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A particle of mass m moves in 1-dimensional potential (x), which vanishes at infinity. The exact ground state eigenfunction is y(x) = A sech (lambda x) where A and lambda are constant. The ground state energy eigenvalue of this system is,?
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A particle of mass m moves in 1-dimensional potential (x), which vanis...
Ground State Eigenfunction
The ground state eigenfunction of a particle in a potential \( V(x) \) is given by:
- \( y(x) = A \, \text{sech}(\lambda x) \)
Here, \( A \) and \( \lambda \) are constants, and the function vanishes at infinity, indicating it is normalizable.
Energy Eigenvalue Calculation
To find the ground state energy eigenvalue, we start with the time-independent Schrödinger equation:
- \( -\frac{\hbar^2}{2m} \frac{d^2 y}{dx^2} + V(x) y = E y \)
Assuming \( V(x) \) approaches zero at infinity, we can analyze the potential's effect on the wave function.
Second Derivative of the Eigenfunction
Calculating the second derivative of \( y(x) \):
- \( \frac{d^2 y}{dx^2} = -2\lambda^2 A \, \text{sech}(\lambda x) \, \text{tanh}(\lambda x) \)
Substituting back into the Schrödinger equation yields:
- \( -\frac{\hbar^2}{2m} \left( -2\lambda^2 A \, \text{sech}(\lambda x) \, \text{tanh}(\lambda x) \right) + 0 = E A \, \text{sech}(\lambda x) \)
Finding the Ground State Energy
To satisfy the equation, we obtain:
- \( \frac{\hbar^2 \lambda^2}{m} = E \)
This means that the ground state energy eigenvalue \( E \) can be expressed as:
- \( E = \frac{\hbar^2 \lambda^2}{m} \)
Conclusion
The energy eigenvalue \( E \) is dependent on the parameter \( \lambda \), which is typically determined by boundary conditions or specific forms of the potential \( V(x) \).
The ground state eigenfunction of a particle in a potential \( V(x) \) is given by:
- \( y(x) = A \, \text{sech}(\lambda x) \)
Here, \( A \) and \( \lambda \) are constants, and the function vanishes at infinity, indicating it is normalizable.
Energy Eigenvalue Calculation
To find the ground state energy eigenvalue, we start with the time-independent Schrödinger equation:
- \( -\frac{\hbar^2}{2m} \frac{d^2 y}{dx^2} + V(x) y = E y \)
Assuming \( V(x) \) approaches zero at infinity, we can analyze the potential's effect on the wave function.
Second Derivative of the Eigenfunction
Calculating the second derivative of \( y(x) \):
- \( \frac{d^2 y}{dx^2} = -2\lambda^2 A \, \text{sech}(\lambda x) \, \text{tanh}(\lambda x) \)
Substituting back into the Schrödinger equation yields:
- \( -\frac{\hbar^2}{2m} \left( -2\lambda^2 A \, \text{sech}(\lambda x) \, \text{tanh}(\lambda x) \right) + 0 = E A \, \text{sech}(\lambda x) \)
Finding the Ground State Energy
To satisfy the equation, we obtain:
- \( \frac{\hbar^2 \lambda^2}{m} = E \)
This means that the ground state energy eigenvalue \( E \) can be expressed as:
- \( E = \frac{\hbar^2 \lambda^2}{m} \)
Conclusion
The energy eigenvalue \( E \) is dependent on the parameter \( \lambda \), which is typically determined by boundary conditions or specific forms of the potential \( V(x) \).
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A particle of mass m moves in 1-dimensional potential (x), which vanishes at infinity. The exact ground state eigenfunction is y(x) = A sech (lambda x) where A and lambda are constant. The ground state energy eigenvalue of this system is,?
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