# Tensor and vector product for Quantum

• A
• MrMuscle
In summary: Thank you!In summary, the author is first applying the I-operator to the qbit zero and the environment to find the new density matrix, but then he is distributing <0| . He suggests that we shouldn't expect the result to be a single number, but that it will be a matrix again, albeit a smaller one.
MrMuscle
TL;DR Summary
Hello, is there an operational order between tensor and matrix multiplications?
Hello, I am calculating the krauss operators to find the new density matrix after the interaction between environment and the qubit.
My question is: Is there an operational order between matrix multiplication and tensor product? Because apparently author is first applying I on |0> and X on |0> and obtain
I|0> =|0>, and X|0>=|1>. instead of first calculating U.
Then instead of making the tensor product he is distributing <0| .
Can someone please explain me why that's the case?

On attachment you can find more detailed explanation.

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MrMuscle said:
Summary: Hello, is there an operational order between tensor and matrix multiplications?

Hello, I am calculating the krauss operators to find the new density matrix after the interaction between environment and the qubit.
My question is: Is there an operational order between matrix multiplication and tensor product? Because apparently author is first applying I on |0> and X on |0> and obtain
I|0> =|0>, and X|0>=|1>. instead of first calculating U.
Then instead of making the tensor product he is distributing <0| .
Can someone please explain me why that's the case?

Please give a reference to the book or paper at which you are looking, as I am not sure about the notation used, and I need more context. For example, what does ##\left| 0_E \right>## mean? Something about a qbit zero and the environment ##E##?

MrMuscle said:
Because apparently author is first applying I on |0> and X on |0> and obtain
I|0> =|0>, and X|0>=|1>. instead of first calculating U.
You can't simply apply the ##I##-part of the operator product to ##|0_E \rangle##. This would lead to the invalid expression ##\langle 0_E|P_0 \otimes |0_E\rangle##.

$$\langle 0_E|P_0 \otimes I |0_E\rangle + \langle 0_E|P_1 \otimes X|0_E\rangle \\ P_0 \langle 0_E| I |0_E\rangle + P_1 \langle 0_E| X|0_E\rangle$$

which is the proper way to do things if ##|0_E \rangle## and ##I## refer to the environement and ##P_0## and ##P_1## refer to the system (as George said, you should provide a tangible reference for context).

George Jones said:
Please give a reference to the book or paper at which you are looking, as I am not sure about the notation used, and I need more context. For example, what does ##\left| 0_E \right>## mean? Something about a qbit zero and the environment ##E##?
Hello, I hope the picture on the attachment helps!
It is from a book named "Quantum Computer Explained" by David McMahon.
Chapter 12, Quantum Noise and Error Correction

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kith said:
You can't simply apply the ##I##-part of the operator product to ##|0_E \rangle##. This would lead to the invalid expression ##\langle 0_E|P_0 \otimes |0_E\rangle##.

$$\langle 0_E|P_0 \otimes I |0_E\rangle + \langle 0_E|P_1 \otimes X|0_E\rangle \\ P_0 \langle 0_E| I |0_E\rangle + P_1 \langle 0_E| X|0_E\rangle$$

which is the proper way to do things if ##|0_E \rangle## and ##I## refer to the environement and ##P_0## and ##P_1## refer to the system (as George said, you should provide a tangible reference for context).
Thanks for the answer. How can you take out P0 and P1?

Are you familiar with the partial trace?

kith said:
Are you familiar with the partial trace?
Kind of yes. I know how to calculate it but don't understand the concept behind.

Ok. Both the partial trace and your example involve expressions like
$$\langle \psi_E | A_S \otimes B_E | \psi_E \rangle$$
(where ##S## stands for system and ##E## stands for environment).

On a first glance, this looks like an ordinary expectation value which can be evaluated by applying the operator / matrix to the vector followed by evaluating the inner product. On a second glance, this doesn't work because the dimensions don't match: the vector lives in the Hilbert space of the environment which has a smaller dimension than the composite space (at least for finite-dimensional Hilbert spaces) on which the operator acts. So it isn't a priori clear whether this expression is meaningful.

What we can say is that we shouldn't expect it to yielad a single number. A matrix which represents the operator ##A_S \otimes B_E## contains more rows than the number of components of the vector. So what we should expect is that each component gets multiplied by multiple matrix entries. This suggests that the result has to be again a matrix, albeit a smaller one.

I don't have authorative references handy but check these two links for how the partial trace works and why it is meaningful to define it this way:
https://physics.stackexchange.com/q...ake-the-partial-trace-to-describe-a-subsystem

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## 1. What is the difference between a tensor and a vector in quantum mechanics?

A tensor is a mathematical object that represents the physical properties of a quantum system, such as spin or polarization. It has multiple components and can be transformed under certain operations. A vector, on the other hand, is a simpler mathematical object with only magnitude and direction, and it represents a state or observable in quantum mechanics.

## 2. How are tensor and vector products used in quantum mechanics?

In quantum mechanics, tensor and vector products are used to describe the interactions between multiple quantum systems. For example, the tensor product of two state vectors represents the combined state of two particles, while the vector product of two operators represents the combined operation on those particles.

## 3. Can tensors and vectors be used interchangeably in quantum mechanics?

No, tensors and vectors have different mathematical properties and cannot be used interchangeably in quantum mechanics. Tensors have multiple components and can represent higher-order properties, while vectors have only magnitude and direction and represent simpler properties.

## 4. How do tensor and vector products relate to entanglement in quantum mechanics?

Tensor products are used to describe the entangled state of two or more particles in quantum mechanics. Entanglement occurs when the state of one particle cannot be described without considering the state of the other particles, and tensor products are used to represent this interconnectedness.

## 5. Are there any practical applications of tensor and vector products in quantum mechanics?

Yes, tensor and vector products are used in various practical applications in quantum mechanics, such as quantum computing and quantum cryptography. They are also used in experiments to study entanglement and other quantum phenomena.

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