1

We have p = c(A + B) and [x, p] = iħ.

xp - px = iħ
cx(A + B) - c(A + B)x = iħ
x(A + B) - (A + B)x = iħ/c | multiply on A from the left
Ax(A + B) - A(A + B)x = iħ/c A | A(A + B) = 1 + (A + B)A
Ax(A + B) - (1 + (A + B)A)x = iħ/c A
Ax(A + B) - x - (A + B)Ax = iħ/c A
[Ax, A + B] - x = iħ/c A
x = -iħ/c A

Checking: [x, p] = xp - px = (-iħ/c A)(c(A + B)) - (c(A + B))(-iħ/c A) = -iħ A(A + B) + iħ (A + B) A = -ih

Ouch! Like always accurate within a sign... :cry:

Anyway, arkajad, many thanks for guiding me. I didn't have enough sleep last night. Maybe tomorrow I'll find the missing sign...
 
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  • #4
Get a sleep and you will have it. But, perhaps, first prove for yourself a very useful identity: for any number a, and any operators A,B,C we have that

[A,B+C]=[A,B]+[A,C]
[A+B,C]=[A,C]+[B,C]
[aA,B]=a[A,B]
[A,aB]=a[A,B]

To save some work use [A,B]=-[B,A] as many times as useful. Once you are done - remember these rules, as they will be handy in the future.
 
  • #5


I would first clarify the notation used in the problem. The operator for position is commonly denoted as \hat{x}, while the operator for momentum is denoted as \hat{p}. It is also important to note that the commutator of two operators is defined as \left [ \hat{A}, \hat{B} \right ] = \hat{A}\hat{B} - \hat{B}\hat{A}.

Now, to find the operator for position, we can use the canonical commutation relation, \left [ \hat{ x }, \hat{ p } \right ] = i \hbar. Substituting the given operator for momentum, we have:

\left [ \hat{ x }, \left(\hbar/2m\right)^{1/2}\left(A + B\right) \right ] = i \hbar

Expanding the commutator, we get:

\hat{x} \left(\hbar/2m\right)^{1/2}\left(A + B\right) - \left(\hbar/2m\right)^{1/2}\left(A + B\right)\hat{x} = i \hbar

Simplifying, we get:

\hat{x}A + \hat{x}B - A\hat{x} - B\hat{x} = i \hbar

Using the given information that all other commutators are zero, we can simplify further to:

\hat{x}A - A\hat{x} = i \hbar

We can now solve for the operator for position, \hat{x}:

\hat{x} = \frac{i \hbar}{A - A}

However, this leads to an undefined result. This could be because the given operator for momentum, \left(\hbar/2m\right)^{1/2}\left(A + B\right), is not a valid operator for momentum. In quantum mechanics, the operators must be Hermitian and satisfy other important properties. Without more information about the operators A and B, it is not possible to find the operator for position x.
 

FAQ: What Is the Position Operator if Momentum Is Given by a Specific Operator?

1. What are operators in quantum mechanics?

Operators in quantum mechanics are mathematical objects that represent physical quantities, such as position, momentum, and energy. They act on quantum states to produce new states, and are essential for describing the behavior of particles at the quantum level.

2. How do operators act on quantum states?

Operators act on quantum states by performing mathematical operations on the wave function, which describes the state of a particle. The result of the operation is a new wave function, which represents the new state of the particle.

3. What is the significance of operators in quantum mechanics?

Operators are central to the mathematical formalism of quantum mechanics. They allow us to calculate the probabilities of different outcomes when measuring physical quantities, and they also help us understand the behavior of particles in various physical systems.

4. What is the difference between Hermitian and non-Hermitian operators?

Hermitian operators are self-adjoint, meaning that they are equal to their own adjoint operations. This property ensures that they have real eigenvalues and that their eigenstates form a complete set. Non-Hermitian operators do not have these properties, and are often used to describe systems with dissipation or gain.

5. How are operators related to observables in quantum mechanics?

Operators represent physical observables in quantum mechanics, such as position, momentum, and energy. When an operator acts on a quantum state, the result is an eigenvalue, which represents the value of the observable in that state. This allows us to make predictions about the behavior of particles in different states.

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