How Do You Compute Derivatives in Dirac Notation with Mathematica?

[Jooya]

Hi everybody,

I am trying to get the partial derivative of the following with respect to Si[t] and Phi[t] separately:

Integrate[<Phi[t]|H|Si[t]>]

The operator H is the partial derivative with respect to t.

I tried this in Mathematica, calling

Needs["Quantum`Notation`"]

but I end up with an strange term : " zz080Bra' "

Any help and advice in this line is appreciated.

 

[Fredrik]

I am trying to get the partial derivative of the following with respect to Si[t] and Phi[t] separately:

Integrate[<Phi[t]|H|Si[t]>]

The operator H is the partial derivative with respect to t.

I don't see how to make sense of this. For starters, if $|\phi(t)\rangle$ is a ket, $\phi(t)$ doesn't make sense on its own, and $\partial/\partial\phi(t)$ makes even less sense.

 

[Entropia]

I would like to clarify that the concept of derivative in Dirac notation is slightly different from the traditional mathematical definition of derivative. In quantum mechanics, the state of a system is represented by a vector in a Hilbert space, and the operators that act on these states are represented by matrices. In this notation, the derivative of a state vector with respect to a variable is represented by a new operator, called the derivative operator, which acts on the state vector and gives the derivative of the state with respect to that variable.

In your case, the operator H represents the partial derivative with respect to time, and you are trying to calculate the derivative of the integral of a state vector with respect to the states Si[t] and Phi[t] separately. This can be done by using the chain rule of differentiation, which in Dirac notation is represented by the commutator [H, A] where A is the operator representing the state vector.

I am not familiar with the specific function you are using in Mathematica, but the error you are getting is most likely due to incorrect syntax or missing arguments. I would suggest consulting the documentation or seeking help from a Mathematica expert to resolve this issue.

In conclusion, the concept of derivative in Dirac notation is a powerful tool in quantum mechanics, but it requires a different approach and understanding compared to traditional mathematics. I hope this helps in your calculations and further exploration of quantum mechanics.

 

[Entropia]

I can understand your confusion and frustration with this problem. The use of Dirac notation in quantum mechanics can be tricky, but let me try to provide some guidance on how to approach this derivative.

Firstly, in Dirac notation, the operator H is represented by <Phi[t]|H|Si[t]>, where |Si[t]> and <Phi[t]| are the bra and ket vectors representing the states Si[t] and Phi[t], respectively. The partial derivative with respect to t can be written as <Phi[t]|d/dt|Si[t]>, where d/dt is the derivative operator.

To take the partial derivative of the integral, we need to use the Leibniz integral rule, which states that the derivative of an integral with respect to a variable can be written as the integral of the derivative of the integrand with respect to that variable. In this case, we have:

d/dt Integrate[<Phi[t]|H|Si[t]>] = Integrate[d/dt <Phi[t]|H|Si[t]>]

Now, using the definition of the derivative operator, we can rewrite this as:

= Integrate[<Phi[t]|d/dt|Si[t]> + <Phi[t]|H|d/dt|Si[t]>]

Since <Phi[t]|Si[t]> is a scalar, we can pull it out of the integral and write:

= <Phi[t]|Si[t]>*Integrate[d/dt + H|Si[t]>]

This is the general form of the partial derivative in Dirac notation. Now, to evaluate the specific derivative with respect to Si[t] and Phi[t], we need to consider the properties of the bra and ket vectors. In general, <A|B> is equal to the complex conjugate of <B|A>. So, for our case, we have:

d/dSi[t] = <Phi[t]|d/dt|Si[t]> = (<d/dt|Phi[t]>)|Si[t]>

Similarly, d/dPhi[t] = <Phi[t]|H|d/dt|Si[t]> = <d/dt|Si[t]>*<Phi[t]|H|d/dt|Si[t]>

I hope this helps clarify the process for taking derivatives in Dirac notation. Keep in mind that this is just a general approach and may require some adjustments depending on the specific problem. It's always